Trigonometric Functions Class 11 Notes CBSE Maths Chapter 3 [Free PDF Download]

Section–A (1 Mark Questions)

1. If $tan\Theta =\frac{-4}{3}$ , then find $sin\Theta$.

Ans. Since, $\tan \theta=-\frac{4}{3}$ is negative, $\theta$ lies either in second quadrant or in fourth quadrant.

$$ \begin{aligned} & \because 1+\cot ^2 \theta=\operatorname{cosec}^2 \theta \\ & \Rightarrow 1+\frac{1}{\left(-\frac{4}{3}\right)^2}=\frac{1}{\sin ^2 \theta} \\ & \Rightarrow \sin ^2 \theta=\left(\frac{4}{5}\right)^2 \end{aligned} $$

Thus $\sin \theta=\frac{4}{5}$ if $\theta$ lies in the second quadrant or $\sin \theta=-\frac{4}{5}$, if $\theta$ lies in the fourth quadrant.

2. Find the greatest value of sin x cos x.

Ans. $\sin x \cos x=\frac{1}{2}(2 \sin x \cos x)=\frac{1}{2} \sin 2 x$ Greatest value of $\sin 2 x=1$

$\therefore$ Greatest value of $\dfrac{1}{2} \sin 2 x=\dfrac{1}{2} \times 1=\dfrac{1}{2}$

3. Find the degree measure of angle $\left ( \frac{\pi }{8} \right )^{c}$  radian.

Ans. We know that, $\pi$ radians $=180^{\circ}$

$$ \begin{aligned} & \Rightarrow \frac{\pi}{8} \text { radians }=22.5^{\circ} \\ & =22^{\circ}+0.5^{\circ}=22^{\circ}+30^{\prime}=22^{\circ} 30^{\prime} \end{aligned} $$

4. Evaluate: $sin \left ( \frac{-15\pi }{4} \right )$ .

Ans.

$$ \begin{aligned} & \text { 4. } \sin \left(\frac{-15 \pi}{4}\right)=\sin \left(\frac{-16 \pi+\pi}{4}\right) \\ & =\sin \left(-4 \pi+\frac{\pi}{4}\right) \\ & =\sin \left(\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}} \end{aligned} $$

5. The minute hand of a watch is 1.5 cm long. How far does it tip move in 40 minutes? (Use $\pi$ = 3.14).

Ans. Angle rotated by minute hand in 60 minutes = $2\pi$    radians 

Therefore, angle rotated by minute hand in 40 minutes = $\frac{40}{60}\times2\pi =\frac{4\pi }{3}$  radians.

Hence, the required distance travelled is given by

$l=r\Theta =1.5\times\frac{4\pi }{3}cm=2\pi\;cm$

$=2\times3.14\;cm=6.28\;cm$

Section–B (2 Marks Questions)

6. Prove that cot2x cotx - cot3x cot2x - cot3x cotx = 1

Ans. We have, cot 3x=cot (2x+x)

$\Rightarrow cot\;3x=\dfrac{cot\;2x\;cot\;x -1}{cot\;2x+cot\;x}$

$\Rightarrow$ Cot 3x cot 2x + cot 3x cot x = cot 2x cot x-1

$\Rightarrow$ cot 2x cot x - cot 3x cot 2x - cot 3x xot x=1

7. Evaluate: $\frac{1-\tan^2{15^\circ}}{1+\tan^2{15^\circ}}$ 

Ans. Given that: $\frac{1-\tan^2{15^\circ}}{1+\tan^2{15^\circ}}$

Let $\Theta =15^{\circ}$

$\therefore 2\Theta =30^{\circ}$

We know that, $\therefore cos\;2\Theta =\frac{1-\tan^2{15^\circ}}{1+\tan^2{15^\circ}}$

$\Rightarrow  cos\;30 =\frac{1-\tan^2{15^\circ}}{1+\tan^2{15^\circ}}$

$\frac{1-\tan^2{15^\circ}}{1+\tan^2{15^\circ}}=\frac{\sqrt{3}}{3}$

8. If for real value of x, $cos\Theta =x+\frac{1}{x}$ then what can you say about $\Theta$   ?

Ans. Given that: $cos\Theta =x+\frac{1}{x}$ 

$\Rightarrow cos\Theta =x+\frac{1}{x}$

$$ \begin{aligned} & \Rightarrow x^2+1=x \cos \theta \\ & \Rightarrow x^2-x \cos \theta+1=0 \end{aligned} $$

For real value of $x, b^2-4 a c \geq 0$

$$ \begin{aligned} & \Rightarrow(-\cos \theta)^2-4 \times 1 \times 1 \geq 0 \\ & \Rightarrow \cos ^2 \theta-4 \geq 0 \\ & \Rightarrow \cos ^2 \theta \geq 4 \\ & \Rightarrow \cos \theta \geq \pm 2 \quad[\because-1 \leq \cos \theta \leq 1] \end{aligned} $$

So, the value of $\theta$ is not possible.

9. Find the value of $sin\left ( \frac{\pi }{4}+\Theta  \right )-cos \left ( \frac{\pi }{4}-\Theta  \right )$ .

Ans. Given expression :

$$ \begin{aligned} & \sin \left(\frac{\pi}{4}+\theta\right)-\cos \left(\frac{\pi}{4}-\theta\right) \\ & \sin \left(\frac{\pi}{4}+\theta\right)=\sin \left(\frac{\pi}{4}\right) \cos \theta+\cos \left(\frac{\pi}{4}\right) \sin \theta \\ & =\frac{1}{\sqrt{2}} \cos \theta+\frac{1}{\sqrt{2}} \sin \theta \\ & \cos \left(\frac{\pi}{4}-\theta\right)=\cos \left(\frac{\pi}{4}\right) \cos \theta+\sin \left(\frac{\pi}{4}\right) \sin \theta \\ & =\frac{1}{\sqrt{2}} \cos \theta+\frac{1}{\sqrt{2}} \sin \theta \\ & \therefore \sin \left(\frac{\pi}{4}+\theta\right)-\cos \left(\frac{\pi}{4}-\theta\right) \\ & =\frac{1}{\sqrt{2}} \cos \theta+\frac{1}{\sqrt{2}} \sin \theta-\frac{1}{\sqrt{2}} \cos \theta-\frac{1}{\sqrt{2}} \sin \theta \\ & =0 \end{aligned} $$

10. Find the value of $tan\;\frac{\pi }{12}$ .

Ans. $\tan \frac{\pi}{12}=\tan \left(\frac{\pi}{4}-\frac{\pi}{6}\right)$

$$ \begin{aligned} & =\frac{\tan \frac{\pi}{4}-\tan \frac{\pi}{6}}{1+\tan \frac{\pi}{4} \cdot \tan \frac{\pi}{6}}=\frac{1-\frac{1}{\sqrt{3}}}{1+1 \cdot \frac{1}{\sqrt{3}}}=\frac{\sqrt{3}-1}{\sqrt{3}+1} \\ & =\frac{(\sqrt{3}-1)(\sqrt{3}-1)}{(\sqrt{3}+1)(\sqrt{3}-1)}=\frac{4-2 \sqrt{3}}{2}=2-\sqrt{3} \end{aligned} $$

11. Prove that : $tan\;225^{\circ}\;cot\;405^{\circ}+tan\;765^{\circ}\;cot\;675^{\circ}=0$ 

Ans.

$$ \begin{aligned} & \text { 11. LHS }= \\ & \tan 225^{\circ} \cot 405^{\circ}+\tan 765^{\circ} \cot 675^{\circ} \\ & =\tan \left(180^{\circ}+45^{\circ}\right) \cot \left(360^{\circ}+45^{\circ}\right) \\ & \quad \quad \quad \quad \tan \left(360^{\circ} \times 2+45^{\circ}\right) \cot \left(360^{\circ} \times 2-45^{\circ}\right) \\ & =\tan 45^{\circ} \cot 45^{\circ}+\tan 45^{\circ}\left[-\cot 45^{\circ}\right] \\ & =1 \times 1-1 \times 1 \\ & =0=\text { RHS } \end{aligned} $$

Hence proved.

12. For $0<x<\frac{\pi }{2}$ , show that $\sqrt{\frac{(1-cos2x)}{1+cos2x}}=tan\;x$ 

Ans.

$$ \begin{aligned} & \text { LHS }=\sqrt{\frac{(1-\cos 2 x)}{(1+\cos 2 x)}} \\ & =\sqrt{\frac{1-\left(1-2 \sin ^2 x\right)}{1+\left(2 \cos ^2 x-1\right)}}=\sqrt{\frac{2 \sin ^2 x}{2 \cos ^2 x}} \\ & =|\tan x|=\tan x \quad\left\{\because 0<x<\frac{\pi}{2}\right\} \\ & =\text { RHS } \end{aligned} $$

13. Prove that: $\frac{cos(2\pi +x)cosec(2\pi +x)tan\left ( \frac{\pi }{2}+x \right )}{sec\left ( \frac{\pi }{2}+x \right )cos\;x\;cot(\pi +x)}=1$ 

Ans. LHS $=$

$$\cos (2 \pi+x) \operatorname{cosec}(2 \pi+x) \tan \left(\frac{\pi}{2}+x\right)$$

$$ \begin{aligned} & \sec \left(\frac{\pi}{2}+x\right) \cos x \cot (\pi+x) \\ \Rightarrow & \frac{-\cos x \operatorname{cosec} x \cot x}{-\operatorname{cosec} x \cos x \cot x}=1 \end{aligned} $$

Hence proved

PDF Summary - Class 11 Maths Trigonometric FunctionsNotes (Chapter 3)

1. The Meaning of Trigonometry

${{\text{Tri }}}{{\text{ Gon }}}{{\text{ Metron }}}$

$\downarrow\quad\;\;\;\;\downarrow\quad\;\;\;\;\downarrow$

$3\quad{{\text{ sides }}}{{\text{ Measure }}}$

As a result, this area of mathematics was established in the ancient past to measure a triangle's three sides, three angles, and six components. Time-trigonometric functions are utilised in a variety of ways nowadays. The sine and cosine of an angle in a right-angled triangle are the two fundamental functions, and there are four more derivative functions.

2. Basic Trigonometric Identities

(a) ${\sin ^2}\theta  + {\cos ^2}\theta  = 1: - 1 \leqslant \sin \theta  \leqslant 1; - 1 \leqslant \cos \theta  \leqslant 1\forall \theta  \in {\text{R}}$

(b) ${\sec ^2}\theta  - {\tan ^2}\theta  = 1:|\sec \theta | \geqslant 1\forall \theta  \in {\text{R}}$

(c) ${\operatorname{cosec} ^2}\theta  - {\cot ^2}\theta  = 1:|\operatorname{cosec} \theta | \geqslant 1\forall \theta  \in {\text{R}}$

Trigonometric Ratios of Standard Angles:

The relation between these trigonometric identities with the sides of the triangles can be given as follows:

• Sine θ $=$ Opposite/Hypotenuse

• Cos θ  $=$ Adjacent/Hypotenuse

• Tan θ  $=$ Opposite/Adjacent

• Cot θ $=$ Adjacent/Opposite

• Cosec θ  = Hypotenuse/Opposite

• Sec θ  = Hypotenuse/Adjacent

The following are the signs of trigonometric ratios in different quadrants:

3. Trigonometric Ratios of Allied Angles

We might calculate the trigonometric ratios of angles of any value using the trigonometric ratio of allied angles.

1. Sin(–θ)=–Sinθ

2. Cos(–θ)=Cosθ

3. Tan(–θ)=–Tanθ

4. Sin(90o–θ)=Cosθ

5. Cos(90o–θ)=Sinθ

6. Tan(90o–θ)=Cotθ

7. Sin(180o–θ)=Sinθ

8. Cos(180o–θ)=–Cosθ

9. Tan(180o–θ)=–Tanθ

10. Sin(270o–θ)=–Cosθ

11. Cos(270o–θ)=–Sinθ

12. Tan(270o–θ)=Cotθ

13. Sin(90o+θ)=Cosθ

14. Cos(90o+θ)=–Sinθ

15. Tan(90o+θ)=–Cotθ

16. Sin(180o+θ)=–Sinθ

17. Cos(180o+θ)=–Cosθ

18. Tan(180o+θ)=Tanθ

19. Sin(270o+θ)=–Cosθ

20. Cos(270o+θ)=Sinθ

21. Tan(270o+θ)=–Cotθ

4. Trigonometric Functions of Sum or Difference of Two Angles

(a) $\sin ({\text{A}} + {\text{B}}) = \sin {\text{A}}\cos {\text{B}} + \cos {\text{A}}\sin {\text{B}}$

(b) $\sin ({\text{A}} - {\text{B}}) = \sin {\text{A}}\cos {\text{B}} - \cos {\text{A}}\sin {\text{B}}$

(c) $\cos (A + B) = \cos A\cos B - \sin A\sin B$

(d) $\cos ({\text{A}} - {\text{B}}) = \cos {\text{A}}\cos {\text{B}} + \sin {\text{A}}\sin {\text{B}}$

(e) $\tan (A + B) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}$

(f) $\tan ({\text{A}} - {\text{B}}) = \dfrac{{\tan {\text{A}} - \tan {\text{B}}}}{{1 + \tan {\text{A}}\tan {\text{B}}}}$

(g) $\cot (A + B) = \dfrac{{\cot A\cot B - 1}}{{\cot B + \cot A}}$

(f) $\cot ({\text{A}} - {\text{B}}) = \dfrac{{\cot {\text{A}}\cot {\text{B}} + 1}}{{\cot {\text{B}} - \cot {\text{A}}}}$

(h) ${\sin ^2}\;{\text{A}} - {\sin ^2}\;{\text{B}} = {\cos ^2}\;{\text{B}} - {\cos ^2}\;{\text{A}} = \sin ({\text{A}} + {\text{B}}) \cdot \sin ({\text{A}} - {\text{B}})$

(i) ${\cos ^2}\;{\text{A}} - {\sin ^2}\;{\text{B}} = {\cos ^2}\;{\text{B}} - {\sin ^2}\;{\text{A}} = \cos ({\text{A}} + {\text{B}}) \cdot \cos ({\text{A}} - {\text{B}})$

(j) $\tan (A + B + C) = \dfrac{{\tan A + \tan B + \tan C - \tan A\tan B\tan C}}{{1 - \tan A\tan B - \tan B\tan C - \tan C\tan A}}$

5. Multiple Angles and Half Angles

(a) $\sin 2\;{\text{A}} = 2\sin {\text{A}}\cos {\text{A}};\quad \sin \theta  = 2\sin \dfrac{\theta }{2}\cos \dfrac{\theta }{2}$

(b) $\cos 2\;{\text{A}} = {\cos ^2}\;{\text{A}} - {\sin ^2}\;{\text{A}} = 2{\cos ^2}\;{\text{A}} - 1 = 1 - 2{\sin ^2}\;{\text{A}}$ 

$2{\cos ^2}\dfrac{\theta }{2} = 1 + \cos \theta ,2{\sin ^2}\dfrac{\theta }{2} = 1 - \cos \theta $

(c) $\tan 2\;{\text{A}} = \dfrac{{2\tan {\text{A}}}}{{1 - {{\tan }^2}\;{\text{A}}}};\tan \theta  = \dfrac{{2\tan \dfrac{\theta }{2}}}{{1 - {{\tan }^2}\dfrac{\theta }{2}}}$

(d) $\sin 2\;{\text{A}} = \dfrac{{2\tan {\text{A}}}}{{1 - {{\tan }^2}\;{\text{A}}}};\cos 2\;{\text{A}} = \dfrac{{1 - {{\tan }^2}\;{\text{A}}}}{{1 + {{\tan }^2}\;{\text{A}}}}$

(e) $\sin 3\;{\text{A}} = 3\sin {\text{A}} - 4{\sin ^3}\;{\text{A}}$

(f) $\cos 3\;{\text{A}} = 4{\cos ^3}\;{\text{A}} - 3\cos {\text{A}}$

(g) $\tan 3\;{\text{A}} = \dfrac{{3\tan {\text{A}} - {{\tan }^3}\;{\text{A}}}}{{1 - 3{{\tan }^2}\;{\text{A}}}}$

6. Transformation of Products into Sum or Difference of Sines & Cosines

(a) $2\sin {\text{A}}\cos {\text{B}} = \sin ({\text{A}} + {\text{B}}) + \sin ({\text{A}} - {\text{B}})$

(b) $2\cos {\text{A}}\sin {\text{B}} = \sin ({\text{A}} + {\text{B}}) - \sin ({\text{A}} - {\text{B}})$

(c) $2\cos {\text{A}}\cos {\text{B}} = \cos ({\text{A}} + {\text{B}}) + \cos ({\text{A}} - {\text{B}})$

(d) $2\sin {\text{A}}\sin {\text{B}} = \cos ({\text{A}} - {\text{B}}) - \cos ({\text{A}} + {\text{B}})$

7. Factorisation of the Sum or Difference of Two Sines or Cosines

(a) $\sin {\text{C}} + \sin {\text{D}} = 2\sin \dfrac{{{\text{C}} + {\text{D}}}}{2}\cos \dfrac{{{\text{C}} - {\text{D}}}}{2}$

(b) $\sin {\text{C}} - \sin {\text{D}} = 2\cos \dfrac{{{\text{C}} + {\text{D}}}}{2}\sin \dfrac{{{\text{C}} - {\text{D}}}}{2}$

(c) $\cos C + \cos D = 2\cos \dfrac{{C + D}}{2}\cos \dfrac{{C - D}}{2}$

(d) $\cos {\text{C}} - \cos {\text{D}} =  - 2\sin \dfrac{{{\text{C}} + {\text{D}}}}{2}\sin \dfrac{{{\text{C}} - {\text{D}}}}{2}$

8. Important Trigonometric Ratios

(a) $\sin {\text{n}}\pi  = 0;\cos {\text{n}}\pi  = {( - 1)^{\text{n}}};\tan {\text{n}}\pi  = 0$ where ${\text{n}} \in {\text{Z}}$

(b) $\sin {15^\circ }$ or $\sin \dfrac{\pi }{{12}} = \dfrac{{\sqrt 3  - 1}}{{2\sqrt 2 }} = \cos {75^\circ }$ or $\cos \dfrac{{5\pi }}{{12}}$; 

$\cos {15^\circ }$ or $\cos \dfrac{\pi }{{12}} = \dfrac{{\sqrt 3  + 1}}{{2\sqrt 2 }} = \sin {75^\circ }$ or $\sin \dfrac{{5\pi }}{{12}}$ 

$\tan {15^\circ } = \dfrac{{\sqrt 3  - 1}}{{\sqrt 3  + 1}} = 2 - \sqrt 3  = \cot {75^\circ }$ 

$\tan {75^\circ } = \dfrac{{\sqrt 3  + 1}}{{\sqrt 3  - 1}} = 2 + \sqrt 3  = \cot {15^\circ }$

(c) $\sin \dfrac{\pi }{{10}}$ or $\sin {18^\circ } = \dfrac{{\sqrt 5  - 1}}{4}$ & $\cos {36^\circ }$ or $\cos \dfrac{\pi }{5} = \dfrac{{\sqrt 5  + 1}}{4}$

9. Conditional Identities

If ${\text{A}} + {\text{B}} + {\text{C}} = \pi $ then :

(i) $\sin 2\;{\text{A}} + \sin 2\;{\text{B}} + \sin 2{\text{C}} = 4\sin {\text{A}}\sin {\text{B}}\sin {\text{C}}$

(ii) $\sin {\text{A}} + \sin {\text{B}} + \sin {\text{C}} = 4\cos \dfrac{{\text{A}}}{2}\cos \dfrac{{\text{B}}}{2}\cos \dfrac{{\text{C}}}{2}$

(iii) $\cos 2\;{\text{A}} + \cos 2\;{\text{B}} + \cos 2{\text{C}} =  - 1 - 4\cos {\text{A}}\cos {\text{B}}\cos {\text{C}}$

(iv) $\cos {\text{A}} + \cos {\text{B}} + \cos {\text{C}} = 1 + 4\sin \dfrac{{\text{A}}}{2}\sin \dfrac{{\text{B}}}{2}\sin \dfrac{{\text{C}}}{2}$

(v) $\tan {\text{A}} + \tan {\text{B}} + \tan {\text{C}} = \tan {\text{A}}\tan {\text{B}}\tan {\text{C}}$

(vi) $\tan \dfrac{{\text{A}}}{2}\tan \dfrac{{\text{B}}}{2} + \tan \dfrac{{\text{B}}}{2}\tan \dfrac{{\text{C}}}{2} + \tan \dfrac{{\text{C}}}{2}\tan \dfrac{{\text{A}}}{2} = 1$

(vii) $\cot \dfrac{{\text{A}}}{2} + \cot \dfrac{{\text{B}}}{2} + \cot \dfrac{{\text{C}}}{2} = \cot \dfrac{{\text{A}}}{2} \cdot \cot \dfrac{{\text{B}}}{2} \cdot \cot \dfrac{{\text{C}}}{2}$

(viii) $\cot A\cot B + \cot B\cot C + \cot C\cot A = 1$

10. Range of Trigonometric Expression

${{\text{E}} = {\text{a}}\sin \theta  + {\text{b}}\cos \theta }$

${{\text{E}} = \sqrt {{{\text{a}}^2} + {{\text{b}}^2}} \sin (\theta  + \alpha ),\left( {{\text{ where }}\tan \alpha  = \dfrac{{\text{b}}}{{\text{a}}}} \right)}$ 

${{\text{E}} = \sqrt {{{\text{a}}^2} + {{\text{b}}^2}} \cos (\theta  - \beta ),\left( {{\text{ where }}\tan \beta  = \dfrac{{\text{a}}}{{\text{b}}}} \right)}$

Hence for any real value of $\theta , - \sqrt {{a^2} + {b^2}}  \leqslant E \leqslant \sqrt {{a^2} + {b^2}} $

The trigonometric functions are very important for studying triangles, light, sound or waves. The values of these trigonometric functions in different domains and ranges can be used from the following table:

Trigonometric Functions in Different Domains and Ranges

11. Sine and Cosine Series

(a) $\quad \sin \alpha  + \sin (\alpha  + \beta ) + \sin (\alpha  + 2\beta ) +  \ldots . + \sin (\alpha  + \overline {n - 1} \beta )$

$ = \dfrac{{\sin \dfrac{{{\text{n}}\beta }}{2}}}{{\sin \dfrac{\beta }{2}}}\sin \left( {\alpha  + \dfrac{{{\text{n}} - 1}}{2}\beta } \right)$

(b) 

${\cos \alpha  + \cos (\alpha  + \beta ) + \cos (\alpha  + 2\beta ) +  \ldots  + \cos (\alpha  + \overline {n - 1} \beta )}$

${ = \dfrac{{\sin \dfrac{{n\beta }}{2}}}{{\sin \dfrac{\beta }{2}}}\cos \left( {\alpha  + \dfrac{{n - 1}}{2}\beta } \right)}$ 

12. Graphs of Trigonometric Functions 

(a). ${y = \sin x,}$

${x \in R;y \in [ - 1,1]}$ 

(b). $y = \cos x$

$x \in R;y \in [ - 1,1]$ 

(c) ${y = \tan x}$

${x \in R - \left\{ {(2n + 1)\dfrac{\pi }{2};n \in Z} \right\};y \in R}$ 

(d) ${y = \cot x}$

${x \in R - \{ n\pi ;n \in z\} ;y \in R}$ 

(e) ${y = \operatorname{cosec} x}$ 

${x \in R - \{ n\pi ;n \in Z\} ;y \in ( - \infty , - 1] \cup [1,\infty )}$

(f) $y = \sec x\quad $

$x \in R - \left\{ {(2n + 1)\dfrac{\pi }{2};n \in Z} \right\};y \in ( - \infty , - 1] \cup [1,\infty )$ 

Trigonometric Equations

13. Trigonometric Equations

Trigonometric equations are equations using trigonometric functions with unknown angles. 

e.g.,  $\cos \theta  = 0,{\cos ^2}\theta  - 4\cos \theta  = 1$.

The value of the unknown angle that satisfies a trigonometric equation is called a solution.

e.g., $\quad \sin \theta  = \dfrac{1}{{\sqrt 2 }} \Rightarrow \theta  = \dfrac{\pi }{4}$ or $\theta  = \dfrac{\pi }{4},\dfrac{{3\pi }}{4},\dfrac{{9\pi }}{4},\dfrac{{11\pi }}{4}, \ldots $

As a result, the trigonometric equation can have an unlimited number of solutions and is categorised as follows:

(i). Principal Solution

As we know, the values of $\sin x$ and $\cos x$ will get repeated after an interval of $2 \pi$. In the same way, the values of $\tan x$ will get repeated after an interval of $\pi$. 

So, if the equation has a variable $0 \leq \mathrm{x}<2 \pi$, then the solutions will be termed as principal solutions. 

Example:

Find the principal solutions of the equation $\sin x=\dfrac{\sqrt{3}}{2}$.

Solution: We know that, $\sin \dfrac{\pi}{3}=\dfrac{\sqrt{3}}{2}$

Also, $\sin \dfrac{2 \pi}{3}=\sin \left(\pi-\dfrac{\pi}{3}\right)$

Now, we know that $\sin (\pi-x)=\sin x$. 

Hence, $\sin \dfrac{2 \pi}{3}=\sin \dfrac{\pi}{3}=\dfrac{\sqrt{3}}{2}$

Therefore, the principal solutions of $\sin x=\dfrac{\sqrt{3}}{2}$ are $\mathrm{x}=\dfrac{\pi}{3}$ and $\dfrac{2 \pi}{3}$.

(ii). General solution

A general solution is one that involves the integer 'n' and yields all trigonometric equation solutions. Also, the character ' $\mathrm{Z}$ ' is used to denote the set of integers.

Find the solution of $\sin x=-\dfrac{\sqrt{3}}{2}$.

Solution: We know that $\sin \dfrac{\pi}{3}=\dfrac{\sqrt{3}}{2}$. Therefore, $\sin x=-\dfrac{\sqrt{3}}{2}=-\sin \dfrac{\pi}{3}$

Using the unit circle properties, we get $\sin x=-\sin \dfrac{\pi}{3}=\sin \left(\pi+\dfrac{\pi}{3}\right)=\sin \dfrac{4 \pi}{3}$ Hence, $\sin x=\sin \dfrac{4 \pi}{3}$

Since, we know that for any real numbers $x$ and $y, \sin x=\sin y$ implies $x=n \pi+(-1)^{n} y$, where $n \in Z$.

So, we get, $x=n \pi+(-1)^{\mathrm{n}}\left(\dfrac{4 \pi}{3}\right)$

14.1 Results

1. $\quad \sin \theta  = 0 \Leftrightarrow \theta  = \operatorname{n} \pi $

2. $\cos \theta  = 0 \Leftrightarrow \theta (2{\text{n}} + 1)\dfrac{\pi }{2}$

3. $\tan \theta  = 0 \Leftrightarrow \theta  = {\text{n}}\pi $

4. $\sin \theta  = \sin \alpha  \Leftrightarrow \theta  = {\text{n}}\pi  + {( - 1)^{\text{n}}}\alpha $, where $\alpha  \in \left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right]$

5. $\cos \theta  = \cos \alpha  \Leftrightarrow \theta  = 2{\text{n}}\pi  \pm \alpha $, where $\alpha  \in [0,\pi ]$

6. $\tan \theta  = \tan \alpha  \Leftrightarrow \theta  = {\text{n}}\pi  + \alpha $, where $\alpha  \in \left( { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right)$

7. ${\sin ^2}\theta  = {\sin ^2}\alpha  \Leftrightarrow \theta  = {\text{n}}\pi  \pm \alpha $.

8. ${\cos ^2}\theta  = {\cos ^2}\alpha  \Leftrightarrow \theta  = $ n $\pi  \pm \alpha $.

9. ${\tan ^2}\theta  = {\tan ^2}\alpha  \Leftrightarrow \theta  = {\text{n}}\pi  \pm \alpha $.

10. $\sin \theta  = 1 \Leftrightarrow \theta  = (4{\text{n}} + 1)\dfrac{\pi }{2}$

11. $\cos \theta  = 1 \Leftrightarrow \theta  = 2{\text{n}}\pi $

12. $\cos \theta  =  - 1 \Leftrightarrow \theta  = (2{\text{n}} + 1)\pi $.

13. $\sin \theta  = \sin \alpha $ and $\cos \theta  = \cos \alpha  \Leftrightarrow \theta  = 2{\text{n}}\pi  + \alpha $

Steps to Solve Trigonometric Functions:

The following are the stages of solving trigonometric equations:

Step 1: Decompose the trigonometric equation into a single trigonometric ratio, preferably the sine or cos function.

Step 2: Factor the trigonometric polynomial given in terms of the ratio.

Step 3: Write down the general solution after solving for each factor.

Note:

1. Unless otherwise stated, is treated as an integer throughout this chapter.

2. Unless the answer is required in a specific interval or range, the general solution should be supplied.

3. The angle's main value is regarded as $\alpha $. (The main value is the angle with the least numerical value.)

Download Free Trigonometric Functions Class 11 Notes PDF

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A Few Glimpses of Class 11 Chapter 3 Trigonometric Functions

The word trigonometry is derived from the Greek word 'trignon' and 'metron' which implies 'measuring the slides of a triangle'. This topic was originally developed to solve the geometrical problem including triangles. Trigonometry was used by engineers, sea captains for navigation purposes, surveyors to find out new lands, etc. Presently, trigonometry is used in many areas such as for estimating the heights of tides in the ocean, designing electric current, the science of seismology, etc.

In your earlier classes, you must have studied trigonometric ratios of an acute angle as a ratio of the sides of the right-angle triangle. You must have also studied trigonometric identities and applications of trigonometric ratios.

In this chapter, you will extrapolate the concepts of trigonometry ratios to trigonometry functions and study their properties.

Trigonometric Functions

Trigonometry functions also known as circular functions and states the relationship between sides and angles of a right-angle triangle. It implies that the relationship between sides and angle of a right angle triangle is derived by these trigonometric functions. The angles of sine, cosine, and tangent are the primary classification of trigonometric functions. And, the other three functions such as cotangent, secant, and cosecant are derived from the primary trigonometric functions. There exists an inverse trigonometric function for each of the above-mentioned trigonometry functions.

Topic and Subtopics Covered in Class 11 Chapter 3 Trigonometric Functions

Let us know the different topics and subtopics covered in Class 11 Chapter 3 Trigonometric Functions.

3.1: Introduction to Chapter

3.2: Angles

3.2.1: Degree Measure

3.2.2: Radian Measure

3.2.3: Relation between radian and real numbers

3.2.4: Relation between degree and radian

3.3: Trigonometric Functions

3.3.1: Sign of Trigonometric Functions

3.3.2: Domain And Range of Trigonometric Functions

3.4: Trigonometric Functions of Sum and Difference of Two Angles

3.5: Trigonometric Equations

Download free Trigonometric Functions Class 11 Notes pdf now to get brief information on all the topics and subtopics covered in the chapter. With the help of these revision notes, you will understand all the above-discussed topics in a better way as these notes are properly arranged for easy preparation of the chapter. These revision notes are of utmost importance to refer while preparing for the exam. So download it now and minimize your stress.

Benefits of by Vedantu Class 11 Maths Notes Chapter 3

Some of the benefits of Class 11 Maths Notes Chapter 3 offered by Vedantu are discussed below:

• Class 11 Maths Notes Chapter 3 provides an outline of the chapter in a short and precise manner.

• Students can easily revise the important formulas and theorems related to the trigonometric function quickly and efficiently.

• Students can save their valuable time by referring to the Mathematics notes for Class 11 Chapter 3.

• Students can immediately recall all the important concepts of the chapter just by having a look at the revision notes.

• Students can use the Class 11 Maths Notes Chapter 3 to revise the topic trigonometry rigorously just before the exam.