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# NCERT Solutions for Class 11 Maths Miscellaneous Exercise Chapter 3 - Trigonometric Functions

Last updated date: 02nd Aug 2024
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## NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Miscellaneous Exercise

NCERT Solutions for Class 11 Maths Chapter 3, Trigonometric Functions, provides answers to all Miscellaneous Exercise problems. These solutions are important for understanding the main ideas in Chapter 3, which are important for CBSE Board exams and competitive tests. Practising NCERT Solutions for Maths Class 11 problems will help you understand trigonometric functions better.

Table of Content
1. NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Miscellaneous Exercise
2. Access NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions
2.1Miscellaneous Exercise
3. Class 11 Maths Chapter 3: Exercises Breakdown
4. CBSE Class 11 Maths Chapter 3 Other Study Materials
5. Chapter-Specific NCERT Solutions for Class 11 Maths
FAQs

Focus on learning the different trigonometric identities and formulas in this chapter. Knowing these concepts is key to solving problems effectively. To do well in your exams, download the CBSE Class 11 Maths Syllabus and practice them offline regularly. This will help you prepare thoroughly and improve your confidence for the exams.

Competitive Exams after 12th Science

## Access NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions

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

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

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

## Conclusion

NCERT Class 11th Maths Chapter 3 Miscellaneous Exercise is crucial for understanding various concepts thoroughly. It covers diverse problems that require the application of multiple formulas and techniques. It's important to focus on understanding the underlying principles behind each question rather than just memorizing solutions. Remember to understand the theory behind each concept, practice regularly, and refer to solved examples to master this exercise effectively.

## Class 11 Maths Chapter 3: Exercises Breakdown

 Exercise Number of Questions Exercise 3.1 7 Questions & Solutions Exercise 3.2 10 Questions & Solutions Exercise 3.3 25 Questions & Solutions

## CBSE Class 11 Maths Chapter 3 Other Study Materials

 S. No Important Links for Chapter 3 Trigonometric Functions 1 Class 11 Trigonometric Functions Important Questions 2 Class 11 Trigonometric Functions Revision Notes 3 Class 11 Trigonometric Functions Important Formulas 4 Class 11 Trigonometric Functions NCERT Exemplar Solution 5 Class 11 Trigonometric Functions RD Sharma Solutions 6 Class 11 Trigonometric Functions RS Aggarwal Solutions

## Chapter-Specific NCERT Solutions for Class 11 Maths

Given below are the chapter-wise NCERT Solutions for Class 11 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

## FAQs on NCERT Solutions for Class 11 Maths Miscellaneous Exercise Chapter 3 - Trigonometric Functions

1. What is the focus of the NCERT Solutions of Miscellaneous Exercise Class 11 Chapter 3?

The Miscellaneous Exercise Class 11 Chapter 3 covers a variety of problems that integrate all concepts from the chapter. It focuses on complex applications of trigonometric functions, which require a solid understanding of the basics. These problems are designed to test and improve your problem-solving skills and conceptual clarity. Practising these questions helps in mastering the chapter thoroughly.

2. How important is NCERT Solutions of Miscellaneous Exercise Class 11 Chapter 3?

NCERT Solutions of Class 11 Maths Miscellaneous Exercise Class 11 Chapter 3 is very important for exams as it includes higher-order thinking questions. These questions test a deep understanding of trigonometric functions, which are crucial for scoring well. By solving these, students can prepare for a wide range of questions that might appear in exams. These problems often mirror the complexity and style of exam questions.

3. Which key concepts should I focus on in NCERT Class 11 Maths Ch 3 Miscellaneous Exercise Solutions?

In NCERT Class 11 Maths Ch 3 Miscellaneous Exercise Solutions, focus on trigonometric identities, transformations, and solving trigonometric equations. These concepts are the foundation for solving complex problems in trigonometry. Mastery of these areas is essential, as they frequently appear in both miscellaneous exercises and exams. Understanding these topics will help in tackling various types of trigonometric problems efficiently.

4. How many questions are there in the NCERT Class 11 Maths Ch 3 Miscellaneous Exercise Solutions?

There are 10 questions in NCERT Class 11 Maths Ch 3 Miscellaneous Exercise Solutions. These questions cover a wide range of applications of trigonometric functions. The variety of problems helps in thoroughly testing and reinforcing the concepts learned in the chapter. Practising all these questions ensures comprehensive preparation.

5. What type of questions are included in NCERT Class 11 Maths Ch 3 Miscellaneous Exercise Solutions?

NCERT Class 11 Maths Ch 3 Miscellaneous Exercise Solutions includes questions such as proving trigonometric identities, solving trigonometric equations, and applying trigonometric functions to real-world problems. These problems require a good grasp of the basic and advanced concepts of trigonometry. They help in enhancing problem-solving skills and application-based understanding.

6. What should I focus on while practising in NCERT Miscellaneous Exercise Chapter 3 Class 11?

While practising on NCERT Miscellaneous Exercise Chapter 3 Class 11, focus on accuracy and speed. Make sure you understand each step of the solutions to build a strong conceptual foundation. Regular practice and review of mistakes will help in improving precision and speed. This approach is crucial for performing well in exams, where time management and accuracy are key.

7. Does Class 11 Trigonometry Miscellaneous Exercise come on board exams?

Yes, questions from the Class 11 Trigonometry Miscellaneous Exercise in the NCERT textbook often appear in board exams. These exercises cover a range of important concepts and problem types that are crucial for thorough exam preparation. Practising Miscellaneous Exercises helps students understand and apply the concepts more effectively, increasing their chances of performing well in the exams.

8. Does the Class 11th Maths Chapter 3 Miscellaneous Exercise have important questions?

The Class 11th Maths Chapter 3 Miscellaneous Exercise contains important questions that are beneficial for understanding and practising the concepts covered in the chapter. Chapter 3 typically deals with Trigonometric Functions, which are fundamental for higher-level maths and various competitive exams. The miscellaneous exercise usually includes a mix of problems that cover different aspects of the chapter.

9. What is the Class 11 Chapter 3 Miscellaneous Exercise?

The Class 11 Chapter 3 Miscellaneous Exercise is a set of diverse and challenging problems found at the end of each chapter in the NCERT textbook. It includes questions that cover all the key concepts of the chapter, providing comprehensive practice for students.

10. How do Class 11 Maths Chapter 3 Miscellaneous Exercise Solutions help with the Miscellaneous Exercise?

NCERT Solutions provides detailed answers for each question. They show you the steps to solve problems, making it easier to understand and learn the methods.