NCERT Solutions for Class 11 Maths Chapter 3 - Trigonometric Functions

Exercise 3.1: In this exercise, students are introduced to trigonometric ratios of acute angles and their applications in solving problems related to heights and distances. The exercise covers basic concepts such as the definition of trigonometric ratios, Pythagoras theorem, and the concept of complementary angles.

Exercise 3.2: This exercise focuses on the evaluation of trigonometric ratios of special angles such as 0, 30, 45, 60, and 90 degrees. The exercise also includes the derivation of trigonometric ratios of 30, 45, and 60 degrees using the concept of the unit circle.

Exercise 3.3: This exercise covers the trigonometric ratios of angles between 0 and 90 degrees. The exercise includes problems on finding the values of trigonometric ratios using various techniques such as the use of Pythagoras theorem, the double-angle formula, and the half-angle formula.

Exercise 3.4: In this exercise, students learn about the trigonometric ratios of complementary angles and the concept of co-functions. The exercise covers problems on finding the values of trigonometric ratios using the co-function identity.

Miscellaneous Exercise: This exercise includes a variety of problems that cover different topics such as the use of trigonometric ratios in solving real-life problems, the use of trigonometric identities to simplify expressions, and the solution of trigonometric equations. This exercise provides an opportunity for students to apply the concepts learned in the chapter to solve more complex problems.

Access NCERT Solution for Class 11 Maths Chapter 3 - Trigonometric Functions

Exercise 3.1

1. Find the radian measures corresponding to the following degree measures:

(i) $\text{2}{{\text{5}}^{\text{o}}}$

Ans: We know that $\text{18}{{\text{0}}^{\text{o}}}\text{= }\!\!\pi\!\!\text{ }$ radian

Therefore  ${{1}^{\circ }}=\dfrac{\pi }{180}$ radian 

hence, 

$\text{2}{{\text{5}}^{\text{o}}}\text{=}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{180}}\text{ }\!\!\times\!\!\text{ 25}$ radian  

      $\text{=}\dfrac{\text{5 }\!\!\pi\!\!\text{ }}{\text{36}}$radian

(ii) $\text{-4}{{\text{7}}^{\text{o}}}\text{30 }\!\!'\!\!\text{ }$

Ans: Here we have,

  $\text{-4}{{\text{7}}^{\text{o}}}\text{30 }\!\!'\!\!\text{ =-47}{{\dfrac{\text{1}}{\text{2}}}^{\text{o}}}$   

           $\text{=-}\dfrac{\text{95}}{\text{2}}$ degree

Since we know that, $\text{18}{{\text{0}}^{\text{o}}}\text{= }\!\!\pi\!\!\text{ }$ radian

Therefore  ${{\text{1}}^{\text{o}}}\text{=}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{180}}$ radian 

Hence,

$\text{-}\dfrac{\text{95}}{\text{2}}$ degree$\text{=}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{180}}\text{ }\!\!\times\!\!\text{ }\left( \dfrac{\text{-95}}{\text{2}} \right)$ radian

                  $\text{=}\left( \dfrac{\text{-19}}{\text{36 }\!\!\times\!\!\text{ 2}} \right)\text{ }\!\!\pi\!\!\text{ }$  radian

                  $\text{=}\dfrac{\text{-19}}{\text{72}}\text{ }\!\!\pi\!\!\text{ }$radian

Therefore,

$\text{-4}{{\text{7}}^{\text{o}}}\text{30 }\!\!'\!\!\text{ =-}\dfrac{\text{19}}{\text{72}}\text{ }\!\!\pi\!\!\text{ }$ radian

(iii) $\text{24}{{\text{0}}^{\text{o}}}$

Ans: We know that,

$\text{18}{{\text{0}}^{\text{o}}}\text{= }\!\!\pi\!\!\text{ }$ radian

Therefore  ${{\text{1}}^{\text{o}}}\text{=}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{180}}$ radian 

Hence,

$\text{24}{{\text{0}}^{\text{o}}}\text{=}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{180}}\text{ }\!\!\times\!\!\text{ 240}$ radian

       $\text{=}\dfrac{\text{4}}{\text{3}}\text{ }\!\!\pi\!\!\text{ }$radian

(iv) $\text{52}{{\text{0}}^{\text{o}}}$

Ans: We know that,

$\text{18}{{\text{0}}^{\text{o}}}\text{= }\!\!\pi\!\!\text{ }$ radian

Therefore  ${{\text{1}}^{\text{o}}}\text{=}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{180}}$ radian 

Hence,

$\text{52}{{\text{0}}^{\text{o}}}\text{=}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{180}}\text{ }\!\!\times\!\!\text{ 520}$ radian

       $\text{=}\dfrac{\text{26 }\!\!\pi\!\!\text{ }}{\text{9}}$radian

2. Find the degree measures corresponding to the following radian measures

(Use$\text{ }\!\!\pi\!\!\text{ =}\dfrac{\text{22}}{\text{7}}$ ) 

(i) $\dfrac{\text{11}}{\text{16}}$

Ans: We know that,

 $\text{ }\!\!\pi\!\!\text{ }$ radian$\text{=18}{{\text{0}}^{\text{o}}}$

Therefore  $\text{1 radian =}{{\dfrac{\text{180}}{\text{ }\!\!\pi\!\!\text{ }}}^{\text{o}}}$ 

Hence,

$\dfrac{\text{11}}{\text{16}}$ radian$\text{=}\dfrac{\text{180}}{\text{ }\!\!\pi\!\!\text{ }}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{11}}{\text{16}}$ degree

              $\text{=}\dfrac{\text{45 }\!\!\times\!\!\text{ 11}}{\text{ }\!\!\pi\!\!\text{  }\!\!\times\!\!\text{ 4}}$degree

              \[\text{=}\dfrac{\text{45 }\!\!\times\!\!\text{ 11 }\!\!\times\!\!\text{ 7}}{\text{22 }\!\!\times\!\!\text{ 4}}\] degree

              $\text{=}\dfrac{\text{315}}{\text{8}}$degree

Further computing,

$\dfrac{\text{11}}{\text{16}}$ radian$\text{=39}\dfrac{\text{3}}{\text{8}}$ degree

                $\text{=3}{{\text{9}}^{\text{o}}}\text{+}\dfrac{\text{3 }\!\!\times\!\!\text{ 60}}{\text{8}}$ minutes

Since ${{\text{1}}^{\text{o}}}\text{=60 }\!\!'\!\!\text{ }$

    $\dfrac{\text{11}}{\text{16}}$ radian $\text{=3}{{\text{9}}^{\text{o}}}\text{+22 }\!\!'\!\!\text{ +}\dfrac{\text{1}}{\text{2}}$minutes

Since  $\text{1 }\!\!'\!\!\text{ =60''}$ 

$\dfrac{\text{11}}{\text{16}}$ radian$\text{ }\!\!~\!\!\text{ =3}{{\text{9}}^{\text{o}}}\text{22 }\!\!'\!\!\text{ 30''}$

(ii) $\text{-4}$

Ans: We know that,

$\text{ }\!\!\pi\!\!\text{ }$ radian$\text{=18}{{\text{0}}^{\text{o}}}$

Therefore  $\text{1 radian =}{{\dfrac{\text{180}}{\text{ }\!\!\pi\!\!\text{ }}}^{\text{o}}}$ 

Hence,

$\text{-4}$ radian$\text{=}\dfrac{\text{180}}{\text{ }\!\!\pi\!\!\text{ }}\text{ }\!\!\times\!\!\text{ }\left( \text{-4} \right)$ degree

                $\text{=}\dfrac{\text{180 }\!\!\times\!\!\text{ 7}\left( \text{-4} \right)}{\text{22}}$degree

                $\text{=}\dfrac{\text{-2520}}{\text{11}}$ degree

                $\text{=-229}\dfrac{\text{1}}{\text{11}}$degree

Since ${{\text{1}}^{\text{o}}}\text{=60 }\!\!'\!\!\text{ }$

We have,

$\text{-4}$ radian$\text{=-22}{{\text{9}}^{\text{o}}}\text{+}\dfrac{\text{1 }\!\!\times\!\!\text{ 60}}{\text{11}}$ minutes                        

               $\text{=-22}{{\text{9}}^{\text{o}}}\text{+5 }\!\!'\!\!\text{ +}\dfrac{\text{5}}{\text{11}}$ minutes

Since $\text{1 }\!\!'\!\!\text{ =60 }\!\!'\!\!\text{  }\!\!'\!\!\text{ }$

$\text{-4}$ radian$\text{=-22}{{\text{9}}^{\text{o}}}\text{5 }\!\!'\!\!\text{ 27''}$

(iii) $\dfrac{\text{5 }\!\!\pi\!\!\text{ }}{\text{3}}$

Ans: We know that,

$\text{ }\!\!\pi\!\!\text{ }$ radian$\text{=18}{{\text{0}}^{\text{o}}}$

Therefore  $\text{1 radian =}{{\dfrac{\text{180}}{\text{ }\!\!\pi\!\!\text{ }}}^{\text{o}}}$ 

Hence,

$\dfrac{\text{5 }\!\!\pi\!\!\text{ }}{\text{3}}$ radian$\text{=}\dfrac{\text{180}}{\text{ }\!\!\pi\!\!\text{ }}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{5 }\!\!\pi\!\!\text{ }}{\text{3}}$ degree

                 $\text{=30}{{\text{0}}^{\text{o}}}$

(iv)$\dfrac{\text{7 }\!\!\pi\!\!\text{ }}{\text{6}}$

Ans: We know that,

 $\pi $ radian$\text{=18}{{\text{0}}^{\text{o}}}$

Therefore  $\text{1 radian =}{{\dfrac{\text{180}}{\text{ }\!\!\pi\!\!\text{ }}}^{\text{o}}}$ 

Hence,

$\dfrac{\text{7 }\!\!\pi\!\!\text{ }}{\text{6}}$ radian$\text{=}\dfrac{\text{180}}{\text{ }\!\!\pi\!\!\text{ }}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{7 }\!\!\pi\!\!\text{ }}{\text{6}}$

                 $\text{=21}{{\text{0}}^{\text{o}}}$

3. A wheel makes $\text{360}$ revolutions in one minute. Through how many radians does it turn in one second?

Ans: Number of revolutions the wheel makes in $\text{1}$ minute$\text{=360}$

 Number of revolutions the wheel make in $\text{1}$ second$\text{=}\dfrac{\text{360}}{\text{60}}$

                                                                                   $\text{=6}$

In one complete revolution, the wheel turns an angle of \[\text{2 }\!\!\pi\!\!\text{ }\] radian.

Hence, it will turn an angle of $\text{6 }\!\!\times\!\!\text{ 2 }\!\!\pi\!\!\text{ =12 }\!\!\pi\!\!\text{ }$ radian, in $\text{6}$ complete revolutions.

Therefore, the wheel turns an angle of $\text{12 }\!\!\pi\!\!\text{ }$ radian in one second.

4. Find the degree measure of the angle subtended at the centre of a circle of radius $\text{100}$cm by an arc of length  $\text{22}$  cm.

(Use$\text{ }\!\!\pi\!\!\text{ =}\dfrac{\text{22}}{\text{7}}$ ) 

Ans: We know that,

 in a circle of radius $\text{r}$ unit, if  an angle $\text{ }\!\!\theta\!\!\text{ }$ radian at the centre is subtended by an arc of length $\text{l}$ unit then

$\text{ }\!\!\theta\!\!\text{ =}\dfrac{\text{l}}{\text{r}}$                                ……(1)

Therefore, 

Substituting $\text{r=100cm}$ ,\[\text{l=22cm}\] in the formula (1) , we have,

$\text{ }\!\!\theta\!\!\text{ =}\dfrac{\text{22}}{\text{100}}$ radian

Since $\text{1 radian=}\dfrac{\text{180}}{\text{ }\!\!\pi\!\!\text{ }}$ 

Therefore,

 $\text{ }\!\!\theta\!\!\text{ =}\dfrac{\text{180}}{\text{ }\!\!\pi\!\!\text{ }}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{22}}{\text{100}}$ degree

     $\text{=}\dfrac{\text{180 }\!\!\times\!\!\text{ 7 }\!\!\times\!\!\text{ 22}}{\text{22 }\!\!\times\!\!\text{ 100}}$degree

     $\text{=}\dfrac{\text{63}}{\text{5}}$ degree

    $\text{=12}\dfrac{\text{3}}{\text{5}}$degree

Since  ${{\text{1}}^{\text{o}}}\text{=60 }\!\!'\!\!\text{ }$ , we have,

$\text{ }\!\!\theta\!\!\text{ =1}{{\text{2}}^{\text{o}}}\text{36 }\!\!'\!\!\text{ }$

Hence , the required angle is $\text{1}{{\text{2}}^{\text{o}}}\text{36 }\!\!'\!\!\text{ }$.

5. In a circle of diameter $\text{40}$ cm, the length of a chord is $\text{20}$ cm. Find the length of minor arc of the chord.

Ans: Given that, diameter of the circle$=40$ cm

 Hence Radius $\left( r \right)$ of the circle$=\dfrac{40}{2}cm$

                                                   $=20cm$

Let $\text{AB}$ be a chord  of the circle whose length is $20$ cm.

(Image will be uploaded soon)

In $\Delta \text{OAB,}$

$\text{OA=OB}$ 

     $=$ Radius of the circle

     $\text{=20cm}$

Now also, $\text{AB=20cm}$

Therefore, $\Delta \text{OAB}$ is an equilateral triangle.

\[\therefore \text{ }\!\!\theta\!\!\text{ =6}{{\text{0}}^{\text{o}}}\]

     $\text{=}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}$ radian

We know that,

 in a circle of radius $\text{r}$ unit, if  an angle $\text{ }\!\!\theta\!\!\text{ }$ radian at the centre is subtended by an arc of length $\text{l}$ unit then

$\text{ }\!\!\theta\!\!\text{ =}\dfrac{\text{l}}{\text{r}}$                                ……(1)

Substituting $\text{ }\!\!\theta\!\!\text{ =}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}$  in the formula (1),

       $\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}\text{=}\dfrac{\text{arc AB}}{\text{20}}$

$\text{arc AB=}\dfrac{\text{20 }\!\!\pi\!\!\text{ }}{\text{3}}\text{cm}$

Therefore, the length of the minor arc of the chord is $\dfrac{\text{20 }\!\!\pi\!\!\text{ }}{\text{3}}\text{cm}$ .

6. If in two circles, arcs of the same length subtend angles $\text{6}{{\text{0}}^{\text{o}}}$ and $\text{7}{{\text{5}}^{\text{o}}}$ at the centre, find the ratio of their radii.

Ans: Let the radii of the two circles be ${{\text{r}}_{\text{1}}}$ and ${{\text{r}}_{\text{2}}}$ . Let an arc of length ${{\text{l}}_{\text{1}}}$ subtends an angle of $\text{6}{{\text{0}}^{\text{o}}}$ at the centre of the circle of radius ${{\text{r}}_{\text{1}}}$ , whereas let an arc of length ${{\text{l}}_{\text{2}}}$ subtends an angle of $\text{7}{{\text{5}}^{\text{o}}}$ at the centre of the circle of radius ${{\text{r}}_{\text{2}}}$ .

Now, we have,

$\text{6}{{\text{0}}^{\text{o}}}\text{=}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}$ radian  and 

${{75}^{\circ }}=\dfrac{5\pi }{12}$ radian

We know that,

 in a circle of radius $\text{r}$ unit, if  an angle $\text{ }\!\!\theta\!\!\text{ }$ radian at the centre is subtended by an arc of length $\text{l}$ unit then

\[\text{ }\!\!\theta\!\!\text{ =}\dfrac{\text{l}}{\text{r}}\]  

$\text{l=r }\!\!\theta\!\!\text{ }$          

Hence we obtain,                   

$\text{l=}\dfrac{{{\text{r}}_{\text{1}}}\text{ }\!\!\pi\!\!\text{ }}{\text{3}}$ 

and  $\text{l=}\dfrac{{{\text{r}}_{\text{2}}}\text{5 }\!\!\pi\!\!\text{ }}{\text{12}}$

according to the ${{\text{l}}_{\text{1}}}\text{=}{{\text{l}}_{\text{2}}}$ 

thus we have,

$\dfrac{{{\text{r}}_{\text{1}}}\text{ }\!\!\pi\!\!\text{ }}{\text{3 }\!\!\pi\!\!\text{ }}\text{=}\dfrac{{{\text{r}}_{\text{2}}}\text{5 }\!\!\pi\!\!\text{ }}{\text{12}}$

  ${{\text{r}}_{\text{1}}}\text{=}\dfrac{{{\text{r}}_{\text{2}}}\text{5}}{\text{4}}$

 $\dfrac{{{\text{r}}_{\text{1}}}}{{{\text{r}}_{\text{2}}}}\text{=}\dfrac{\text{5}}{\text{4}}$

Hence , the ratio of the radii is $\text{5:4}$ .

7. Find the angle in radian through which a pendulum swings if its length is $\text{75}$ cm and the tip describes an arc of length.

(i)  $\text{10}$ cm

Ans: We know that,

 in a circle of radius $\text{r}$ unit, if  an angle $\text{ }\!\!\theta\!\!\text{ }$ radian at the centre is subtended by an arc of length $\text{l}$ unit then

$\text{ }\!\!\theta\!\!\text{ =}\dfrac{\text{l}}{\text{r}}$  

 Given that $\text{r=75cm}$

And here, $\text{l=10cm}$

Hence substituting the values in the formula,

$\text{ }\!\!\theta\!\!\text{ =}\dfrac{\text{10}}{\text{75}}$ radian

  $\text{=}\dfrac{\text{2}}{\text{15}}$radian

(ii) $\text{15}$ cm

Ans: We know that,

 in a circle of radius $\text{r}$ unit, if  an angle $\text{ }\!\!\theta\!\!\text{ }$ radian at the centre is subtended by an arc of length $\text{l}$ unit then

$\text{ }\!\!\theta\!\!\text{ =}\dfrac{\text{l}}{\text{r}}$  

 Given that $\text{r=75cm}$

And here, $\text{l=15cm}$  

Hence substituting the values in the formula,

$\text{ }\!\!\theta\!\!\text{ =}\dfrac{\text{15}}{\text{75}}$ radian

       $\text{=}\dfrac{\text{1}}{\text{5}}$radian

(iii) $\text{21}$ cm

Ans: We know that,

 in a circle of radius $\text{r}$ unit, if  an angle $\text{ }\!\!\theta\!\!\text{ }$ radian at the centre is subtended by an arc of length $\text{l}$ unit then

$\text{ }\!\!\theta\!\!\text{ =}\dfrac{\text{l}}{\text{r}}$  

And here, $\text{l=21cm}$

Hence substituting the values in the formula,

$\text{ }\!\!\theta\!\!\text{ =}\dfrac{\text{21}}{\text{75}}$ radian

       $\text{=}\dfrac{\text{7}}{\text{25}}$radian

Exercise 3.2

1. Find the values of the other five trigonometric functions if  $\text{cos x=-}\dfrac{\text{1}}{\text{2}}$ , $x$ lies in the third quadrant.

Ans: Here given that,  $\text{cos x=-}\dfrac{\text{1}}{\text{2}}$

Therefore we have,

$\text{sec x=}\dfrac{\text{1}}{\text{cos x}}$

          $\text{=}\dfrac{\text{1}}{\left( \text{-}\dfrac{\text{1}}{\text{2}} \right)}$

          $\text{=-2}$

Now we know that,$\text{si}{{\text{n}}^{\text{2}}}\text{x+co}{{\text{s}}^{\text{2}}}\text{x=1}$

Therefore we have, $\text{si}{{\text{n}}^{\text{2}}}\text{x=1-co}{{\text{s}}^{\text{2}}}\text{x}$

Substituting  $\text{cos x=-}\dfrac{\text{1}}{\text{2}}$ in the formula, we obtain,

$\text{si}{{\text{n}}^{\text{2}}}\text{x=1-}{{\left( \text{-}\dfrac{\text{1}}{\text{2}} \right)}^{\text{2}}}$

$\text{si}{{\text{n}}^{\text{2}}}\text{x=1-}\dfrac{\text{1}}{\text{4}}$

            $\text{=}\dfrac{\text{3}}{\text{4}}$

   $\text{sin x= }\!\!\pm\!\!\text{ }\dfrac{\sqrt{\text{3}}}{\text{2}}$

Since $\text{x}$ lies in the ${{\text{3}}^{\text{rd}}}$quadrant, the value of $\sin x$ will be negative.

$\text{sin x=-}\dfrac{\sqrt{\text{3}}}{\text{2}}$

Therefore, $\text{cosec x=}\dfrac{\text{1}}{\text{sin x}}$ 

                            $\text{=}\dfrac{\text{1}}{\left( \text{-}\dfrac{\sqrt{\text{3}}}{\text{2}} \right)}$ 

                            $\text{=-}\dfrac{\text{2}}{\sqrt{\text{3}}}$ 

Hence ,

 $\text{tan x=}\dfrac{\text{sin x}}{\text{cos x}}$ 

        $\text{=}\dfrac{\left( \text{-}\dfrac{\sqrt{\text{3}}}{\text{2}} \right)}{\left( \text{-}\dfrac{\text{1}}{\text{2}} \right)}$

        $\text{=}\sqrt{\text{3}}$

And 

$\text{cot x=}\dfrac{\text{1}}{\text{tan x}}$

       $\text{=}\dfrac{\text{1}}{\sqrt{\text{3}}}$

2. Find the values of other five trigonometric functions if $\text{sin  x=}\dfrac{\text{3}}{\text{5}}$  , $\text{x}$ lies in second quadrant.

Ans:

Here given that,  $\text{sin x=}\dfrac{\text{3}}{\text{5}}$

Therefore we have,

$\text{cosec x=}\dfrac{\text{1}}{\text{sin x}}$

          $=\dfrac{1}{\left( \dfrac{3}{5} \right)}$

           $=\dfrac{5}{3}$

Now we know that , ${{\sin }^{2}}x+{{\cos }^{2}}x=1$

Therefore we have, $\text{co}{{\text{s}}^{\text{2}}}\text{x=1-si}{{\text{n}}^{\text{2}}}\text{x}$

Substituting  $\sin x=\dfrac{3}{5}$  in the formula, we obtain,

$\text{co}{{\text{s}}^{\text{2}}}\text{x=1-}{{\left( \dfrac{\text{3}}{\text{5}} \right)}^{\text{2}}}$

$\text{co}{{\text{s}}^{\text{2}}}\text{x=1-}\dfrac{\text{9}}{\text{25}}$

         $\text{=}\dfrac{\text{16}}{\text{25}}$ 

   $\text{cos x= }\!\!\pm\!\!\text{ }\dfrac{\text{4}}{\text{5}}$

Since $x$ lies in the ${{2}^{nd}}$quadrant, the value of $\cos x$ will be negative.

$\text{cos x=-}\dfrac{\text{4}}{\text{5}}$

Therefore, $sec x=\dfrac{1}{\cos x}$ 

                          $\text{=}\dfrac{\text{1}}{\left( \text{-}\dfrac{\text{4}}{\text{5}} \right)}$ 

                          $\text{=-}\dfrac{\text{5}}{\text{4}}$ 

Hence ,

 $\tan x=\dfrac{\sin  x}{\cos x}$ 

        $\text{=}\dfrac{\left( \dfrac{\text{3}}{\text{5}} \right)}{\left( \text{-}\dfrac{\text{4}}{\text{5}} \right)}$

        $\text{=-}\dfrac{\text{3}}{\text{4}}$

And 

$\cot x=\dfrac{1}{\tan x}$

        $\text{=-}\dfrac{\text{4}}{\text{3}}$

3. Find the values of other five trigonometric functions if   $\text{cot x=}\dfrac{\text{3}}{\text{4}}$ , $\text{x}$ lies in third quadrant.

Ans: Here given that,  $\cot x=\dfrac{3}{4}$

Therefore we have,

$\tan x=\dfrac{1}{\cot x}$

       $=\dfrac{1}{\left( \dfrac{3}{4} \right)}$

       $=\dfrac{4}{3}$

Now we know that , \[\text{se}{{\text{c}}^{\text{2}}}\text{x-ta}{{\text{n}}^{\text{2}}}\text{x=1}\]

Therefore we have, $\text{se}{{\text{c}}^{\text{2}}}\text{x=1+ta}{{\text{n}}^{\text{2}}}\text{x}$

Substituting  $\text{tan x=}\dfrac{\text{4}}{\text{3}}$  in the formula, we obtain,

${{\sec }^{2}}x=1+{{\left( \dfrac{4}{3} \right)}^{2}}$

$\text{se}{{\text{c}}^{\text{2}}}\text{x=1+}\dfrac{\text{16}}{\text{9}}$

             $=\dfrac{25}{9}$ 

   $\text{sec x= }\!\!\pm\!\!\text{ }\dfrac{\text{5}}{\text{3}}$

Since $x$ lies in the ${{3}^{rd}}$quadrant, the value of $\sec x$ will be negative.

$\text{sec x=-}\dfrac{\text{5}}{\text{3}}$

Therefore, $\cos  x=\dfrac{1}{\sec x}$ 

                         $\text{=}\dfrac{\text{1}}{\left( \text{-}\dfrac{\text{5}}{\text{3}} \right)}$ 

                          $\text{=-}\dfrac{\text{3}}{\text{5}}$ 

Now  , $\text{tan x=}\dfrac{\text{sin x}}{\text{cos x}}$ 

Therefore, $\text{sin x=tan xcos x}$ 

 Hence we have, $\text{sin x=}\dfrac{\text{4}}{\text{3}}\text{ }\!\!\times\!\!\text{ }\left( \text{-}\dfrac{\text{3}}{\text{5}} \right)$   

                                    \[\text{=}\left( \text{-}\dfrac{\text{4}}{\text{5}} \right)\] 

 And 

$\text{cosec x=}\dfrac{\text{1}}{\text{sin x}}$

          $\text{=-}\dfrac{\text{5}}{\text{4}}$

4. Find the values of other five trigonometric functions if $\text{sec  x=}\dfrac{\text{13}}{\text{5}}$  , $\text{x}$ lies in fourth quadrant.

Ans: Here given that,  $\sec x=\dfrac{13}{5}$

Therefore we have,

$\cos x=\dfrac{1}{\sec x}$

        $=\dfrac{1}{\left( \dfrac{13}{5} \right)}$

        $=\dfrac{5}{13}$

Now we know that , $\text{se}{{\text{c}}^{\text{2}}}\text{x-ta}{{\text{n}}^{\text{2}}}\text{x=1}$

Therefore we have, $\text{ta}{{\text{n}}^{\text{2}}}\text{x=se}{{\text{c}}^{\text{2}}}\text{x-1}$

Substituting  \[\text{sec x=}\dfrac{\text{13}}{\text{5}}\]  in the formula, we obtain,

\[\text{ta}{{\text{n}}^{\text{2}}}\text{x=}{{\left( \dfrac{\text{13}}{\text{5}} \right)}^{\text{2}}}\text{-1}\]

$\text{ta}{{\text{n}}^{\text{2}}}\text{x=}\dfrac{\text{169}}{\text{25}}\text{-1}$

         $\text{=}\dfrac{\text{144}}{\text{25}}$ 

   \[\text{tanx= }\!\!\pm\!\!\text{ }\dfrac{\text{12}}{\text{5}}\]

Since $x$ lies in the ${{4}^{th}}$ quadrant, the value of $\tan x$ will be negative.

$\text{tan x=-}\dfrac{\text{12}}{\text{5}}$

Therefore, \[\text{cot x=}\dfrac{\text{1}}{\text{tan x}}\] 

                          $\text{=-}\dfrac{\text{5}}{\text{12}}$ 

Now  , $\text{tan x=}\dfrac{\text{sin x}}{\text{cos x}}$ 

Therefore, $\text{sin x=tan xcos x}$ 

 Hence we have, $\text{sin x=}\dfrac{\text{5}}{\text{13}}\text{ }\!\!\times\!\!\text{ }\left( \text{-}\dfrac{\text{12}}{\text{5}} \right)$   

                                  $\text{=}\left( \text{-}\dfrac{\text{12}}{\text{13}} \right)$ 

 And 

$\text{cosec x=}\dfrac{\text{1}}{\text{sin x}}$

            $\text{=-}\dfrac{\text{13}}{\text{12}}$

5. Find the values of other five trigonometric functions if  $\text{tan x=-}\dfrac{\text{5}}{\text{12}}$ , $\text{x}$ lies in second quadrant.

Ans: Here given that,  $\text{tan x=-}\dfrac{\text{5}}{\text{12}}$

Therefore we have,

$\text{cot x=}\dfrac{\text{1}}{\text{tan x}}$

       $\text{=}\dfrac{\text{1}}{\left( \text{-}\dfrac{\text{5}}{\text{12}} \right)}$

       $\text{=-}\dfrac{\text{12}}{\text{5}}$

Now we know that , $\text{se}{{\text{c}}^{\text{2}}}\text{x-ta}{{\text{n}}^{\text{2}}}\text{x=1}$

Therefore we have, $\text{se}{{\text{c}}^{\text{2}}}\text{x=1+ta}{{\text{n}}^{\text{2}}}\text{x}$

Substituting  $\text{tan x=-}\dfrac{\text{5}}{\text{12}}$  in the formula, we obtain,

$\text{se}{{\text{c}}^{\text{2}}}\text{x=1+}{{\left( \text{-}\dfrac{\text{5}}{\text{12}} \right)}^{\text{2}}}$

$\text{se}{{\text{c}}^{\text{2}}}\text{x=1+}\dfrac{\text{25}}{\text{144}}$

             $=\dfrac{169}{144}$ 

   $\text{sec x= }\!\!\pm\!\!\text{ }\dfrac{\text{13}}{\text{12}}$

Since $x$ lies in the ${{2}^{nd}}$ quadrant, the value of $\sec x$ will be negative.

$\text{sec x=-}\dfrac{\text{13}}{\text{12}}$

Therefore, $\text{cos x=}\dfrac{\text{1}}{\text{sec x}}$ 

                          $\text{=-}\dfrac{\text{12}}{\text{13}}$ 

Now  , $\text{tan x=}\dfrac{\text{sin x}}{\text{cos x}}$ 

Therefore, $\text{sin x=tan xcos x}$ 

 Hence we have, $\text{sin x=}\left( \text{-}\dfrac{\text{5}}{\text{12}} \right)\text{ }\!\!\times\!\!\text{ }\left( \text{-}\dfrac{\text{12}}{\text{13}} \right)$   

                                    $=\left( \dfrac{5}{13} \right)$ 

 And 

$\text{cosec x=}\dfrac{\text{1}}{\text{sin x}}$

           $\text{=}\dfrac{\text{13}}{\text{5}}$

6. Find the value of the trigonometric function $\text{sin76}{{\text{5}}^{\text{o}}}$ .

Ans: We know that the values of $\sin x$ repeat after an interval of $2\pi $ or ${{360}^{\circ }}$ .

Therefore we can write,

$\text{sin76}{{\text{5}}^{\text{o}}}\text{=sin}\left( \text{2 }\!\!\times\!\!\text{ 36}{{\text{0}}^{\text{o}}}\text{+4}{{\text{5}}^{\text{o}}} \right)$

            $\text{=sin4}{{\text{5}}^{\text{o}}}$

            $\text{=}\dfrac{\text{1}}{\sqrt{\text{2}}}\text{.}$

7. Find the value of the trigonometric function $\text{cosec}\left( \text{-141}{{\text{0}}^{\text{o}}} \right)$  

Ans: We  know that the values of $\text{cosec x}$ repeat after an interval of $\text{2 }\!\!\pi\!\!\text{ }$ or ${{360}^{\circ }}$ .

Therefore we can write,

           $\text{cosec}\left( \text{-141}{{\text{0}}^{\text{o}}} \right)\text{=cosec}\left( \text{-141}{{\text{0}}^{\text{o}}}\text{+4 }\!\!\times\!\!\text{ 36}{{\text{0}}^{\text{o}}} \right)$

                               $\text{=cosec}\left( \text{-141}{{\text{0}}^{\text{o}}}\text{+144}{{\text{0}}^{\text{o}}} \right)$

                               $\text{=cosec3}{{\text{0}}^{\text{o}}}$

                               $=2$ 

8. Find the value of the trigonometric function  $\text{tan}\dfrac{\text{19 }\!\!\pi\!\!\text{ }}{\text{3}}$ .

Ans: We know that the values of $\text{tan x}$ repeat after an interval of $\text{ }\!\!\pi\!\!\text{ }$ or \[{{180}^{\circ }}\].

Therefore we can write,

$\text{tan}\dfrac{\text{19 }\!\!\pi\!\!\text{ }}{\text{3}}\text{=tan6}\dfrac{\text{1}}{\text{3}}\text{ }\!\!\pi\!\!\text{ }$

            $\text{=tan}\left( \text{6 }\!\!\pi\!\!\text{ +}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}} \right)$

            \[\text{=tan}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}\]

            $\text{=}\sqrt{\text{3}}$      

9. Find the value of the trigonometric function $\text{sin}\left( \text{-}\dfrac{\text{11 }\!\!\pi\!\!\text{ }}{\text{3}} \right)$ 

Ans: We know that the values of $\text{sin x}$ repeat after an interval of $\text{2 }\!\!\pi\!\!\text{ }$ or ${{360}^{\circ }}$ .

Therefore we can write,

$\text{sin}\left( \text{-}\dfrac{\text{11 }\!\!\pi\!\!\text{ }}{\text{3}} \right)\text{=sin}\left( \text{-}\dfrac{\text{11 }\!\!\pi\!\!\text{ }}{\text{3}}\text{+2 }\!\!\times\!\!\text{ 2 }\!\!\pi\!\!\text{ } \right)$ 

                   $\text{=sin}\left( \dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}} \right)$

                    $=\dfrac{\sqrt{3}}{2}$

10. Find the value of the trigonometric function $\text{cot}\left( \text{-}\dfrac{\text{15 }\!\!\pi\!\!\text{ }}{\text{4}} \right)$ 

Ans: We  know that the values of $\text{cot x}$ repeat after an interval of $\text{ }\!\!\pi\!\!\text{ }$ or \[{{180}^{\circ }}\].

Therefore we can write,

$\text{cot}\left( \text{-}\dfrac{\text{15 }\!\!\pi\!\!\text{ }}{\text{4}} \right)\text{=cot}\left( \text{-}\dfrac{\text{15 }\!\!\pi\!\!\text{ }}{\text{4}}\text{+4 }\!\!\pi\!\!\text{ } \right)$ 

                   $\text{=cot}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}$

              $=1$

Exercise 3.3

1. Prove that $\text{si}{{\text{n}}^{\text{2}}}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}}\text{+co}{{\text{s}}^{\text{2}}}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}\text{-ta}{{\text{n}}^{\text{2}}}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\text{=-}\dfrac{\text{1}}{\text{2}}$

Ans: Substituting the values of  \[\text{sin}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}}\text{,cos}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}}\text{,tan}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\] on left hand side,

$\text{si}{{\text{n}}^{\text{2}}}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}}\text{+co}{{\text{s}}^{\text{2}}}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}\text{-ta}{{\text{n}}^{\text{2}}}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\text{=}{{\left( \dfrac{\text{1}}{\text{2}} \right)}^{\text{2}}}\text{+}{{\left( \dfrac{\text{1}}{\text{2}} \right)}^{\text{2}}}\text{-}{{\left( \text{1} \right)}^{\text{2}}}$

                                  \[\text{=}\dfrac{\text{1}}{\text{4}}\text{+}\dfrac{\text{1}}{\text{4}}\text{-1}\]

                                  $=-\dfrac{1}{2}$

                                  $=$ R.H.S.

Hence proved.

2. Prove that $\text{2si}{{\text{n}}^{\text{2}}}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}}\text{+cose}{{\text{c}}^{\text{2}}}\dfrac{\text{7 }\!\!\pi\!\!\text{ }}{\text{6}}\text{co}{{\text{s}}^{\text{2}}}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}\text{=}\dfrac{\text{3}}{\text{2}}$

Ans: Substituting the values of  $\text{sin}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}}\text{,cosec}\dfrac{\text{7 }\!\!\pi\!\!\text{ }}{\text{6}}\text{,cos}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}$ on left hand side,

L.H.S.$\text{=2si}{{\text{n}}^{\text{2}}}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}}\text{+cose}{{\text{c}}^{\text{2}}}\dfrac{\text{7 }\!\!\pi\!\!\text{ }}{\text{6}}\text{co}{{\text{s}}^{\text{2}}}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}$

           $\text{=2}{{\left( \dfrac{\text{1}}{\text{2}} \right)}^{\text{2}}}\text{+cose}{{\text{c}}^{\text{2}}}\left( \text{ }\!\!\pi\!\!\text{ +}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}} \right){{\left( \dfrac{\text{1}}{\text{2}} \right)}^{\text{2}}}$

           $\text{=2 }\!\!\times\!\!\text{ }\dfrac{\text{1}}{\text{4}}\text{+}{{\left( \text{-cosec}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}} \right)}^{\text{2}}}\left( \dfrac{\text{1}}{\text{4}} \right)$

          $\text{=}\dfrac{\text{1}}{\text{2}}\text{+}{{\left( \text{-2} \right)}^{\text{2}}}\left( \dfrac{\text{1}}{\text{4}} \right)$

Since $\text{cosec x}$ repeat its value after an interval of $\text{2 }\!\!\pi\!\!\text{ }$ , 

we have, $\text{cosec}\dfrac{\text{7 }\!\!\pi\!\!\text{ }}{\text{6}}\text{=-cosec}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}}$  

L.H.S  $=\dfrac{1}{2}+\dfrac{4}{4}$

           $=\dfrac{3}{2}$ 

          $=$ R.H.S.

Hence proved.

3. Prove that $\text{co}{{\text{t}}^{\text{2}}}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}}\text{+cosec}\dfrac{\text{5 }\!\!\pi\!\!\text{ }}{\text{6}}\text{+3ta}{{\text{n}}^{\text{2}}}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}}\text{=6}$

Ans: Substituting the values of  $\text{cot}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}}\text{,cosec}\dfrac{\text{5 }\!\!\pi\!\!\text{ }}{\text{6}}\text{,tan}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}}$ on left hand side,

L.H.S.$\text{=co}{{\text{t}}^{\text{2}}}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}}\text{+cosec}\dfrac{\text{5 }\!\!\pi\!\!\text{ }}{\text{6}}\text{+3ta}{{\text{n}}^{\text{2}}}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}}$

           $\text{=}{{\left( \sqrt{\text{3}} \right)}^{\text{2}}}\text{+cosec}\left( \text{ }\!\!\pi\!\!\text{ -}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}} \right)\text{+3}{{\left( \dfrac{\text{1}}{\sqrt{\text{3}}} \right)}^{\text{2}}}$

         $\text{=3+cosec}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}}\text{+3 }\!\!\times\!\!\text{ }\dfrac{\text{1}}{\text{3}}$

Since $\text{cosec x}$ repeat its value after an interval of $\text{2 }\!\!\pi\!\!\text{ }$ , 

we have, $\text{cosec}\dfrac{\text{5 }\!\!\pi\!\!\text{ }}{\text{6}}\text{=cosec}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}}$  

L.H.S $=3+2+1$ 

          $=1$ 

          $=$ R.H.S.

Hence proved.

4. Prove that $\text{2si}{{\text{n}}^{\text{2}}}\dfrac{\text{3 }\!\!\pi\!\!\text{ }}{\text{4}}\text{+2co}{{\text{s}}^{\text{2}}}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\text{+2se}{{\text{c}}^{\text{2}}}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}\text{=10}$

Ans:

Substituting the values of  $\text{sin}\dfrac{\text{3 }\!\!\pi\!\!\text{ }}{\text{4}}\text{,cos}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\text{,sec}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}$ on left hand side,

L.H.S.$\text{=2si}{{\text{n}}^{\text{2}}}\dfrac{\text{3 }\!\!\pi\!\!\text{ }}{\text{4}}\text{+2co}{{\text{s}}^{\text{2}}}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\text{+2se}{{\text{c}}^{\text{2}}}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}$

           $\text{=2}{{\left\{ \text{sin}\left( \text{ }\!\!\pi\!\!\text{ -}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}} \right) \right\}}^{\text{2}}}\text{+2}{{\left( \dfrac{\text{1}}{\sqrt{\text{2}}} \right)}^{\text{2}}}\text{+2}{{\left( \text{2} \right)}^{\text{2}}}$

           $\text{=2}{{\left\{ \text{sin}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}} \right\}}^{\text{2}}}\text{+2 }\!\!\times\!\!\text{ }\dfrac{\text{1}}{\text{2}}\text{+8}$

Since $\text{sin x}$ repeat its value after an interval of \[\text{2 }\!\!\pi\!\!\text{ }\] , 

we have, \[\text{sin}\dfrac{\text{3 }\!\!\pi\!\!\text{ }}{\text{4}}\text{=sin}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\]  

 L.H.S  $=1+1+8$ 

            $=10$

            $=$ R.H.S.

Hence proved.

5. Find the value of :

(i) $\text{sin7}{{\text{5}}^{\text{o}}}$

Ans: We have,

$\text{sin7}{{\text{5}}^{\text{o}}}\text{=sin(4}{{\text{5}}^{\text{o}}}\text{+3}{{\text{0}}^{\text{o}}}\text{)}$

          $\text{=sin4}{{\text{5}}^{\text{o}}}\text{cos3}{{\text{0}}^{\text{o}}}\text{+cos4}{{\text{5}}^{\text{o}}}\text{sin3}{{\text{0}}^{\text{o}}}$

Since we know that, $\text{sin}\left( \text{x+y} \right)\text{=sin x cos y+cos x sin y}$ 

Therefore we have,

$Sin 75^o = \dfrac{1}{\sqrt 2}\times \dfrac{\sqrt 3}{2} + \dfrac{1}{\sqrt 2}\times\dfrac{1}{2}$

$Sin 75^o = \dfrac{{\sqrt 3}+1}{2\sqrt 2}$

(ii) $\text{tan1}{{\text{5}}^{\text{o}}}$

Ans: We have,

\[\text{tan1}{{\text{5}}^{\text{o}}}\text{=tan}\left( \text{4}{{\text{5}}^{\text{o}}}\text{-3}{{\text{0}}^{\text{o}}} \right)\]

        $\text{=}\dfrac{\text{tan4}{{\text{5}}^{\text{o}}}\text{-tan3}{{\text{0}}^{\text{o}}}}{\text{1+tan4}{{\text{5}}^{\text{o}}}\text{tan3}{{\text{0}}^{\text{o}}}}$

Since we know, $\text{tan}\left( \text{x-y} \right)\text{=}\dfrac{\text{tan x-tan y}}{\text{1+tan x tan y}}$

Therefore we have,

$\text{tan1}{{\text{5}}^{\text{o}}}\text{=}\dfrac{\text{1-}\dfrac{\text{1}}{\sqrt{\text{3}}}}{\text{1+1}\left( \dfrac{\text{1}}{\sqrt{\text{3}}} \right)}$

           $\text{=}\dfrac{\dfrac{\sqrt{\text{3}}\text{-1}}{\sqrt{\text{3}}}}{\dfrac{\sqrt{\text{3}}\text{+1}}{\sqrt{\text{3}}}}$

         $\text{=}\dfrac{\sqrt{\text{3}}\text{-1}}{\sqrt{\text{3}}\text{+1}}$

         $\text{=}\dfrac{{{\left( \sqrt{\text{3}}\text{-1} \right)}^{\text{2}}}}{\left( \sqrt{\text{3}}\text{+1} \right)\left( \sqrt{\text{3}}\text{-1} \right)}$ 

Further computing we have,

$\text{tan1}{{\text{5}}^{\text{o}}}\text{=}\dfrac{\text{3+1-2}\sqrt{\text{3}}}{{{\left( \sqrt{\text{3}} \right)}^{\text{2}}}\text{-}{{\left( \text{1} \right)}^{\text{2}}}}$ 

           \[\text{=}\dfrac{\text{4-2}\sqrt{\text{3}}}{\text{3-1}}\]

           \[\text{=2-}\sqrt{\text{3}}\]

6. Prove that $\text{cos}\left( \dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\text{-x} \right)\text{cos}\left( \dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\text{-y} \right)\text{-sin}\left( \dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\text{-x} \right)\text{sin}\left( \dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\text{-y} \right)\text{=sin}\left( \text{x+y} \right)$

Ans: We know that, $\text{cos}\left( \text{x+y} \right)\text{=cos xcos y-sin xsin y}$ 

$\text{cos}\left( \dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\text{-x} \right)\text{cos}\left( \dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\text{-y} \right)\text{-sin}\left( \dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\text{-x} \right)\text{sin}\left( \dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\text{-y} \right)\text{=cos}\left[ \dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\text{-x+}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\text{-y} \right]$

$\text{=cos}\left[ \dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}\text{-}\left( \text{x+y} \right) \right]$

$\text{=sin}\left( \text{x+y} \right)$ 

L.H.S  $=$ R.H.S.

Hence  proved.

7. Prove that $\dfrac{\text{tan}\left( \dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\text{+x} \right)}{\text{tan}\left( \dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\text{-x} \right)}\text{=}{{\left( \dfrac{\text{1+tanx}}{\text{1-tanx}} \right)}^{\text{2}}}$

Ans: We  know that ,$\text{tan}\left( \text{A+B} \right)\text{=}\dfrac{\text{tan A+tan B}}{\text{1-tan Atan B}}$

and $\text{tan}\left( \text{A-B} \right)\text{=}\dfrac{\text{tan A-tan B}}{\text{1+tan Atan B}}$

L.H.S.$\text{=}\dfrac{\text{tan}\left( \dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\text{+x} \right)}{\text{tan}\left( \dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\text{-x} \right)}$

Using the above formula,

          $\text{L}\text{.H}\text{.S=}\dfrac{\left( \dfrac{\text{tan}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\text{+tanx}}{\text{1-tan}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\text{tanx}} \right)}{\dfrac{\text{tan}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\text{-tanx}}{\text{1+tan}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\text{tanx}}}$

                    $\text{=}\dfrac{\left( \dfrac{\text{1+tan x}}{\text{1-tan x}} \right)}{\left( \dfrac{\text{1-tan x}}{\text{1+tan x}} \right)}$      [ substituting $\text{tan}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\text{=1}$ ]

                    $\text{=}{{\left( \dfrac{\text{1+tan x}}{\text{1-tan x}} \right)}^{\text{2}}}$

                    $=$ R.H.S.

Hence proved.

8.  Prove that  $\dfrac{\text{cos}\left( \text{ }\!\!\pi\!\!\text{ +x} \right)\text{cos}\left( \text{-x} \right)}{\text{sin}\left( \text{ }\!\!\pi\!\!\text{ -x} \right)\text{cos}\left( \dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}\text{+x} \right)}\text{=co}{{\text{t}}^{\text{2}}}\text{x}$

Ans: Observe that $\text{cos x}$ repeats same value after an interval $\text{2 }\!\!\pi\!\!\text{ }$ 

and $\text{sin x}$ repeats same value after an interval  $\text{2 }\!\!\pi\!\!\text{ }$.

L.H.S.$\text{=}\dfrac{\text{cos}\left( \text{ }\!\!\pi\!\!\text{ +x} \right)\text{cos}\left( \text{-x} \right)}{\text{sin}\left( \text{ }\!\!\pi\!\!\text{ -x} \right)\text{cos}\left( \dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}\text{+x} \right)}$

           $\text{=}\dfrac{\left[ \text{-cos x} \right]\left[ \text{cos x} \right]}{\left( \text{sin x} \right)\left( \text{-sin x} \right)}$

           $\text{=}\dfrac{\text{-co}{{\text{s}}^{\text{2}}}\text{x}}{\text{-si}{{\text{n}}^{\text{2}}}\text{x}}$

           $\text{=co}{{\text{t}}^{\text{2}}}\text{x}$

Hence proved.

9. Prove that,

$\text{Cos}\left( \dfrac{\text{3 }\!\!\pi\!\!\text{ }}{\text{2}}\text{+x} \right)\text{Cos}\left( \text{2 }\!\!\pi\!\!\text{ +x} \right)\left[ \text{cot}\left( \dfrac{\text{3 }\!\!\pi\!\!\text{ }}{\text{2}}\text{-x} \right)\text{+cot}\left( \text{2 }\!\!\pi\!\!\text{ +x} \right) \right]\text{=1}$

Ans: We know that $\text{cot x}$  repeats same value after an interval $2\pi $ .

L.H.S.$=Cos\left( \dfrac{3\pi }{2}+x \right)Cos\left( 2\pi +x \right)\left[ cot\left( \dfrac{3\pi }{2}-x \right)+cot\left( 2\pi +x \right) \right]$

           $\text{=sin x cos x}\left[ \text{tan x+cot x} \right]$

Substituting $\text{tan x=}\dfrac{\text{sin x}}{\text{cos x}}$ and

$\text{cot x=}\dfrac{\text{cos x}}{\text{sin x}}$ ,

$\text{L}\text{.H}\text{.S=sin xcos x}\left( \dfrac{\text{sin x}}{\text{cos x}}\text{+}\dfrac{\text{cos x}}{\text{sin x}} \right)$

         $\text{=}\left( \text{sin x cos x} \right)\left[ \dfrac{\text{si}{{\text{n}}^{\text{2}}}\text{x+co}{{\text{s}}^{\text{2}}}\text{x}}{\text{sin x cos x}} \right]$

         $\text{=1}$ 

        $\text{=}$ R.H.S.

Hence proved.

10. Prove that $\text{sin}\left( \text{n+1} \right)\text{xsin}\left( \text{n+2} \right)\text{x+cos (n+1)x cos (n+2)x=cos x}$

Ans: We know that , $\text{cos}\left( \text{x-y} \right)\text{=cosxcosy+sinxsiny}$ 

L.H.S.$\text{=sin}\left( \text{n+1} \right)\text{xsin}\left( \text{n+2} \right)\text{x+cos (n+1)x cos (n+2)x}$

           $\text{=cos}\left[ \left( \text{n+1} \right)\text{x-}\left( \text{n+2} \right)\text{x} \right]$ 

           $\text{=cos}\left( \text{-x} \right)$ 

           $\text{=cosx}$ 

           $=$  R.H.S.

11. Prove that $\text{cos}\left( \dfrac{\text{3 }\!\!\pi\!\!\text{ }}{\text{4}}\text{+x} \right)\text{-cos}\left( \dfrac{\text{3 }\!\!\pi\!\!\text{ }}{\text{4}}\text{-x} \right)\text{=-}\sqrt{\text{2}}\text{sinx}$

Ans: We  know that , $\text{cos A-cos B=-2sin}\left( \dfrac{\text{A+B}}{\text{2}} \right)\text{.sin}\left( \dfrac{\text{A-B}}{\text{2}} \right)$

$\therefore $ L.H.S.$\text{=cos}\left( \dfrac{\text{3 }\!\!\pi\!\!\text{ }}{\text{4}}\text{+x} \right)\text{-cos}\left( \dfrac{\text{3 }\!\!\pi\!\!\text{ }}{\text{4}}\text{-x} \right)$

               $\text{=-2sin}\left\{ \dfrac{\left( \dfrac{\text{3 }\!\!\pi\!\!\text{ }}{\text{4}}\text{+x} \right)\text{+}\left( \dfrac{\text{3 }\!\!\pi\!\!\text{ }}{\text{4}}\text{-x} \right)}{\text{2}} \right\}\text{.sin}\left\{ \dfrac{\left( \dfrac{\text{3 }\!\!\pi\!\!\text{ }}{\text{4}}\text{+x} \right)\text{-}\left( \dfrac{\text{3 }\!\!\pi\!\!\text{ }}{\text{4}}\text{-x} \right)}{\text{2}} \right\}$

               $\text{=-2sin}\left( \dfrac{\text{3 }\!\!\pi\!\!\text{ }}{\text{4}} \right)\text{sin x}$ 

Since $\text{sin x}$ repeats the same value after an interval $\text{2 }\!\!\pi\!\!\text{ }$ , 

we have, $\text{sin}\left( \dfrac{\text{3 }\!\!\pi\!\!\text{ }}{\text{4}} \right)\text{=sin}\left( \text{ }\!\!\pi\!\!\text{ -}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}} \right)$ 

therefore, 

$\text{L}\text{.H}\text{.S=-2sin}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\text{sin x}$ 

           $\text{=-2 }\!\!\times\!\!\text{ }\dfrac{\text{1}}{\sqrt{\text{2}}}\text{ }\!\!\times\!\!\text{ sinx}$

           $\text{=-}\sqrt{\text{2}}\text{sin x}$

           $\text{=}$ R.H.S.

Hence proved.

12. Prove that $\text{si}{{\text{n}}^{\text{2}}}\text{6x-si}{{\text{n}}^{\text{2}}}\text{4x=sin 2x sin 10x}$

Ans: We know that,$\text{sinA+sinB=2sin}\left( \dfrac{\text{A+B}}{\text{2}} \right)\text{cos}\left( \dfrac{\text{A-B}}{\text{2}} \right)$

And $\text{sin A-sin B=2cos}\left( \dfrac{\text{A+B}}{\text{2}} \right)\text{sin}\left( \dfrac{\text{A-B}}{\text{2}} \right)$

$\therefore $ L.H.S.$\text{=si}{{\text{n}}^{\text{2}}}\text{6x-si}{{\text{n}}^{\text{2}}}\text{4xa}$

               $\text{=}\left( \text{sin 6x+sin 4x} \right)\left( \text{sin 6x-sin 4x} \right)$

               $\text{=}\left[ \text{2sin}\left( \dfrac{\text{6x+4x}}{\text{2}} \right)\text{cos}\left( \dfrac{\text{6x-4x}}{\text{2}} \right) \right]\left[ \text{2cos}\left( \dfrac{\text{6x+4x}}{\text{2}} \right)\text{.sin}\left( \dfrac{\text{6x-4x}}{\text{2}} \right) \right]$

               $\text{=}\left( \text{2sin 5x cos x} \right)\left( \text{2cos 5x sin x} \right)$

Now we know that, $\text{sin 2x=2sin x cos x}$ ,

Therefore we have,

$\text{L}\text{.H}\text{.S=}\left( \text{2sin 5x cos 5x} \right)\left( \text{2sin x cos x} \right)$ 

          $\text{=sin 10x sin 2x}$

          $\text{=}$ R.H.S.

13. Prove that $\text{co}{{\text{s}}^{\text{2}}}\text{2x-co}{{\text{s}}^{\text{2}}}\text{6x=sin 4x sin 8x}$

Ans: We  know that,

$\text{cos A+cos B=2cos}\left( \dfrac{\text{A+B}}{\text{2}} \right)\text{cos}\left( \dfrac{\text{A-B}}{\text{2}} \right)$

And $\text{cos A-cos B=-2sin}\left( \dfrac{\text{A+B}}{\text{2}} \right)\text{sin}\left( \dfrac{\text{A-B}}{\text{2}} \right)$

 L.H.S.$\text{=co}{{\text{s}}^{\text{2}}}\text{2x-co}{{\text{s}}^{\text{2}}}\text{6x}$

            $\text{=}\left( \text{cos 2x+cos 6x} \right)\left( \text{cos 2x-6x} \right)$

            $\text{=}\left[ \text{2cos}\left( \dfrac{\text{2x+6x}}{\text{2}} \right)\text{cos}\left( \dfrac{\text{2x-6x}}{\text{2}} \right) \right]\left[ \text{-2sin}\left( \dfrac{\text{2x+6x}}{\text{2}} \right)\text{sin}\left( \dfrac{\text{2x-6x}}{\text{2}} \right) \right]$

Further computing, we have,

$\text{L}\text{.H}\text{.S=}\left[ \text{2cos 4x cos}\left( \text{-2x} \right) \right]\left[ \text{-2sin 4xsin}\left( \text{-2x} \right) \right]$

         $\text{=}\left[ \text{2cos  4x cos 2x} \right]\left[ \text{-2sin 4x}\left( \text{-sin 2x} \right) \right]$

         $\text{=}\left( \text{2sin 4x cos 4x} \right)\left( \text{2sin 2xcos 2x} \right)$

Now we know that, $\text{sin 2x=2sin x cos x}$ 

Therefore we have,                

$\text{L}\text{.H}\text{.S=sin 8x sin 4x}$

         $\text{=}$  R.H.S.

.Hence proved.

14. Prove that $\text{sin 2x+2sin 4x+sin6=4co}{{\text{s}}^{\text{2}}}\text{xsin 4x}$

Ans: We know that, \[\text{sin A+sin B=2sin}\left( \dfrac{\text{A+B}}{\text{2}} \right)\text{cos}\left( \dfrac{\text{A-B}}{\text{2}} \right)\]

L.H.S.$\text{=sin 2x+2sin 4x+sin 6x}$

.           $\text{=}\left[ \text{sin 2x+sin 6x} \right]\text{+2sin 4x}$

           $\text{=}\left[ \text{2sin}\left( \dfrac{\text{2x+6x}}{\text{2}} \right)\text{cos}\left( \dfrac{\text{2x-6x}}{\text{2}} \right) \right]\text{+2sin4x}$

           $\text{=2sin 4xcos}\left( \text{-2x} \right)\text{+2sin 4x}$

Further computing,

We have, $\text{L}\text{.H}\text{.S=2sin 4x cos 2x+2sin 4x}$ 

                         $\text{=2sin 4x}\left( \text{cos 2x+1} \right)$ 

Now we know that, $\text{cos 2x+1=2co}{{\text{s}}^{\text{2}}}\text{x}$ 

Therefore we have,

$\text{L}\text{.H}\text{.S=2sin 4x}\left( \text{2co}{{\text{s}}^{\text{2}}}\text{x} \right)$ 

         $\text{=4co}{{\text{s}}^{\text{2}}}\text{xsin 4x}$

 = R.H.S.

Hence proved.

15. Prove that $\text{cot 4x}\left( \text{sin 5x+sin 3x} \right)\text{=cot x}\left( \text{sin 5x-sin 3x} \right)$

Ans: We know that, $\text{sin A+sin B=2sin}\left( \dfrac{\text{A+B}}{\text{2}} \right)\text{cos}\left( \dfrac{\text{A-B}}{\text{2}} \right)$

L.H.S.$\text{=cot 4x}\left( \text{sin 5x+sin 3x} \right)$

          $\text{=}\dfrac{\text{cot 4x}}{\text{sin 4x}}\left[ \text{2sin}\left( \dfrac{\text{5x+3x}}{\text{2}} \right)\text{cos}\left( \dfrac{\text{5x+3x}}{\text{2}} \right) \right]$

          $\text{=}\left( \dfrac{\text{cos 4x}}{\text{sin 4x}} \right)\left[ \text{2sin 4x cos x} \right]$

          $\text{=2cos 4x cos x}$

Now also ,we know that, $\text{sin A-sin B=2cos}\left( \dfrac{\text{A+B}}{\text{2}} \right)\text{sin}\left( \dfrac{\text{A-B}}{\text{2}} \right)$

 R.H.S.$\text{=cot x}\left( \text{sin 5x-sin 3x} \right)$

            $\text{=}\dfrac{\text{cos x}}{\text{sin x}}\left[ \text{2cos}\left( \dfrac{\text{5x+3x}}{\text{2}} \right)\text{sin}\left( \dfrac{\text{5x-3x}}{\text{2}} \right) \right]$

            $\text{=}\dfrac{\text{cos x}}{\text{sin x}}\left[ \text{2cos 4x sin x} \right]$

            $\text{=2cos 4x cos x}$

Therefore , we can conclude that,

L.H.S.=R.H.S.

Hence proved.

16. Prove that $\dfrac{\text{cos 9x-cos 5x}}{\text{sin 17x-sin 3x}}\text{=-}\dfrac{\text{sin 2x}}{\text{cos 10x}}$

Ans: We know that,

$\text{cos A-cos B=-2sin}\left( \dfrac{\text{A+B}}{\text{2}} \right)\text{sin}\left( \dfrac{\text{A-B}}{\text{2}} \right)$

And  $\text{sin A-sin B=2cos}\left( \dfrac{\text{A+B}}{\text{2}} \right)\text{sin}\left( \dfrac{\text{A-B}}{\text{2}} \right)$

 L.H.S.$\text{=}\dfrac{\text{cos 9x-cos 5x}}{\text{sin 17x-sin 3x}}$

            \[\text{=}\dfrac{\text{-2sin}\left( \dfrac{\text{9x+5x}}{\text{2}} \right)\text{.sin}\left( \dfrac{\text{9x-5x}}{\text{2}} \right)}{\text{2cos}\left( \dfrac{\text{17x+3x}}{\text{2}} \right)\text{.sin}\left( \dfrac{\text{17x-3x}}{\text{2}} \right)}\]                    (following the formula)

            $\text{=}\dfrac{\text{-2sin 7x}\text{.sin 2x}}{\text{2cos 10x}\text{.sin 7x}}$

            $\text{=-}\dfrac{\text{sin 2x}}{\text{cos 10x}}$ 

            $=$ R.H.S.

Hence proved.

17. Prove that:$\dfrac{\text{sin 5x+sin 3x}}{\text{cos 5x+cos 3x}}\text{=tan 4x}$

Ans:

We  know that

$\text{sin A+sin B=2sin}\left( \dfrac{\text{A+B}}{\text{2}} \right)\text{cos}\left( \dfrac{\text{A-B}}{\text{2}} \right)\text{,}$

$\text{cos A+cos B=2cos}\left( \dfrac{\text{A+B}}{\text{2}} \right)\text{cos}\left( \dfrac{\text{A-B}}{\text{2}} \right)$

Now , L.H.S.$\text{=}\dfrac{\text{sin 5x+sin 3x}}{\text{cos 5x+cos 3x}}$

                      $\text{=}\dfrac{\text{2sin}\left( \dfrac{\text{5x+3x}}{\text{2}} \right)\text{cos}\left( \dfrac{\text{5x-3x}}{\text{2}} \right)}{\text{2cos}\left( \dfrac{\text{5x+3x}}{\text{2}} \right)\text{cos}\left( \dfrac{\text{5x-3x}}{\text{2}} \right)}$                  (using the formula)

                      $\text{=}\dfrac{\text{2sin}\left( \dfrac{\text{5x+3x}}{\text{2}} \right)\text{cos}\left( \dfrac{\text{5x-3x}}{\text{2}} \right)}{\text{2cos}\left( \dfrac{\text{5x+3x}}{\text{2}} \right)\text{cos}\left( \dfrac{\text{5x-3x}}{\text{2}} \right)}$

                      $\text{=}\dfrac{\text{2sin 4x cos x}}{\text{2cos 4x cos x}}$

Further computing we have,

$\text{L}\text{.H}\text{.S=tan 4x}$ 

          $=$ R.H.S.

18. Prove that \[\dfrac{\text{sin x-sin y}}{\text{cos x+cos y}}\text{=tan}\dfrac{\text{x-y}}{\text{2}}\]

Ans: We  know that,

$\text{sin A-sin B=2cos}\left( \dfrac{\text{A+B}}{\text{2}} \right)\text{sin}\left( \dfrac{\text{A-B}}{\text{2}} \right)\text{,}$

.$\text{cos A+cos B=2cos}\left( \dfrac{\text{A+B}}{\text{2}} \right)\text{cos}\left( \dfrac{\text{A-B}}{\text{2}} \right)$

  L.H.S.\[\text{=}\dfrac{\text{sin x-sin y}}{\text{cosx+cosy}}\]

               $\text{=}\dfrac{\text{2cos}\left( \dfrac{\text{x+y}}{\text{2}} \right)\text{.sin}\left( \dfrac{\text{x-y}}{\text{2}} \right)}{\text{2cos}\left( \dfrac{\text{x+y}}{\text{2}} \right)\text{.cos}\left( \dfrac{\text{x-y}}{\text{2}} \right)}$

               $\text{=}\dfrac{\text{sin}\left( \dfrac{\text{x-y}}{\text{2}} \right)}{\text{cos}\left( \dfrac{\text{x-y}}{\text{2}} \right)}$

              $\text{=tan}\left( \dfrac{\text{x-y}}{\text{2}} \right)$

Therefore $\text{L}\text{.H}\text{.S=R}\text{.H}\text{.S}$ 

Hence proved.

19. Prove that $\dfrac{\text{sin x+sin 3x}}{\text{cos x+cos 3x}}\text{=tan 2x}$

Ans: We  know that

$\text{sin A+sin B=2sin}\left( \dfrac{\text{A+B}}{\text{2}} \right)\text{cos}\left( \dfrac{\text{A+B}}{\text{2}} \right)\text{,}$

.\[\text{cos A+cos B=2cos}\left( \dfrac{\text{A+B}}{\text{2}} \right)\text{cos}\left( \dfrac{\text{A-B}}{\text{2}} \right)\]

Now , L.H.S.$\text{=}\dfrac{\text{sinx+sin3x}}{\text{cos x+cos 3x}}$

                      $\text{=}\dfrac{\text{2sin}\left( \dfrac{\text{x+3x}}{\text{2}} \right)\text{cos}\left( \dfrac{\text{x-3x}}{\text{2}} \right)}{\text{2cos}\left( \dfrac{\text{x+3x}}{\text{2}} \right)\text{cos}\left( \dfrac{\text{x-3x}}{\text{2}} \right)}$                   (using the formula)

                      $\text{=}\dfrac{\text{sin 2x}}{\text{cos 2x}}$

                      $\text{=tan 2x}$

Therefore  L.H.S$=$ R.H.S.

Hence proved.

20. Prove that $\dfrac{\text{sin x-sin 3x}}{\text{si}{{\text{n}}^{\text{2}}}\text{x-co}{{\text{s}}^{\text{2}}}\text{x}}\text{=2sin x}$

Ans: We know that,

$\text{sin A-sin B=2cos}\left( \dfrac{\text{A+B}}{\text{2}} \right)\text{sin}\left( \dfrac{\text{A-B}}{\text{2}} \right)$

And \[\text{co}{{\text{s}}^{\text{2}}}\text{A-si}{{\text{n}}^{\text{2}}}\text{A=cos 2A}\]

 L.H.S.$\text{=}\dfrac{\text{sin x-sin 3x}}{\text{si}{{\text{n}}^{\text{2}}}\text{x-co}{{\text{s}}^{\text{2}}}\text{x}}$

       \[\text{=}\dfrac{\text{2cos}\left( \dfrac{\text{x+3x}}{\text{2}} \right)\text{sin}\left( \dfrac{\text{x-3x}}{\text{2}} \right)}{\text{-cos2x}}\]

            $\text{=}\dfrac{\text{2cos2xsin}\left( \text{-x} \right)}{\text{-cos 2x}}$

            $\text{=-2 }\!\!\times\!\!\text{ }\left( \text{-sinx} \right)$

Therefore , we have,

$\text{L}\text{.H}\text{.S=2sin x}$

        $=$ R.H.S.

Hence proved.

21. Prove that $\dfrac{\text{cos 4x+cos 3x+cos 2x}}{\text{sin 4x+sin 3x+sin 2x}}\text{=cot 3x}$

Ans: We know that,

$\text{cos A+cos B=2cos}\left( \dfrac{\text{A+B}}{\text{2}} \right)\text{cos}\left( \dfrac{\text{A-B}}{\text{2}} \right)$

And, $\text{sin A+sin B=2sin}\left( \dfrac{\text{A+B}}{\text{2}} \right)\text{cos}\left( \dfrac{\text{A-B}}{\text{2}} \right)$

Now, L.H.S.$\text{=}\dfrac{\text{cos 4x+cos 3x+cos 2x}}{\text{sin 4x+sin 3x+sin 2x}}$

                     \[\text{=}\dfrac{\left( \text{cos 4x+cos 2x} \right)\text{+cos 3x}}{\left( \text{sin4x+sin2x} \right)\text{+sin 3x}}\]

                     \[\text{=}\dfrac{\text{2cos}\left( \dfrac{\text{4x+2x}}{\text{2}} \right)\text{cos}\left( \dfrac{\text{4x-2x}}{\text{2}} \right)\text{+cos3x}}{\text{2sin}\left( \dfrac{\text{4x+2x}}{\text{2}} \right)\text{cos}\left( \dfrac{\text{4x-2x}}{\text{2}} \right)\text{+sin 3x}}\]   (using the formulas)

                      \[\text{=}\dfrac{\text{2cos 3x cos x+cos 3x}}{\text{2sin 3x cos x+sin 3x}}\]

Further computing, we obtain,

L.H.S$\text{=}\dfrac{\text{cos 3x}\left( \text{2cos x+1} \right)}{\text{sin 3x}\left( \text{2cos x+1} \right)}$ 

          \[\text{=cot 3x}\] 

          $=$ R.H.S.

Hence proved.

22. Prove that \[\text{cot x cot 2x-cot 2x cot 3x-cot 3x cot x=1}\]

Ans:

We know that, \[\text{cot}\left( \text{A+B} \right)\text{=}\dfrac{\text{cotAcotB-1}}{\text{cot A+cot B}}\]

Now , L.H.S.$\text{=cot xcot 2x-cot 2x cot 3x-cot 3x cot x}$

                      \[\text{=cot x cot 2x-cot 3x}\left( \text{cot 2x+cot x} \right)\]

                      \[\text{=cot x cot 2x-cot}\left( \text{2x+x} \right)\left( \text{cot 2x+cot x} \right)\]

                      \[\text{=cot x cot 2x-}\left[ \dfrac{\text{cot 2x cot x-1}}{\text{cot x+cot 2x}} \right]\left( \text{cot 2x+cot x} \right)\]

Further computing we obtain,

$\text{L}\text{.H}\text{.S=cot x cot 2x-}\left( \text{cot 2x cot x-1} \right)$ 

         \[\text{=1}\]

         $\text{=}$ R.H.S.

Hence proved.

23. Prove that $\text{tan 4x=}\dfrac{\text{4tan x}\left( \text{1-ta}{{\text{n}}^{\text{2}}}\text{x} \right)}{\text{1-6ta}{{\text{n}}^{\text{2}}}\text{x+ta}{{\text{n}}^{\text{4}}}\text{x}}$

Ans: We  know that $\text{tan 2A=}\dfrac{\text{2tan A}}{\text{1-ta}{{\text{n}}^{\text{2}}}\text{A}}$

L.H.S.$\text{=tan 4x}$

           $\text{=tan2}\left( \text{2x} \right)$

           \[\text{=}\dfrac{\text{2tan 2x}}{\text{1-ta}{{\text{n}}^{\text{2}}}\left( \text{2x} \right)}\][using the formula]

           $\text{=}\dfrac{\left( \dfrac{\text{4tan x}}{\text{1-ta}{{\text{n}}^{\text{2}}}\text{x}} \right)}{\left[ \text{1-}\dfrac{\text{4ta}{{\text{n}}^{\text{2}}}\text{x}}{{{\left( \text{1-ta}{{\text{n}}^{\text{2}}}\text{x} \right)}^{\text{2}}}} \right]}$

Further  computing, we obtain,

L.H.S $\text{=}\dfrac{\left( \dfrac{\text{4tan x}}{\text{1-ta}{{\text{n}}^{\text{2}}}\text{x}} \right)}{\left[ \dfrac{{{\left( \text{1-ta}{{\text{n}}^{\text{2}}}\text{x} \right)}^{\text{2}}}\text{4ta}{{\text{n}}^{\text{2}}}\text{x}}{{{\left( \text{1-ta}{{\text{n}}^{\text{2}}}\text{x} \right)}^{\text{2}}}} \right]}$$$$$

          $\text{=}\dfrac{\text{4tan x}\left( \text{1-ta}{{\text{n}}^{\text{2}}}\text{x} \right)}{\text{1+ta}{{\text{n}}^{\text{4}}}\text{x-2ta}{{\text{n}}^{\text{2}}}\text{x-4ta}{{\text{n}}^{\text{2}}}\text{x}}$

          $\text{=}\dfrac{\text{4tan x}\left( \text{1-ta}{{\text{n}}^{\text{2}}}\text{x} \right)}{\text{1-6ta}{{\text{n}}^{\text{2}}}\text{x+ta}{{\text{n}}^{\text{4}}}\text{x}}$ 

          $=$ R.H.S.

Hence  proved.

24. Prove that $\text{cos 4x=1-8si}{{\text{n}}^{\text{2}}}\text{xco}{{\text{s}}^{\text{2}}}\text{x}$

Ans: We know that, $\text{cos 2x=1-2si}{{\text{n}}^{\text{2}}}\text{x}$ 

And $\text{sin 2x=2sin x cos x}$ 

L.H.S.\[\text{=cos 4x}\]

           $\text{=cos 2}\left( \text{2x} \right)$

           $\text{=1-2si}{{\text{n}}^{\text{2}}}\text{2x}$

           $\text{=1-2}{{\left( \text{2sin x cos x} \right)}^{\text{2}}}$

Further computing we get,

L.H.S$\text{=1-8si}{{\text{n}}^{\text{2}}}\text{xco}{{\text{s}}^{\text{2}}}\text{x}$ 

          $=$R.H.S.

Hence  proved.

25. Prove that $\text{cos 6x=32xco}{{\text{s}}^{\text{6}}}\text{x-48co}{{\text{s}}^{\text{4}}}\text{x+18co}{{\text{s}}^{\text{2}}}\text{x-1}$

Ans: We know that, $\text{cos 3A=4co}{{\text{s}}^{\text{3}}}\text{A-3cosA}$

and  $\text{cos 2x=1-2si}{{\text{n}}^{\text{2}}}\text{x}$

L.H.S.$\text{=cos 6x}$

           $\text{=cos 3}\left( \text{2x} \right)$

           \[\text{=4co}{{\text{s}}^{\text{3}}}\text{2x-3cos 2x}\]

           \[\text{=4}\left[ {{\left( \text{2co}{{\text{s}}^{\text{2}}}\text{x-1} \right)}^{\text{3}}}\text{-3}\left( \text{2co}{{\text{s}}^{\text{2}}}\text{x-1} \right) \right]\]

Further computing,

L.H.S$\text{=4}\left[ {{\left( \text{2co}{{\text{s}}^{\text{2}}}\text{x} \right)}^{\text{3}}}\text{-}{{\left( \text{1} \right)}^{\text{3}}}\text{-3}\left( \text{2co}{{\text{s}}^{\text{2}}}\text{x} \right) \right]\text{-6co}{{\text{s}}^{\text{2}}}\text{x+3}$

         $\text{=4}\left[ {{\left( \text{2co}{{\text{s}}^{\text{2}}}\text{x} \right)}^{\text{3}}}\text{-}{{\left( \text{1} \right)}^{\text{3}}}\text{-3}{{\left( \text{2co}{{\text{s}}^{\text{2}}}\text{x} \right)}^{\text{2}}}\text{+3}\left( \text{2co}{{\text{s}}^{\text{2}}}\text{x} \right) \right]\text{-6co}{{\text{s}}^{\text{2}}}\text{x+3}$

         $\text{=4}\left[ \text{8co}{{\text{s}}^{\text{6}}}\text{x-1-12co}{{\text{s}}^{\text{4}}}\text{x+6co}{{\text{s}}^{\text{2}}}\text{x} \right]\text{-6co}{{\text{s}}^{\text{2}}}\text{x+3}$

         $\text{=32co}{{\text{s}}^{\text{6}}}\text{x-48co}{{\text{s}}^{\text{4}}}\text{x+18co}{{\text{s}}^{\text{2}}}\text{x-1}$

Therefore we have,

L.H.S $=$ R.H.S.

Hence  proved.

Exercise 3.4

1. Find the principal and general solutions of the \[\text{tan x=}\sqrt{\text{3}}\].

Ans: Here given that,

\[\text{tan x=}\sqrt{\text{3}}\]

We know that $\text{tan}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}\text{=}\sqrt{\text{3}}$

and $\text{tan}\left( \dfrac{\text{4 }\!\!\pi\!\!\text{ }}{\text{3}} \right)\text{=tan}\left( \text{ }\!\!\pi\!\!\text{ +}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}} \right)\text{=tan}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}\text{=}\sqrt{\text{3}}$

Therefore, the principal solutions are\[\text{x=}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}\] and \[\dfrac{\text{4 }\!\!\pi\!\!\text{ }}{\text{3}}\] .

Now, \[\text{tan x=tan}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}\]

Which implies,

$\text{x=n }\!\!\pi\!\!\text{ +}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}$ , where \[\text{n}\in \text{Z}\]

Therefore, the general solution is \[\text{x=n }\!\!\pi\!\!\text{ +}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}\], where $\text{n}\in \text{Z}$ .

2. Find the principal and general solutions of the equation $\text{secx=2}$

Ans: Here it is given that,

$\text{sec x=2}$

Now we know that

$\text{sec}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}\text{=2}$ and 

$\text{sec}\dfrac{\text{5 }\!\!\pi\!\!\text{ }}{\text{3}}\text{=sec}\left( \text{2 }\!\!\pi\!\!\text{ -}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}} \right)$

          $\text{=sec}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}$

          $\text{=2}$ 

Therefore, the principal solutions are$\text{x=}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}$ and \[\dfrac{\text{5 }\!\!\pi\!\!\text{ }}{\text{3}}\].

Now, $\text{sec x=sec}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}$

and we know , 

$\sec x=\dfrac{1}{\cos  x}$

Therefore , we have,

\[\text{cos x=cos}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}\]

Which implies,

$\text{x=2n }\!\!\pi\!\!\text{  }\!\!\pm\!\!\text{ }\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}$ , where $\text{n}\in \text{Z}$ .

Therefore, the general solution is $\text{x=2n }\!\!\pi\!\!\text{  }\!\!\pm\!\!\text{ }\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}$ , where $n\in Z$ .

3. Find the principal and general solutions of the equation $\text{cot x=-}\sqrt{\text{3}}$

Ans: Here it is given that,

\[\text{cot x=-}\sqrt{\text{3}}\]

Now we know that $\text{cot}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}}\text{=}\sqrt{\text{3}}$

And 

\[\text{cot}\left( \text{ }\!\!\pi\!\!\text{ -}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}} \right)\text{=-cot}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}}\]

                 \[\text{=-}\sqrt{\text{3}}\]

and  $\text{cot}\left( \text{2 }\!\!\pi\!\!\text{ -}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}} \right)\text{=-cot}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}}$

                          \[\text{=-}\sqrt{\text{3}}\]

Therefore we have,

$\text{cot}\dfrac{\text{5 }\!\!\pi\!\!\text{ }}{\text{6}}\text{=-}\sqrt{\text{3}}$ 

and $\text{cot}\dfrac{\text{11 }\!\!\pi\!\!\text{ }}{\text{6}}\text{=-}\sqrt{\text{3}}$

Therefore, the principal solutions are $\text{x=}\dfrac{\text{5 }\!\!\pi\!\!\text{ }}{\text{6}}$ and $\dfrac{\text{11 }\!\!\pi\!\!\text{ }}{\text{6}}$.

.Now, $\text{cot x=cot}\dfrac{\text{5 }\!\!\pi\!\!\text{ }}{\text{6}}$

And we know \[\text{cot x=}\dfrac{\text{1}}{\text{tan x}}\]

Therefore we have,

$\text{tan x=tan}\dfrac{\text{5 }\!\!\pi\!\!\text{ }}{\text{6}}$

Which implies,

$\text{x=n }\!\!\pi\!\!\text{ +}\dfrac{\text{5 }\!\!\pi\!\!\text{ }}{\text{6}}$ , where $\text{n}\in \text{Z}$

Therefore, the general solution is $\text{x=n }\!\!\pi\!\!\text{ +}\dfrac{\text{5 }\!\!\pi\!\!\text{ }}{\text{6}}$ , where $\text{n}\in \text{Z}$ .

4. Find the general solution of $\text{cosec x=-2}$

Ans: Here it is given that,

$\text{cosec x=-2}$

Now we know that

$\text{cosec}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}}\text{=2}$

and

$\text{cosec}\left( \text{ }\!\!\pi\!\!\text{ +}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}} \right)\text{=-cosec}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}}\text{a}$ 

                     $\text{=-2}$

and \[\text{cosec}\left( \text{2 }\!\!\pi\!\!\text{ -}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}} \right)\text{=-cosec}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}}\]

                                        \[\text{=-2}\]

therefore we have,

\[\text{cosec}\dfrac{\text{7 }\!\!\pi\!\!\text{ }}{\text{6}}\text{=-2}\]and $\text{cosec}\dfrac{\text{11 }\!\!\pi\!\!\text{ }}{\text{6}}\text{=-2}$

Hence , the principal solutions are$\text{x=}\dfrac{\text{7 }\!\!\pi\!\!\text{ }}{\text{6}}\,$ and $\text{ }\dfrac{11\pi }{6}$.

Now, $\text{cosec x=cosec}\dfrac{\text{7 }\!\!\pi\!\!\text{ }}{\text{6}}$

And we know, \[\text{cosec x=}\dfrac{\text{1}}{\text{sin x}}\]

Therefore , we have,

$\text{sin x=sin}\dfrac{\text{7 }\!\!\pi\!\!\text{ }}{\text{6}}$

Which implies,

$\text{x=n }\!\!\pi\!\!\text{ +}{{\left( \text{-1} \right)}^{\text{n}}}\dfrac{\text{7 }\!\!\pi\!\!\text{ }}{\text{6}}$  

,where $\text{n}\in \text{Z}$.

Therefore, the general solution is $\text{x=n }\!\!\pi\!\!\text{ +}{{\left( \text{-1} \right)}^{\text{n}}}\dfrac{\text{7 }\!\!\pi\!\!\text{ }}{\text{6}}\text{ }$ ,where $\text{n}\in \text{Z}$.

5. Find the general solution of the equation $\text{cos 4x=cos 2x}$

Ans: Here it is given that, $\text{cos 4x=cos 2x}$

Which implies,

$\text{cos 4x-cos 2x=0}$

Now we know that, $\text{cos A-cos B=-2sin}\left( \dfrac{\text{A+B}}{\text{2}} \right)\text{sin}\left( \dfrac{\text{A-B}}{\text{2}} \right)$ 

Therefore we have,

$\text{-2sin}\left( \dfrac{\text{4x+2x}}{\text{2}} \right)\text{sin}\left( \dfrac{\text{4x-2x}}{\text{2}} \right)\text{=0}$

                            $\text{sin 3x sin x=0}$

Hence we have, $\text{sin 3x=0}\,\,$

Or, $\text{ sin x=0}$

Therefore, $\text{3x=n }\!\!\pi\!\!\text{ }$

 or    $\text{x=n }\!\!\pi\!\!\text{ }$    ,where $\text{ n}\in \text{Z}$

 therefore, \[\text{x=}\dfrac{\text{n }\!\!\pi\!\!\text{ }}{\text{3}}\]     

 or  $\text{x=n }\!\!\pi\!\!\text{ }$  ,where $\text{ n}\in \text{Z}$.

6. Find the general solution of the equation $\text{cos 3x+cos x-cos 2x=0}$.

Ans: Here given that,

$\text{cos 3x+cos x-cos 2x=0}$

Now we know that, $\text{cos A+cos B=2cos}\left( \dfrac{\text{A+B}}{\text{2}} \right)\text{cos}\left( \dfrac{\text{A-B}}{\text{2}} \right)$

Therefore  $\text{cos 3x+cos x-cos 2x=0}$ implies

$\text{2cos}\left( \dfrac{\text{3x+x}}{\text{2}} \right)\text{cos}\left( \dfrac{\text{3x-x}}{\text{2}} \right)\text{-cos 2x=0}$

                     $\text{2cos 2x cos x-cos 2x=0}$

                        $\text{cos 2x}\left( \text{2cos x-1} \right)\text{=0}$

Hence we have, 

 Either $\text{cos 2x=0}$

Or $\text{cos x=}\dfrac{\text{1}}{\text{2}}$ 

Which in turn implies that,

Either $\text{2x=}\left( \text{2n+1} \right)\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}\,$

Or,  $\text{cos x=cos}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}$    , where  $\text{n}\in \text{Z}$

Therefore, 

Either  $\text{x=}\left( \text{2n+1} \right)\dfrac{\text{ }\!\!\pi\!\!\text{ }}{4}\,\,$

Or,  $\text{x=2n }\!\!\pi\!\!\text{  }\!\!\pm\!\!\text{ }\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}$  ,where  \[\text{n}\in \text{Z}\].

7. Find the general solution of the equation \[\text{sin 2x+cos x=0}\] .

Ans: Here it is given that,

\[\text{sin 2x cos x=0}\]

Now we know that, $\text{sin 2x=2sin x cos x}$ 

Therefore we have,

\[\text{2sin x cos x+cos x=0}\]

Which implies,

     \[\text{cos x(2sin x+1)= }\!\!~\!\!\text{ 0}\]

Therefore we have,

Either $\text{cos x=0}$ 

Or, $\text{sin x=-}\dfrac{\text{1}}{\text{2}}$  

Hence we have,

 \[\text{x= }\!\!~\!\!\text{ (2n+1)}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}\]   , where $\text{n}\in \text{Z}$ .

Either 

Or,  \[\text{sin x=-}\dfrac{\text{1}}{\text{2}}\]

             $\text{=-sin}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}}$ 

             $\text{=sin}\left( \text{ }\!\!\pi\!\!\text{ -}\dfrac{\text{7 }\!\!\pi\!\!\text{ }}{\text{6}} \right)$ 

             $\text{=sin}\dfrac{\text{7 }\!\!\pi\!\!\text{ }}{\text{6}}$ 

Which implies

$\text{x=n }\!\!\pi\!\!\text{ +}{{\left( \text{-1} \right)}^{\text{n}}}\dfrac{\text{7 }\!\!\pi\!\!\text{ }}{\text{6}}$  , where  $\text{n}\in \text{Z}$

Therefore, the general solution is \[\left( \text{2n+1} \right)\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}\,\] or  \[\text{n }\!\!\pi\!\!\text{ +}{{\left( \text{-1} \right)}^{\text{n}}}\dfrac{\text{7 }\!\!\pi\!\!\text{ }}{\text{6}}\text{,n}\in \text{Z}\].

8. Find the general solution of the equation \[\text{se}{{\text{c}}^{\text{2}}}\text{2x=1-tan 2x}\]

Ans: Here given that , \[\text{se}{{\text{c}}^{\text{2}}}\text{2x=1-tan 2x}\]

Now we know that, $\text{se}{{\text{c}}^{\text{2}}}\text{x-ta}{{\text{n}}^{\text{2}}}\text{x=1}$ 

Therefore we have,

      \[\text{se}{{\text{c}}^{\text{2}}}\text{2x=1-tan 2x}\]  implies

           \[\text{1+ta}{{\text{n}}^{\text{2}}}\text{2x=1-tan 2x}\]

   \[\text{ta}{{\text{n}}^{\text{2}}}\text{2x+tan 2x=0}\]

 \[\text{tan 2x(tan 2x+1)= }\!\!~\!\!\text{ 0}\]

Hence  either $\text{tan 2x=0}$ 

Or, $\text{tan 2x=-1}$ 

Which implies  either  \[\text{x=}\dfrac{\text{n }\!\!\pi\!\!\text{ }}{\text{2}}\]  , where $\text{n}\in \text{Z}$ ,

Or, $\text{tan 2x=-1}$ 

               $\text{=-tan}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}$

               $\text{=tan}\left( \text{ }\!\!\pi\!\!\text{ -}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}} \right)$

                $\text{=tan}\dfrac{\text{3 }\!\!\pi\!\!\text{ }}{\text{4}}$

Which in turn implies that,  

$\text{2x=n }\!\!\pi\!\!\text{ +}\dfrac{\text{3 }\!\!\pi\!\!\text{ }}{\text{4}}\text{,}$ where  $\text{n}\in \text{Z}$

i.e,  $\text{x=}\dfrac{\text{n }\!\!\pi\!\!\text{ }}{\text{2}}\text{+}\dfrac{\text{3 }\!\!\pi\!\!\text{ }}{\text{8}}\text{,}$ where $\text{n}\in \text{Z}$.

Therefore, the general solution is $\dfrac{\text{n }\!\!\pi\!\!\text{ }}{\text{2}}\,\,$ or  $\,\,\dfrac{\text{n }\!\!\pi\!\!\text{ }}{\text{2}}\text{+}\dfrac{\text{3 }\!\!\pi\!\!\text{ }}{\text{8}}\text{,n}\in \text{Z}$.

9. Find the general solution of the equation \[\text{sin x+sin 3x+sin 5x=0}\]

Ans:

Here given that ,\[\text{sin x+sin 3x+sin 5x=0}\]

Now we know that,  $\text{sin A+sin B=2sin}\left( \dfrac{\text{A+B}}{\text{2}} \right)\text{cos}\left( \dfrac{\text{A-B}}{\text{2}} \right)$

Therefore ,

                             \[\text{sin x+sin 3x+sin 5x=0}\]

                          \[\left( \text{sin x+sin 3x} \right)\text{+sin 5x=0}\] 

     $\left[ \text{2sin}\left( \dfrac{\text{x+5x}}{\text{2}} \right)\text{cos}\left( \dfrac{\text{x-5x}}{\text{2}} \right) \right]\text{+sin 3x=0}\,$

                       \[\text{2sin 3x cos (-2x)+sin 3x= }\!\!~\!\!\text{ 0}\]

Simplifying we get,

\[\text{2sin 3xcos 2x+sin 3x=0}\]

     \[\text{sin 3x(2cos 2x+1)= }\!\!~\!\!\text{ 0}\]

Hence either $\text{sin 3x=0}$ 

Or, $\text{cos 2x=-}\dfrac{\text{1}}{\text{2}}$ 

Which implies  $\text{3x=n }\!\!\pi\!\!\text{ }$ , where $\text{n}\in \text{Z}$ 

Or,   $\text{cos 2x=-}\dfrac{\text{1}}{\text{2}}$

                  $\text{=-cos}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}$ 

                  $\text{=cos}\left( \text{ }\!\!\pi\!\!\text{ -}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}} \right)$ 

                     $\text{=cos}\dfrac{\text{2 }\!\!\pi\!\!\text{ }}{\text{3}}$ 

i.e., either $\text{x=}\dfrac{\text{n }\!\!\pi\!\!\text{ }}{\text{3}}$   , where $\text{n}\in \text{Z}$ 

or,  $\text{2x=2n }\!\!\pi\!\!\text{  }\!\!\pm\!\!\text{ }\dfrac{\text{2 }\!\!\pi\!\!\text{ }}{\text{3}}$   ,where $\text{n}\in \text{Z}$ .

Therefore, the general solution is $\dfrac{\text{n }\!\!\pi\!\!\text{ }}{\text{3}}\,$ or $\text{n }\!\!\pi\!\!\text{  }\!\!\pm\!\!\text{ }\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}\text{,n}\in \text{Z}$.

Miscellaneous Exercise

1. Prove that: $\text{2cos}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{13}}\text{cos}\dfrac{\text{9 }\!\!\pi\!\!\text{ }}{\text{13}}\text{+cos}\dfrac{\text{3 }\!\!\pi\!\!\text{ }}{\text{13}}\text{+cos}\dfrac{\text{5 }\!\!\pi\!\!\text{ }}{\text{13}}\text{=0}$

Ans: We know that $\text{cos x+cos y=2cos}\left( \dfrac{\text{x+y}}{\text{2}} \right)\text{cos}\left( \dfrac{\text{x-y}}{\text{2}} \right)$

Now L.H.S.$\text{=2cos}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{13}}\text{cos}\dfrac{\text{9 }\!\!\pi\!\!\text{ }}{\text{13}}\text{+cos}\dfrac{\text{3 }\!\!\pi\!\!\text{ }}{\text{13}}\text{+cos}\dfrac{\text{5 }\!\!\pi\!\!\text{ }}{\text{13}}$

                    $\text{=2cos}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}\text{cos}\dfrac{\text{9 }\!\!\pi\!\!\text{ }}{\text{13}}\text{+2cos}\left( \dfrac{\dfrac{\text{3 }\!\!\pi\!\!\text{ }}{\text{13}}\text{+}\dfrac{\text{5 }\!\!\pi\!\!\text{ }}{\text{13}}}{\text{2}} \right)\text{cos}\left( \dfrac{\dfrac{\text{3 }\!\!\pi\!\!\text{ }}{\text{13}}\text{-}\dfrac{\text{5 }\!\!\pi\!\!\text{ }}{\text{13}}}{\text{2}} \right)$       (using the formula)

                    $\text{=2cos}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{13}}\text{cos}\dfrac{\text{9 }\!\!\pi\!\!\text{ }}{\text{13}}\text{+2cos}\dfrac{\text{4 }\!\!\pi\!\!\text{ }}{\text{13}}\text{cos}\left( \dfrac{\text{- }\!\!\pi\!\!\text{ }}{\text{13}} \right)$

                    $\text{=2cos}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{13}}\text{cos}\dfrac{\text{9 }\!\!\pi\!\!\text{ }}{\text{13}}\text{+2cos}\dfrac{\text{4 }\!\!\pi\!\!\text{ }}{\text{13}}\text{cos}\left( \dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{13}} \right)$

Simplifying, 

$\text{L}\text{.H}\text{.S=2cos}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{13}}\left[ \text{cos}\dfrac{\text{9 }\!\!\pi\!\!\text{ }}{\text{13}}\text{+cos}\dfrac{\text{4 }\!\!\pi\!\!\text{ }}{\text{13}} \right]$

         $\text{=2cos}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{13}}\left[ \text{2cos}\left( \dfrac{\dfrac{\text{9 }\!\!\pi\!\!\text{ }}{\text{13}}\text{+}\dfrac{\text{4 }\!\!\pi\!\!\text{ }}{\text{13}}}{\text{2}} \right)\text{cos}\dfrac{\dfrac{\text{9 }\!\!\pi\!\!\text{ }}{\text{13}}\text{-}\dfrac{\text{4 }\!\!\pi\!\!\text{ }}{\text{13}}}{\text{2}} \right]$

          $\text{=2cos}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{13}}\left[ \text{2cos}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}\text{cos}\dfrac{\text{5 }\!\!\pi\!\!\text{ }}{\text{26}} \right]$

Substituting $\text{cos}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}\text{=0}$ , we get,

$\text{L}\text{.H}\text{.S=2cos}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{13}}\text{ }\!\!\times\!\!\text{ 2 }\!\!\times\!\!\text{ 0 }\!\!\times\!\!\text{ cos}\dfrac{\text{5 }\!\!\pi\!\!\text{ }}{\text{26}}$ 

           $\text{=0}$

           $=\text{R}\text{.H}\text{.S}$

Hence proved.

2. Prove that: $\left( \text{sin 3x+sin x} \right)\text{sin x+}\left( \text{cos 3x-cos x} \right)\text{cos x=0}$

Ans:

We know that, $\text{sin x+sin y=2sin}\left( \dfrac{\text{x+y}}{\text{2}} \right)\text{cos}\left( \dfrac{\text{x-y}}{\text{2}} \right)$ 

And  $\text{cos x-cos y=-2sin}\left( \dfrac{\text{x+y}}{\text{2}} \right)\text{sin}\left( \dfrac{\text{x-y}}{\text{2}} \right)$ 

Now, 

L.H.S.$\text{=}\left( \text{sin 3x+sin x} \right)\text{sin x+}\left( \text{cos 3x-cos x} \right)\text{cos x}$

           $\text{=sin 3x sin x+si}{{\text{n}}^{\text{2}}}\text{x+cos 3x cos x-co}{{\text{s}}^{\text{2}}}\text{x}$    (using the formula)

           $\text{=cos 3x cos x+sin 3x sin x-}\left( \text{co}{{\text{s}}^{\text{2}}}\text{x-si}{{\text{n}}^{\text{2}}}\text{x} \right)$

  Simplifying  we get,       

$\text{L}\text{.H}\text{.S=cos}\left( \text{3x-x} \right)\text{-cos 2x}\,$

         $\text{=cos 2x-cos 2x}$

         $\text{=0}$

         $=\text{R}\text{.H}\text{.S}\text{.}$

3. Prove that: ${{\left( \text{cos x+cos y} \right)}^{\text{2}}}\text{+}{{\left( \text{sin x-sin y} \right)}^{\text{2}}}\text{=4co}{{\text{s}}^{\text{2}}}\dfrac{\text{x+y}}{\text{2}}$

Ans: We know that, $\text{cos}\left( \text{x+y} \right)\text{=cos x cos y-sin xsin y}$

and  L.H.S$\text{=}{{\left( \text{cos x+cos y} \right)}^{\text{2}}}\text{+}{{\left( \text{sin x-sin y} \right)}^{\text{2}}}$

                  \begin{align} & \text{=co}{{\text{s}}^{\text{2}}}\text{x+co}{{\text{s}}^{\text{2}}}\text{y+2cos x cos y+si}{{\text{n}}^{\text{2}}}\text{x+si}{{\text{n}}^{\text{2}}}\text{y-2sin x sin y} \\ &  \\ \end{align}

                 $\text{=}\left( \text{co}{{\text{s}}^{\text{2}}}\text{x+si}{{\text{n}}^{\text{2}}}\text{x} \right)\text{+}\left( \text{co}{{\text{s}}^{\text{2}}}\text{y+si}{{\text{n}}^{\text{2}}}\text{y} \right)\text{+2}\left( \text{cos x cos y-sin x sin y} \right)$

Simplifying and using the formula,

L.H.S$\text{=1+1+2cos}\left( \text{x+y} \right)$ 

         $\text{=2}\left[ \text{1+cos}\left( \text{x+y} \right) \right]$ 

         $\text{=2}\left[ \text{1+2co}{{\text{s}}^{\text{2}}}\dfrac{\left( \text{x+y} \right)}{\text{2}}\text{-1} \right]$ 

 [since $\text{2co}{{\text{s}}^{\text{2}}}\dfrac{\left( \text{x+y} \right)}{\text{2}}\text{-1=cos}\left( \text{x+y} \right)$ ]

          $\text{=4co}{{\text{s}}^{\text{2}}}\left( \text{x+y} \right)$ 

Therefore  L.H.S$=$ R.H.S

Hence proved.

4. Prove that:  ${{\left( \text{cos x-cos y} \right)}^{\text{2}}}\text{+}{{\left( \text{sin x-sin y} \right)}^{\text{2}}}\text{=4si}{{\text{n}}^{\text{2}}}\dfrac{\text{x-y}}{\text{2}}$ 

Ans: We know that, $\text{cos}\left( \text{x-y} \right)\text{=cos x cos y+sin x sin y}$ 

L.H.S.$\text{=}{{\left( \text{cos x-cos y} \right)}^{\text{2}}}\text{+}{{\left( \text{sin x-sin y} \right)}^{\text{2}}}$

           $\text{=co}{{\text{s}}^{\text{2}}}\text{x+co}{{\text{s}}^{\text{2}}}\text{y-2cos x cos y+si}{{\text{n}}^{\text{2}}}\text{x+si}{{\text{n}}^{\text{2}}}\text{y-2sin x sin y}$

           \[\text{=}\left( \text{co}{{\text{s}}^{\text{2}}}\text{x+si}{{\text{n}}^{\text{2}}}\text{x} \right)\text{+}\left( \text{co}{{\text{s}}^{\text{2}}}\text{y+si}{{\text{n}}^{\text{2}}}\text{y} \right)\text{-2}\left[ \text{cos x cos y+sin x sin y} \right]\]

Simplifying and using the formula  we get,

         L.H.S $\text{=1+1-2}\left[ \text{cos}\left( \text{x-y} \right) \right]\,\,$

                   $\text{=2}\left[ \text{1-cos}\left( \text{x-y} \right) \right]$

                   $\text{=2}\left[ \text{1-}\left\{ \text{1-2si}{{\text{n}}^{\text{2}}}\left( \dfrac{\text{x-y}}{\text{2}} \right) \right\} \right]\,$ 

[since  $\text{1-2si}{{\text{n}}^{\text{2}}}\dfrac{\left( \text{x-y} \right)}{\text{2}}\text{=cos}\left( \text{x-y} \right)$ ]

                   $\text{=4si}{{\text{n}}^{\text{2}}}\left( \dfrac{\text{x-y}}{\text{2}} \right)$ 

Therefore  L.H.S$=$ R.H.S

Hence proved.

5. Prove that: $\text{sin x+sin 3x+sin 5x+sin 7x=4cos xcos 2xsin 4x}$

Ans: We  know  that $\text{sin A+sin B=2sin}\left( \dfrac{\text{A+B}}{\text{2}} \right)\text{.cos}\left( \dfrac{\text{A-B}}{\text{2}} \right)$

$\text{L}\text{.H}\text{.S}\text{. =sin x+sin 3x+sin 5x+sin 7x}$

            \[\text{=}\left( \text{sin x+sin 5x} \right)\text{+}\left( \text{sin 3x+sin 7x} \right)\]      

Using the formula and simplifying,

          $\text{=2sin}\left( \dfrac{\text{x+5x}}{\text{2}} \right)\text{.cos}\left( \dfrac{\text{x-5x}}{\text{2}} \right)\text{+2sin}\left( \dfrac{\text{3x+7x}}{\text{2}} \right)\text{cos}\left( \dfrac{\text{3x-7x}}{\text{2}} \right)$       

          \[\text{=2cos 2x}\left[ \text{sin 3x+sin 5x} \right]\]

          \[\text{=2cos 2x}\left[ \text{2sin}\left( \dfrac{\text{3x+5x}}{\text{2}} \right)\text{.cos}\left( \dfrac{\text{3x-5x}}{\text{2}} \right) \right]\] 

          \[\text{=2cos 2x}\left[ \text{2sin 4x}\text{.cos}\left( \text{-x} \right) \right]\] 

Therefore we have,

\[\text{L}\text{.H}\text{.S=4cos 2x sin 4x cos x}\] 

           \[=\text{R}\text{.H}\text{.S}\]

6. Prove that: $\dfrac{\left( \text{sin 7x+sin 5x} \right)\text{+}\left( \text{sin 9x+sin 3x} \right)}{\left( \text{cos 7x+cos 5x} \right)\text{+}\left( \text{cos 9x+cos 3x} \right)}\text{=tan 6x}$

Ans: We  known that,

$\text{sinA+sinB=2sin}\left( \dfrac{\text{A+B}}{\text{2}} \right)\text{.cos}\left( \dfrac{\text{A-B}}{\text{2}} \right)$

And $\text{cos A+cos B=2cos}\left( \dfrac{\text{A+B}}{\text{2}} \right)\text{.cos}\left( \dfrac{\text{A-B}}{\text{2}} \right)$

$\text{L}\text{.H}\text{.S}\text{. =}\dfrac{\left( \text{sin 7x+sin 5x} \right)\text{+}\left( \text{sin9x+sin3x} \right)}{\left( \text{cos 7x+cos 5x} \right)\text{+}\left( \text{cos9x+cos3x} \right)}$

Using the formula and simplifying,

          $\text{=}\dfrac{\left[ \text{2sin}\left( \dfrac{\text{7x+5x}}{\text{2}} \right)\text{.cos}\left( \dfrac{\text{7x-5x}}{\text{2}} \right) \right]\text{+}\left[ \text{2sin}\left( \dfrac{\text{9x+3x}}{\text{2}} \right)\text{.cos}\left( \dfrac{\text{9x-3x}}{\text{2}} \right) \right]}{\left[ \text{2cos}\left( \dfrac{\text{7x+5x}}{\text{2}} \right)\text{.cos}\left( \dfrac{\text{7x-5x}}{\text{2}} \right) \right]\text{+}\left[ \text{2cos}\left( \dfrac{\text{9x+3x}}{\text{2}} \right)\text{.cos}\left( \dfrac{\text{9x-3x}}{\text{2}} \right) \right]}$

          $\text{=}\dfrac{\left[ \text{2sin 6x}\text{.cos x} \right]\text{+}\left[ \text{2sin 6x}\text{.cos 3x} \right]}{\left[ \text{2cos 6x}\text{.cos x} \right]\text{+}\left[ \text{2cos 6x}\text{.cos 6x} \right]}$

          $\text{=}\dfrac{\text{2sin 6x}\left[ \text{cos x+cos 3x} \right]}{\text{2cos 6x}\left[ \text{cos x+cos 3x} \right]}$

           $\text{=tan 6x}$

Therefore L.H.S$=$ R.H.S

Hence  proved.

7. Prove that: 

$\text{sin 3x+sin 2x-sin x=4sin xcos}\dfrac{\text{x}}{\text{2}}\text{cos}\dfrac{\text{3x}}{\text{2}}$

Ans: We know that,

$\text{sin A+sin B=2sin}\left( \dfrac{\text{A+B}}{\text{2}} \right)\text{.cos}\left( \dfrac{\text{A-B}}{\text{2}} \right)$

And  $\text{sin A-sin B=2sin}\left( \dfrac{\text{A-B}}{\text{2}} \right)\text{.cos}\left( \dfrac{\text{A+B}}{\text{2}} \right)$ 

$\text{L}\text{.H}\text{.S}\text{.=sin3x+sin2x-sinx}$

          $\text{=sin 3x+}\left[ \text{2cos}\left( \dfrac{\text{2x+x}}{\text{2}} \right)\text{sin}\left( \dfrac{\text{2x-x}}{\text{2}} \right) \right]\,$

          $\text{=sin 3x+}\left[ \text{2cos}\left( \dfrac{\text{3x}}{\text{2}} \right)\text{sin}\left( \dfrac{\text{x}}{\text{2}} \right) \right]$

Since we know that, $\text{sin 2x=2sin xcos x}$ 

$\text{L}\text{.H}\text{.S=2sin}\dfrac{\text{3x}}{\text{2}}\text{.cos}\dfrac{\text{3x}}{\text{2}}\text{+2cos}\dfrac{\text{3x}}{\text{2}}\text{sin}\dfrac{\text{x}}{\text{2}}\,\,\,\,\,\,$

         $\text{=2cos}\left( \dfrac{\text{3x}}{\text{2}} \right)\left[ \text{sin}\left( \dfrac{\text{3x}}{\text{2}} \right)\text{+sin}\left( \dfrac{\text{x}}{\text{2}} \right) \right]$

         \[\text{=2cos}\left( \dfrac{\text{3x}}{\text{2}} \right)\left[ \text{2sin}\left\{ \dfrac{\left( \dfrac{\text{3x}}{\text{2}} \right)\text{+}\left( \dfrac{\text{x}}{\text{2}} \right)}{\text{2}} \right\}\text{cos}\left\{ \dfrac{\left( \dfrac{\text{3x}}{\text{2}} \right)\text{-}\left( \dfrac{\text{x}}{\text{2}} \right)}{\text{2}} \right\} \right]\]

          $\text{=2cos}\left( \dfrac{\text{3x}}{\text{2}} \right)\text{.2sin xcos}\left( \dfrac{\text{x}}{\text{2}} \right)$

Therefore 

 $\text{L}\text{.H}\text{.S=4sin xcos}\left( \dfrac{\text{x}}{\text{2}} \right)\text{cos}\left( \dfrac{\text{3x}}{\text{2}} \right)$

          $\text{=R}\text{.H}\text{.S}$ 

8.Find $\text{sin}\dfrac{\text{x}}{\text{2}}\text{,cos}\dfrac{\text{x}}{\text{2}}$ and $\text{tan}\dfrac{\text{x}}{\text{2}}$ ,if $\text{tanx=-}\dfrac{\text{4}}{\text{3}}$ , $\text{x}$ lies in 2nd quadrant.   

Ans: Here, $\text{x}$ is in 2nd quadrant.

Therefore ,

$\text{ }\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}\text{x }\!\!\pi\!\!\text{ }$

i.e,  $\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}<\dfrac{\text{x}}{\text{2}}<\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}$

hence $\dfrac{\text{x}}{\text{2}}$ lies in 1st quadrant.

Therefore, \[\text{sin}\dfrac{\text{x}}{\text{2}}\text{,cos}\dfrac{\text{x}}{\text{2}}\,\,\,\] and $\text{tan}\dfrac{\text{x}}{\text{2}}$ are positive.

 Given that $\text{tan x=-}\dfrac{\text{4}}{\text{3}}$

We know that, $\text{se}{{\text{c}}^{\text{2}}}\text{x=1+ta}{{\text{n}}^{\text{2}}}\text{x}$

$\text{se}{{\text{c}}^{\text{2}}}\text{x=1+ta}{{\text{n}}^{\text{2}}}\text{x}$

         $\text{=1+}{{\left( \text{-}\dfrac{\text{4}}{\text{3}} \right)}^{\text{2}}}$ 

         \[\text{=}\dfrac{\text{25}}{\text{9}}\]

As \[\text{x}\] is in 2nd quadrant, $\text{sec x}$ is negative.

Therefore , $\text{secx=-}\dfrac{\text{5}}{\text{3}}$ 

Then $\text{cos x=-}\dfrac{\text{3}}{\text{5}}$ 

Now we know that, $\text{2co}{{\text{s}}^{\text{2}}}\dfrac{\text{x}}{\text{2}}\text{=cos x+1}$ 

Computing we get, $\text{2co}{{\text{s}}^{\text{2}}}\dfrac{\text{x}}{\text{2}}\text{=}\dfrac{\text{2}}{\text{5}}$ 

Hence \[\text{cos}\dfrac{\text{x}}{\text{2}}\text{=}\dfrac{\text{1}}{\sqrt{\text{5}}}\] 

Now we know that, $\text{si}{{\text{n}}^{\text{2}}}\text{x+co}{{\text{s}}^{\text{2}}}\text{x=1}$ 

Therefore substituting $\text{cos}\dfrac{\text{x}}{\text{2}}\text{=}\dfrac{\text{1}}{\sqrt{\text{5}}}$ and computing ,

$\text{sin}\dfrac{\text{x}}{\text{2}}\text{=}\dfrac{\text{2}}{\sqrt{\text{5}}}$ 

Hence ,

$\text{tan}\dfrac{\text{x}}{\text{2}}\text{=}\dfrac{\text{sin}\dfrac{\text{x}}{\text{2}}}{\text{cos}\dfrac{\text{x}}{\text{2}}}$ 

          $=2$ 

Thus, the respective values of$\text{sin}\dfrac{\text{x}}{\text{2}}\text{,cos}\dfrac{\text{x}}{\text{2}}\,\text{,tan}\dfrac{\text{x}}{\text{2}}\,$

are $\,\dfrac{2\sqrt{5}}{5},\dfrac{\sqrt{5}}{5},\,\,2$ .

9.Find $\text{sin}\dfrac{\text{x}}{\text{2}}\text{,cos}\dfrac{\text{x}}{\text{2}}$ and $\text{tan}\dfrac{\text{x}}{\text{2}}$ ,if $\cos x\text{=-}\dfrac{1}{\text{3}}$ , $\text{x}$ lies in 3rd quadrant.   

Ans: Here, $\text{x}$ is in 3rd quadrant.

Therefore ,

$\text{  }\!\!\pi\!\!\text{ x}\dfrac{\text{3 }\!\!\pi\!\!\text{ }}{\text{2}}$

i.e,  $\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}\text{}\dfrac{\text{x}}{\text{2}}\text{}\dfrac{\text{3 }\!\!\pi\!\!\text{ }}{\text{4}}$

hence $\dfrac{\text{x}}{\text{2}}$ lies in 2nd quadrant.

Therefore, $\text{cos}\dfrac{\text{x}}{\text{2}}\,\,\,$ and $\text{tan}\dfrac{\text{x}}{\text{2}}$ are negative and $\text{sin}\dfrac{\text{x}}{\text{2}}$ is positive.

 Given that $\text{cos x=-}\dfrac{1}{\text{3}}$

Now we know that, $\text{2co}{{\text{s}}^{\text{2}}}\dfrac{\text{x}}{\text{2}}\text{=cosx+1}$ 

Computing we get, $\text{2co}{{\text{s}}^{\text{2}}}\dfrac{\text{x}}{\text{2}}\text{=}\dfrac{\text{2}}{\text{3}}$ 

Hence $\text{cos}\dfrac{\text{x}}{\text{2}}\text{=-}\dfrac{\text{1}}{\sqrt{\text{3}}}$ 

Now we know that, $\text{si}{{\text{n}}^{\text{2}}}\text{x+co}{{\text{s}}^{\text{2}}}\text{x=1}$ 

Therefore substituting $\text{cos}\dfrac{\text{x}}{\text{2}}\text{=-}\dfrac{\text{1}}{\sqrt{\text{3}}}$ and computing ,

$\text{sin}\dfrac{\text{x}}{\text{2}}\text{=}\sqrt{\dfrac{\text{2}}{\text{3}}}$ 

Hence ,

$\text{tan}\dfrac{\text{x}}{\text{2}}\text{=}\dfrac{\text{sin}\dfrac{\text{x}}{\text{2}}}{\text{cos}\dfrac{\text{x}}{\text{2}}}$ 

          $\text{=-}\sqrt{\text{2}}$ 

Thus, the respective values of $\text{sin}\dfrac{\text{x}}{\text{2}}\text{,cos}\dfrac{\text{x}}{\text{2}}\,\text{,tan}\dfrac{\text{x}}{\text{2}}\,$

are  $\,\sqrt{\dfrac{\text{2}}{\text{3}}}\text{,-}\dfrac{\text{1}}{\sqrt{\text{3}}}\text{,}\,\text{-}\,\sqrt{\text{2}}$.

10.Find$\text{sin}\dfrac{\text{x}}{\text{2}}\text{,cos}\dfrac{\text{x}}{\text{2}}$ and $\text{tan}\dfrac{\text{x}}{\text{2}}$ ,if $\text{sin x=}\dfrac{1}{4}$ , $\text{x}$ lies in 2nd quadrant.   

Ans: Here, $\text{x}$ lies in 2nd quadrant.

Therefore ,

$\text{ }\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}\text{x }\!\!\pi\!\!\text{ }$

i.e,  $\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}<\dfrac{\text{x}}{\text{2}}<\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}$

hence  $\dfrac{\text{x}}{\text{2}}$ lies in 1st quadrant.

Therefore, $\text{sin}\dfrac{\text{x}}{\text{2}}\text{,cos}\dfrac{\text{x}}{\text{2}}\,\,\,$ and $\text{tan}\dfrac{\text{x}}{\text{2}}$ are positive.

 Given that $\text{sin x=}\dfrac{\text{1}}{\text{4}}$

Now we know that, $\text{si}{{\text{n}}^{\text{2}}}\text{x+co}{{\text{s}}^{\text{2}}}\text{x=1}$ 

Therefore substituting $\text{sin x=}\dfrac{\text{1}}{\text{4}}$ and computing ,

$\text{cos x=-}\dfrac{\sqrt{\text{15}}}{\text{4}}$ 

since $\text{x}$ lies in 2nd quadrant,  $\text{cos x}$ is negative.

Now we know that, $\text{2si}{{\text{n}}^{\text{2}}}\dfrac{\text{x}}{\text{2}}\text{=1-cos x}$ 

Computing we get, $\text{2si}{{\text{n}}^{\text{2}}}\dfrac{\text{x}}{\text{2}}\text{=1+}\dfrac{\sqrt{\text{15}}}{\text{4}}$ 

Hence $\text{sin}\dfrac{\text{x}}{\text{2}}\text{=}\sqrt{\dfrac{\text{4+}\sqrt{\text{15}}}{\text{8}}}$ 

Now we know that, $\text{si}{{\text{n}}^{\text{2}}}\text{x+co}{{\text{s}}^{\text{2}}}\text{x=1}$ 

Therefore substituting $\text{sin}\dfrac{\text{x}}{\text{2}}\text{=}\sqrt{\dfrac{\text{4+}\sqrt{\text{15}}}{\text{8}}}$ and computing ,

$\text{cos}\dfrac{\text{x}}{\text{2}}\text{=}\sqrt{\dfrac{\text{4-}\sqrt{\text{15}}}{\text{8}}}$ 

Hence ,

$\text{tan}\dfrac{\text{x}}{\text{2}}\text{=}\dfrac{\text{sin}\dfrac{\text{x}}{\text{2}}}{\text{cos}\dfrac{\text{x}}{\text{2}}}$ 

          $\text{=}\dfrac{\sqrt{\text{4+}\sqrt{\text{15}}}}{\sqrt{\text{4-}\sqrt{\text{15}}}}$ 

       \[\text{=4+}\sqrt{\text{15}}\] 

Thus, the respective values of $\text{sin}\dfrac{\text{x}}{\text{2}}\text{,cos}\dfrac{\text{x}}{\text{2}}\,\text{,tan}\dfrac{\text{x}}{\text{2}}\,$

are  $\sqrt{\dfrac{\text{4+}\sqrt{\text{15}}}{\text{8}}}\text{,}\sqrt{\dfrac{\text{4-}\sqrt{\text{15}}}{\text{8}}}\text{,4+}\sqrt{\text{15}}$ .

NCERT Solutions for CBSE Class 11 Maths Chapter 3 Exercise 3.2 Trigonometric Functions

NCERT Solutions for Class 11 Maths Chapter 3 – Free PDF Download

The Mathematical jargon in Trigonometry Class 11 is undoubtedly a complex affair for most students to grasp. But at the same time, it is an indispensable topic in 11th standard syllabus. Stronghold on this topic constitutes a very basic understanding of geometric calculations and is crucial for solving multiple numerical problems in other subjects besides Mathematics.

NCERT Solutions for Class 11 Maths Chapter 3 are formulated by experienced and highly learned tutors focusing specifically on the problematic areas students tend to encounter in this chapter. Numerical sums based on the application of trigonometric formulas, graphs of trigonometric functions, conversion of angles from radian to degree and vice versa, and general solutions have been explained in great detail and elaborate steps.

Additionally, these study guides come with topic-wise exercises and probable questions as well, because increased practice will only help students improve their problem-solving speed and accuracy. In order to avail these suggestions offline, you can click on NCERT Solutions for Class 11 Maths Chapter 3 PDF Download option on Vedantu’s website.

NCERT Solutions for Class 11 Maths Chapter 3 Subtopics

Before wandering into the conceptual details of Trigonometric Functions Class 11, you can have a look at the summary of all topics discussed in this chapter.

Section 1: Introduction

In this part, students are introduced to fundamental trigonometric functions and their application. You will be required to perform calculations of distances based on these ratios.

Section 2: Angles

Class 11 Maths Trigonometry comes with four subsections under this topic. Students will learn about measurement of angles in different units, radians and degrees, and relations between them. Solutions to numerical sums involving the conversion of angle measurements from one form to another have been discussed in detail in NCERT Solutions for Class 11 Maths Chapter 3.

Section 3: Trigonometric Functions

This section deals with two topics. First one teaches the signs and symbols of generalised trigonometric functions in all four quadrants of a graph. In the next subsection, students will be acquainted with the domain and range of the different functions, that is, how the value of a function increases or decreases for an increasing angle value. You can find tables and detailed explanations on the working of these concepts in Trigonometric Functions Class 11 NCERT Solutions.

Section 4: Trigonometric Functions of Sum and Difference of Two Angles

The concluding section of Chapter 3 Maths Class 11 PDF focuses on the derivation of formulas and expressions based on trigonometric functions of the sums and differences between two angles. These formulas have been explained with the help of examples. Students must have a clear understanding of these as they make up an important part of geometric evaluations and are applied in a wide range of calculations.

Important Points Covered in Class 11 Maths Chapter 3 - Trigonometric Functions

• The term “Angle” refers to a measurement of rotation of a ray about its fixed end point. The initial position of the ray before starting the rotation is called the initial side of the angle and the final position of the ray after rotation is called the terminal side of the angle.

• In the Sexagesimal system, the angles are measured in degrees. 

1° = 60′ reads as 1 degree is equal to 60 minutes. 

1′ = 60″ reads as 1 minute is equal to 60 seconds.

• In the Circular system, the angles are measured in radians.

1 Radian = 180/π degree and 

1 Degree = π/180 radians

• Functions of negative angles (A)

• sin (-A) = - sinA,

• cosec (-A) = - cosec A,

• tan (-A) = - tan A,

• cos (-A) = cos A,

• cot (-A) = - cot A,

• sec (-A) = sec A.

Note: The trigonometric functions cosine and secant are even functions. The remaining all trigonometric functions are odd functions.

• Trigonometric formulas of compound angles:

• sin (A + B) = sin A cos B + cos A sin B

• sin (A – B) = sin A cos B – cos A sin B

• cos (A + B) = cos A cos B – sin A sin B

• cos (A – B) = cos A cos B + sin A sin B

• 2 sin A cos B = sin (A + B) + sin (A – B)

• 2 cos A sin B = sin (A + B) – sin (A – B)

• 2 cos A cos B = cos (A + B) + cos (A – B)

• 2 sin A sin B = cos (A – B) – cos (A + B)

• Trigonometric equations are the equations which include trigonometric functions of unknown angles.

• Solutions of a trigonometric equation are all the possible values of the unknown angle that satisfies the given equation. These solutions (possible values) can be an infinite number in some cases.

• The general solution of a trigonometric equation is defined as the solution that involves an integer ‘n’ which gives all solutions of the given trigonometric equation.

• The principal solution of a trigonometric equation is defined as the solutions which lie in the interval 0 ≤ x ≤ 2π.

We Cover All The Exercises In The Chapter Given Below

Key Features of NCERT Solutions for Class 11 Maths Chapter 3 - Trigonometric Functions

Key features of NCERT Solutions for Class 11 Maths Chapter 3 are as follows:

• Pythagorean Identities will be used to solve many trigonometric problems in which one trigonometric ratio is given and the other ratios must be calculated.

• We will learn the relationship between the ratios of the sides of a right-angled triangle and their angles.

• We will learn about trigonometric ratios such as sine, cosine, tangent, cotangent, secant, and cosecant. 

• The solutions deal with angle measurement and angle problems. 

• We use fundamental ratios to construct other key trigonometric functions: cotangent, secant, and cosecant. These functions serve as the foundation for all of the essential trigonometry topics.