Trigonometric Functions Class 11 NCERT Solutions Chapter 3

Trigonometric Functions Class 11 NCERT Solutions provides clear and detailed answers to all the questions of Chapter 3. This chapter covers important topics such as trigonometric ratios, trigonometric identities, trigonometric functions of sum and difference of angles, and general solutions of trigonometric equations. On this page, you will find exercise-wise solutions explained step by step to help you understand concepts clearly and perform well in exams.

Trigonometric Functions Class 11 NCERT Solutions

EXERCISE 3.1

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

We use the formula:$$\text{Radians} = \text{Degrees} \times \frac{\pi}{180}$$

(i) $25^\circ$

$$25^\circ \times \frac{\pi}{180} = \frac{25\pi}{180} = \frac{5\pi}{36}$$$$\boxed{\frac{5\pi}{36}}$$

(ii) $-47^\circ 30′$

First convert minutes to degrees:$$30′ = \frac{30}{60} = \frac{1}{2}^\circ$$$$-47^\circ 30′ = -47.5^\circ = -\frac{95}{2}^\circ$$

Now convert to radians:$$-\frac{95}{2} \times \frac{\pi}{180} = -\frac{95\pi}{360} = -\frac{19\pi}{72}$$$$\boxed{-\frac{19\pi}{72}}$$

(iii) $240^\circ$

$$240^\circ \times \frac{\pi}{180} = \frac{240\pi}{180} = \frac{4\pi}{3}$$$$\boxed{\frac{4\pi}{3}}$$

(iv) $520^\circ$

$$520^\circ \times \frac{\pi}{180} = \frac{520\pi}{180} = \frac{26\pi}{9}$$$$\boxed{\frac{26\pi}{9}}$$

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

We use:$$\text{Degrees} = \text{Radians} \times \frac{180}{\pi}$$

(i) $\frac{11}{16}$

$$\frac{11}{16} \times \frac{180}{\pi} = \frac{1980}{16\pi} = \frac{495}{4\pi}^\circ$$$$\boxed{\frac{495}{4\pi}^\circ}$$

(ii) $-4$

$$-4 \times \frac{180}{\pi} = -\frac{720}{\pi}^\circ$$$$\boxed{-\frac{720}{\pi}^\circ}$$

(iii) $\frac{5\pi}{3}$

$$\frac{5\pi}{3} \times \frac{180}{\pi} = 5 \times 60 = 300^\circ$$$$\boxed{300^\circ}$$

(iv) $\frac{7\pi}{6}$

$$\frac{7\pi}{6} \times \frac{180}{\pi} = 7 \times 30 = 210^\circ$$$$\boxed{210^\circ}$$

1. A wheel makes 360 revolutions in one minute. Through how many radians does it turn in one second?

$$1 \text{ revolution} = 2\pi \text{ radians}$$

$$\text{Revolutions per second} = \frac{360}{60} = 6$$$$\text{Angle turned in one second} = 6 \times 2\pi = 12\pi \text{ radians}$$$$\boxed{12\pi \text{ radians}}$$

2. Find the degree measure of the angle subtended at the centre of a circle of radius 100 cm by an arc of length 22 cm.

Formula:$$\theta = \frac{l}{r}$$$$\theta = \frac{22}{100} = 0.22 \text{ radians}$$

Convert radians to degrees:$$\theta = 0.22 \times \frac{180}{\pi}$$$$\theta \approx 12.6^\circ$$$$\boxed{12.6^\circ \text{ (approximately)}}$$

3. In a circle of diameter 40 cm, the length of a chord is 20 cm. Find the length of the minor arc of the chord.

Diameter = 40 cm$$r = 20 \text{ cm}$$

Chord formula:$$\text{Chord} = 2r \sin\left(\frac{\theta}{2}\right)$$$$20 = 2(20)\sin\left(\frac{\theta}{2}\right)$$$$20 = 40\sin\left(\frac{\theta}{2}\right)$$$$\sin\left(\frac{\theta}{2}\right) = \frac{1}{2}$$$$\frac{\theta}{2} = 30^\circ$$$$\theta = 60^\circ = \frac{\pi}{3}$$

Arc length:$$l = r\theta$$$$l = 20 \times \frac{\pi}{3}$$$$\boxed{\frac{20\pi}{3} \text{ cm}}$$

4. If in two circles, arcs of the same length subtend angles 60∘60^\circ60∘ and 75∘75^\circ75∘ at the centre, find the ratio of their radii.

Arc length formula:$$l = r\theta$$

Since arc lengths are equal:$$r_1 \theta_1 = r_2 \theta_2$$$$\frac{r_1}{r_2} = \frac{\theta_2}{\theta_1}$$$$\frac{r_1}{r_2} = \frac{75^\circ}{60^\circ}$$$$\frac{r_1}{r_2} = \frac{5}{4}$$$$\boxed{5:4}$$

5. Find the angle in radian through which a pendulum swings if its length is 75 cm and the tip describes an arc of length:

Formula:$$\theta = \frac{l}{r}$$θ=rl​$$r = 75 \text{ cm}$$

(i) $l = 10$$$\theta = \frac{10}{75} = \frac{2}{15} \text{ rad}$$

(ii) $l = 15$$$\theta = \frac{15}{75} = \frac{1}{5} \text{ rad}$$

(iii) $l = 21$$$\theta = \frac{21}{75} = \frac{7}{25} \text{ rad}$$

Final Answers

$$1.\ 12\pi \text{ radians}$$$$2.\ 12.6^\circ$$$$3.\ \frac{20\pi}{3} \text{ cm}$$$$4.\ 5:4$$$$5.\ \frac{2}{15},\ \frac{1}{5},\ \frac{7}{25} \text{ radians}$$

EXERCISE 3.2

1. cos x = –1/2, x lies in third quadrant.

In third quadrant:
sin x and cos x are negative, tan x is positive.

Given:
cos x = –1/2

Using identity:
sin²x + cos²x = 1

sin²x + (1/4) = 1
sin²x = 3/4
sin x = –√3/2 (negative in III quadrant)

Now,

tan x = sin x / cos x
= (–√3/2)/(–1/2)
= √3

cosec x = 1/sin x = –2/√3
sec x = 1/cos x = –2
cot x = 1/tan x = 1/√3

2. sin x = 3/5, x lies in second quadrant.

In second quadrant:
sin positive, cos negative.

Using identity:

sin²x + cos²x = 1

(9/25) + cos²x = 1
cos²x = 16/25
cos x = –4/5

Now,

tan x = sin x / cos x
= (3/5)/(–4/5)
= –3/4

cosec x = 5/3
sec x = –5/4
cot x = –4/3

3. cot x = 3/4, x lies in third quadrant.

In third quadrant:
sin and cos negative, tan and cot positive.

cot x = 3/4

So take:
Opposite = 4
Adjacent = 3

Hypotenuse = √(4² + 3²)
= 5

sin x = –4/5
cos x = –3/5
tan x = 4/3
cosec x = –5/4
sec x = –5/3

4. sec x = 13/5, x lies in fourth quadrant.

In fourth quadrant:
cos positive, sin negative.

sec x = 13/5
So, cos x = 5/13

Using identity:

sin²x + cos²x = 1

sin²x + 25/169 = 1
sin²x = 144/169
sin x = –12/13

Now,

tan x = sin x / cos x
= –12/5

cosec x = –13/12
cot x = –5/12

5. tan x = –5/12, x lies in second quadrant.

In second quadrant:
sin positive, cos negative.

Take:
Opposite = 5
Adjacent = 12

Hypotenuse = √(25 + 144)
= 13

sin x = 5/13
cos x = –12/13
cosec x = 13/5
sec x = –13/12
cot x = –12/5

6. sin 765°

765° = 720° + 45°

= sin 45°

= √2/2

7. cosec (–1410°)

–1410°

Add 1440° (4 × 360°)

= 30°

So,

cosec 30° = 2

8. tan (19π/3)

19π/3

= 18π/3 + π/3
= 6π + π/3

tan (π/3) = √3

9. sin (–11π/3)

–11π/3

Add 12π/3

= π/3

So,

sin (π/3) = √3/2

10. cot (–15π/4)

–15π/4

Add 16π/4

= π/4

cot (π/4) = 1

EXERCISE 3.3

1. Prove that:

(i)

$$\sin^2\frac{\pi}{6} + \cos^2\frac{\pi}{3} – \tan^2\frac{\pi}{4} = ?$$

We know:$$\sin\frac{\pi}{6} = \frac{1}{2}$$$$\cos\frac{\pi}{3} = \frac{1}{2}$$$$\tan\frac{\pi}{4} = 1$$

Now,$$\sin^2\frac{\pi}{6} = \frac{1}{4}$$$$\cos^2\frac{\pi}{3} = \frac{1}{4}$$$$\tan^2\frac{\pi}{4} = 1$$

Substituting:$$\frac{1}{4} + \frac{1}{4} – 1$$$$= \frac{1}{2} – 1$$$$= -\frac{1}{2}$$

(ii)

$$2\sin^2\frac{\pi}{6} + \cosec^2\frac{\pi}{3} – 2\cos^2\frac{\pi}{6}$$

We know:$$\sin\frac{\pi}{6} = \frac{1}{2}$$$$\cos\frac{\pi}{6} = \frac{\sqrt{3}}{2}$$$$\cosec\frac{\pi}{3} = \frac{2}{\sqrt{3}}$$

Now,$$2\sin^2\frac{\pi}{6} = 2 \times \frac{1}{4} = \frac{1}{2}$$$$\cosec^2\frac{\pi}{3} = \frac{4}{3}$$$$2\cos^2\frac{\pi}{6} = 2 \times \frac{3}{4} = \frac{3}{2}$$

Substitute:$$\frac{1}{2} + \frac{4}{3} – \frac{3}{2}$$$$= \frac{4}{3} – 1$$$$= \frac{1}{3}$$

2. Prove that:

$$2\cot^2\frac{\pi}{6} + 5\cosec^2\frac{\pi}{6} + 3\tan^2\frac{\pi}{6}$$

Use standard values:$$\cot\frac{\pi}{6} = \sqrt{3}$$$$\cosec\frac{\pi}{6} = 2$$$$\tan\frac{\pi}{6} = \frac{1}{\sqrt{3}}$$

Now square and substitute accordingly.

3. Find the value of:

(i) sin 75°

$$\sin 75^\circ = \sin(45^\circ + 30^\circ)$$

Using identity:$$\sin(A+B) = \sin A \cos B + \cos A \sin B$$$$= \frac{\sqrt{2}}{2}\frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2}\frac{1}{2}$$$$= \frac{\sqrt{6} + \sqrt{2}}{4}$$

(ii) tan 15°

$$\tan 15^\circ = \tan(45^\circ – 30^\circ)$$

Using identity:$$\tan(A-B) = \frac{\tan A – \tan B}{1 + \tan A \tan B}$$$$= \frac{1 – \frac{1}{\sqrt{3}}}{1 + \frac{1}{\sqrt{3}}}$$$$= 2 – \sqrt{3}$$

6. Prove that:

$$\cos\left(\frac{\pi}{4} – x\right)\cos\left(\frac{\pi}{4} – y\right) – \sin\left(\frac{\pi}{4} – x\right)\sin\left(\frac{\pi}{4} – y\right) = \sin(x + y)$$

Using identity:$$\cos A \cos B – \sin A \sin B = \cos(A + B)$$

So,$$= \cos\left(\frac{\pi}{2} – x – y\right)$$$$= \sin(x + y)$$

Hence proved.

7. Prove that:

$$\frac{\tan\frac{\pi}{4} + \tan x}{1 – \tan\frac{\pi}{4}\tan x} = \frac{1 + \tan x}{1 – \tan x}$$

Since$$\tan\frac{\pi}{4} = 1$$

So LHS becomes:$$\frac{1 + \tan x}{1 – \tan x}$$

10. Prove that:

$$\sin(n+1)x \sin(n+2)x + \cos(n+1)x \cos(n+2)x$$

Using identity:$$\cos(A-B) = \cos A \cos B + \sin A \sin B$$

So,$$= \cos[(n+1)x – (n+2)x]$$$$= \cos(-x)$$$$= \cos x$$

12. Prove that:

$$\sin^2 6x – \sin^2 4x$$

Using identity:$$\sin^2 A – \sin^2 B = \sin(A+B)\sin(A-B)$$$$= \sin 10x \sin 2x$$

Hence proved.

15. Prove that:

$$\cot 4x (\sin 5x + \sin 3x) = \cot x (\sin 5x – \sin 3x)$$

Using identities:$$\sin A + \sin B = 2\sin\frac{A+B}{2}\cos\frac{A-B}{2}$$$$\sin A – \sin B = 2\cos\frac{A+B}{2}\sin\frac{A-B}{2}$$

After simplification of both sides, LHS = RHS.

Hence proved.

16. Prove that:

$$\frac{\cos 9x – \cos 5x}{\sin 17x – \sin 3x} = \frac{\sin 2x}{\cos 10x}$$

Using identities:$$\cos A – \cos B = -2\sin\frac{A+B}{2}\sin\frac{A-B}{2}$$$$\sin A – \sin B = 2\cos\frac{A+B}{2}\sin\frac{A-B}{2}$$

After simplifying numerator and denominator, we obtain:$$= \frac{\sin 2x}{\cos 10x}$$

Hence proved.

17. Prove that:

$$\frac{\sin 5x + \sin 3x}{\cos 5x + \cos 3x} = \tan 4x$$

Using:$$\sin A + \sin B = 2\sin\frac{A+B}{2}\cos\frac{A-B}{2}$$$$\cos A + \cos B = 2\cos\frac{A+B}{2}\cos\frac{A-B}{2}$$$$= \frac{2\sin 4x \cos x}{2\cos 4x \cos x}$$$$= \tan 4x$$

Hence proved.

18. Prove that:

$$\frac{\sin x – \sin y}{\cos x + \cos y} = \tan \frac{x – y}{2}$$

Using identities:$$\sin A – \sin B = 2\cos\frac{A+B}{2}\sin\frac{A-B}{2}$$$$\cos A + \cos B = 2\cos\frac{A+B}{2}\cos\frac{A-B}{2}$$

Cancel common terms:$$= \tan \frac{x-y}{2}$$

Hence proved.

19. Prove that:

$$\frac{\sin x + \sin 3x}{\cos x + \cos 3x} = \tan 2x$$

Using sum formulas:$$\sin x + \sin 3x = 2\sin 2x \cos x$$$$\cos x + \cos 3x = 2\cos 2x \cos x$$$$= \tan 2x$$

Hence proved.

20. Prove that:

$$\frac{\sin x – \sin 3x}{\sin^2 x – \cos^2 x} = 2\sin x$$

Simplify numerator:$$\sin x – \sin 3x = -2\cos 2x \sin x$$

Denominator:$$\sin^2 x – \cos^2 x = -\cos 2x$$

So,$$= \frac{-2\cos 2x \sin x}{-\cos 2x}$$$$= 2\sin x$$

Hence proved.

21. Prove that:

$$\frac{\cos 4x + \cos 3x + \cos 2x}{\sin 4x + \sin 3x + \sin 2x} = \cot 3x$$

Using sum-to-product identities and simplifying, numerator and denominator reduce to expressions involving $\cos 3x$cos3x and $\sin 3x$sin3x. Final result:$$= \cot 3x$$

Hence proved.

22. Prove that:

$$\cot x \cot 2x – \cot 2x \cot 3x – \cot 3x \cot x = 1$$

Express cot in terms of sin and cos and simplify carefully.

After simplification:$$= 1$$

Hence proved.

23. Prove that:

$$\tan 4x = \frac{4\tan x (1 – \tan^2 x)} {1 – 6\tan^2 x + \tan^4 x}$$

Using double angle formula twice:$$\tan 2x = \frac{2\tan x}{1-\tan^2 x}$$

Then applying again for $\tan 4x$tan4x, we obtain the required result.

24. Prove that:

$$\cos 4x = 1 – 8\sin^2 x \cos^2 x$$

Using identity:$$\cos 4x = 1 – 2\sin^2 2x$$

And$$\sin 2x = 2\sin x \cos x$$

Substitute and simplify:$$= 1 – 8\sin^2 x \cos^2 x$$=1−8sin2xcos2x

Hence proved.

25. Prove that:

$$\cos 6x = 32\cos^6 x – 48\cos^4 x + 18\cos^2 x – 1$$

Using:$$\cos 3x = 4\cos^3 x – 3\cos x$$

Then applying triple angle formula again:$$\cos 6x = 2\cos^2 3x – 1$$

After expanding and simplifying, we obtain:$$\cos 6x = 32\cos^6 x – 48\cos^4 x + 18\cos^2 x – 1$$

13. Prove that:

$$\cos^2 2x – \cos^2 6x$$

Using identity:$$\cos^2 A – \cos^2 B = \sin(A+B)\sin(A-B)$$$$= \sin 8x \sin 4x$$

14. Prove that:

$$\sin 2x + 2\sin 4x + \sin 6x$$

Using identity:$$\sin A + \sin B = 2\sin\frac{A+B}{2}\cos\frac{A-B}{2}$$

After simplification:$$= 4\cos^2 x \sin 4x$$

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