Rbse Solutions for Class 11 maths Chapter 3 Miscellaneous | Trigonometric Functions

Last Updated on November 24, 2025 by Aman Singh

Get detailed, step-by-step solutions for NCERT Class 11 Maths Chapter 3 Miscellaneous Exercise . Master advanced trigonometric proofs involving extensive use of the Sum-to-Product formulas (e.g., $\sin C \pm \sin D$, $\cos C \pm \cos D$) and Half-Angle Identities. Learn how to determine the exact values of $\mathbf{\sin(x/2)}$, $\mathbf{\cos(x/2)}$, and $\mathbf{\tan(x/2)}$ when given $\tan x$, $\cos x$, or $\sin x$ and the corresponding quadrant.

This exercise reviews various trigonometric identities, focusing on Sum-to-Product, Double Angle, and Half-Angle formulas.

Proving Identities (Exercises 1-7)

1. Prove: $\cos 2x \cos \frac{\pi}{13} + \cos \frac{3\pi}{13} + \cos \frac{5\pi}{13} + \cos \frac{9\pi}{13} = 0$

(Note: The question has a typo, the correct identity is $\cos \frac{\pi}{13} + \cos \frac{3\pi}{13} + \cos \frac{5\pi}{13} + \cos \frac{7\pi}{13} + \cos \frac{9\pi}{13} + \cos \frac{11\pi}{13} = 0$. The version presented is likely $\cos \frac{\pi}{13} + \cos \frac{3\pi}{13} + \cos \frac{5\pi}{13} + \cos \frac{7\pi}{13} = 0$. Let’s solve the version provided, assuming $\cos \frac{\pi}{13} + \cos \frac{3\pi}{13} + \cos \frac{5\pi}{13} + \cos \frac{9\pi}{13} = 0$ is the intended question, and use a technique that applies to sequences of cosines.)

This identity involves a series sum of cosines. For a series $\cos \alpha + \cos(\alpha + \beta) + \dots + \cos(\alpha + (n-1)\beta)$, the sum is given by:

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

However, for this specific problem, we look for pairs that sum to $\pi$:

$$\text{LHS} = \cos \frac{\pi}{13} + \cos \frac{3\pi}{13} + \cos \frac{5\pi}{13} + \cos \frac{9\pi}{13}$$

Notice that: $\frac{\pi}{13} + \frac{12\pi}{13} = \pi$.

Let’s rewrite the terms using $\cos(\pi – \theta) = -\cos \theta$.

Since $\frac{13\pi}{13} = \pi$, we know that $\cos \frac{11\pi}{13} = \cos(\pi – \frac{2\pi}{13}) = -\cos \frac{2\pi}{13}$.

Given the problem structure, we look for simplification through the $\cos C + \cos D$ formula. Grouping terms:

$$\text{LHS} = \left(\cos \frac{9\pi}{13} + \cos \frac{\pi}{13}\right) + \left(\cos \frac{5\pi}{13} + \cos \frac{3\pi}{13}\right)$$

Use $\cos C + \cos D = 2 \cos \left(\frac{C+D}{2}\right) \cos \left(\frac{C-D}{2}\right)$:

Group 1:

$$\cos \frac{9\pi}{13} + \cos \frac{\pi}{13} = 2 \cos \left(\frac{10\pi/13}{2}\right) \cos \left(\frac{8\pi/13}{2}\right) = 2 \cos \frac{5\pi}{13} \cos \frac{4\pi}{13}$$

Group 2:

$$\cos \frac{5\pi}{13} + \cos \frac{3\pi}{13} = 2 \cos \left(\frac{8\pi/13}{2}\right) \cos \left(\frac{2\pi/13}{2}\right) = 2 \cos \frac{4\pi}{13} \cos \frac{\pi}{13}$$

$$\text{LHS} = 2 \cos \frac{5\pi}{13} \cos \frac{4\pi}{13} + 2 \cos \frac{4\pi}{13} \cos \frac{\pi}{13}$$

Factor out $2 \cos \frac{4\pi}{13}$:

$$\text{LHS} = 2 \cos \frac{4\pi}{13} \left(\cos \frac{5\pi}{13} + \cos \frac{\pi}{13}\right)$$

Apply $\cos C + \cos D$ to the term in brackets:

$$\cos \frac{5\pi}{13} + \cos \frac{\pi}{13} = 2 \cos \left(\frac{6\pi/13}{2}\right) \cos \left(\frac{4\pi/13}{2}\right) = 2 \cos \frac{3\pi}{13} \cos \frac{2\pi}{13}$$

$$\text{LHS} = 2 \cos \frac{4\pi}{13} \left(2 \cos \frac{3\pi}{13} \cos \frac{2\pi}{13}\right)$$

$$\text{LHS} = 4 \cos \frac{2\pi}{13} \cos \frac{3\pi}{13} \cos \frac{4\pi}{13}$$

This expression does not equal 0. The original question likely intended a symmetric series which cancels out, such as:

$$\cos \frac{\pi}{13} + \cos \frac{3\pi}{13} + \cos \frac{5\pi}{13} + \dots + \cos \frac{11\pi}{13} = 0$$

Let’s assume the question meant to be $\cos \frac{\pi}{13} + \cos \frac{3\pi}{13} + \cos \frac{5\pi}{13} + \cos \frac{7\pi}{13} = 0$.

If $\cos \frac{\pi}{13} + \cos \frac{3\pi}{13} + \cos \frac{5\pi}{13} + \cos \frac{7\pi}{13} = 0$:

$$C = \left(\cos \frac{7\pi}{13} + \cos \frac{\pi}{13}\right) + \left(\cos \frac{5\pi}{13} + \cos \frac{3\pi}{13}\right)$$

• $\cos \frac{7\pi}{13} + \cos \frac{\pi}{13} = 2 \cos \frac{4\pi}{13} \cos \frac{3\pi}{13}$
• $\cos \frac{5\pi}{13} + \cos \frac{3\pi}{13} = 2 \cos \frac{4\pi}{13} \cos \frac{\pi}{13}$$$C = 2 \cos \frac{4\pi}{13} \left(\cos \frac{3\pi}{13} + \cos \frac{\pi}{13}\right)$$$$C = 2 \cos \frac{4\pi}{13} \left(2 \cos \frac{2\pi}{13} \cos \frac{\pi}{13}\right)$$This is still not 0. The identity as written in the book is flawed or a special case is intended that doesn’t simplify.

2. Prove: $(\sin 3x + \sin x) \sin x + (\cos 3x – \cos x) \cos x = 0$

Expand the LHS:

$$\text{LHS} = \sin 3x \sin x + \sin^2 x + \cos 3x \cos x – \cos^2 x$$

Rearrange the terms:

$$\text{LHS} = (\cos 3x \cos x + \sin 3x \sin x) – (\cos^2 x – \sin^2 x)$$

Use the Difference Identity for Cosine: $\cos(A – B) = \cos A \cos B + \sin A \sin B$.

$$\cos 3x \cos x + \sin 3x \sin x = \cos(3x – x) = \cos 2x$$

Use the Double Angle Identity for Cosine: $\cos 2x = \cos^2 x – \sin^2 x$.

$$\text{LHS} = \cos 2x – (\cos 2x)$$

$$\text{LHS} = \mathbf{0} = \text{RHS}$$

3. Prove: $(\cos x + \cos y)^2 + (\sin x – \sin y)^2 = 4 \cos^2 \left(\frac{x + y}{2}\right)$

Expand the squares on the LHS:

$$\text{LHS} = (\cos^2 x + 2 \cos x \cos y + \cos^2 y) + (\sin^2 x – 2 \sin x \sin y + \sin^2 y)$$

Group the squared terms:

$$\text{LHS} = (\cos^2 x + \sin^2 x) + (\cos^2 y + \sin^2 y) + 2 \cos x \cos y – 2 \sin x \sin y$$

Use the Pythagorean Identity ($\cos^2 \theta + \sin^2 \theta = 1$):

$$\text{LHS} = 1 + 1 + 2 (\cos x \cos y – \sin x \sin y)$$

Use the Sum Identity for Cosine: $\cos(A + B) = \cos A \cos B – \sin A \sin B$.

$$\text{LHS} = 2 + 2 \cos(x + y)$$

Factor out 2:

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

Use the Half-Angle Identity form: $1 + \cos 2\theta = 2 \cos^2 \theta$. Here $2\theta = x + y$, so $\theta = \frac{x+y}{2}$.

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

$$\text{LHS} = \mathbf{4 \cos^2 \left(\frac{x + y}{2}\right)} = \text{RHS}$$

4. Prove: $(\cos x – \cos y)^2 + (\sin x – \sin y)^2 = 4 \sin^2 \left(\frac{x – y}{2}\right)$

Expand the squares on the LHS:

$$\text{LHS} = (\cos^2 x – 2 \cos x \cos y + \cos^2 y) + (\sin^2 x – 2 \sin x \sin y + \sin^2 y)$$

Group the squared terms:

$$\text{LHS} = (\cos^2 x + \sin^2 x) + (\cos^2 y + \sin^2 y) – 2 \cos x \cos y – 2 \sin x \sin y$$

Use the Pythagorean Identity ($\cos^2 \theta + \sin^2 \theta = 1$):

$$\text{LHS} = 1 + 1 – 2 (\cos x \cos y + \sin x \sin y)$$

Use the Difference Identity for Cosine: $\cos(A – B) = \cos A \cos B + \sin A \sin B$.

$$\text{LHS} = 2 – 2 \cos(x – y)$$

Factor out 2:

$$\text{LHS} = 2 [1 – \cos(x – y)]$$

Use the Half-Angle Identity form: $1 – \cos 2\theta = 2 \sin^2 \theta$. Here $2\theta = x – y$, so $\theta = \frac{x-y}{2}$.

$$\text{LHS} = 2 \left[2 \sin^2 \left(\frac{x – y}{2}\right)\right]$$

$$\text{LHS} = \mathbf{4 \sin^2 \left(\frac{x – y}{2}\right)} = \text{RHS}$$

5. Prove: $\sin x + \sin 3x + \sin 5x + \sin 7x = 4 \cos x \cos 2x \sin 4x$

Group the terms symmetrically:

$$\text{LHS} = (\sin 7x + \sin x) + (\sin 5x + \sin 3x)$$

Use the $\sin C + \sin D$ Sum-to-Product Formula: $\sin C + \sin D = 2 \sin \left(\frac{C+D}{2}\right) \cos \left(\frac{C-D}{2}\right)$.

Group 1:

$$\sin 7x + \sin x = 2 \sin \left(\frac{8x}{2}\right) \cos \left(\frac{6x}{2}\right) = 2 \sin 4x \cos 3x$$

Group 2:

$$\sin 5x + \sin 3x = 2 \sin \left(\frac{8x}{2}\right) \cos \left(\frac{2x}{2}\right) = 2 \sin 4x \cos x$$

Substitute back into LHS:

$$\text{LHS} = 2 \sin 4x \cos 3x + 2 \sin 4x \cos x$$

Factor out $2 \sin 4x$:

$$\text{LHS} = 2 \sin 4x (\cos 3x + \cos x)$$

Apply the $\cos C + \cos D$ Sum-to-Product Formula to the term in brackets: $\cos C + \cos D = 2 \cos \left(\frac{C+D}{2}\right) \cos \left(\frac{C-D}{2}\right)$.

$$\cos 3x + \cos x = 2 \cos \left(\frac{4x}{2}\right) \cos \left(\frac{2x}{2}\right) = 2 \cos 2x \cos x$$

Substitute back:

$$\text{LHS} = 2 \sin 4x (2 \cos 2x \cos x)$$

$$\text{LHS} = \mathbf{4 \cos x \cos 2x \sin 4x} = \text{RHS}$$

6. Prove: $\frac{(\sin 7x + \sin 5x) + (\sin 9x + \sin 3x)}{(\cos 7x + \cos 5x) + (\cos 9x + \cos 3x)} = \tan 6x$

We apply the Sum-to-Product Formulas to all four groups.

Numerator:

• $\sin 7x + \sin 5x = 2 \sin 6x \cos x$
• $\sin 9x + \sin 3x = 2 \sin 6x \cos 3x$$$\text{Num} = 2 \sin 6x \cos x + 2 \sin 6x \cos 3x = 2 \sin 6x (\cos x + \cos 3x)$$

Denominator:

• $\cos 7x + \cos 5x = 2 \cos 6x \cos x$
• $\cos 9x + \cos 3x = 2 \cos 6x \cos 3x$$$\text{Den} = 2 \cos 6x \cos x + 2 \cos 6x \cos 3x = 2 \cos 6x (\cos x + \cos 3x)$$

LHS:

$$\text{LHS} = \frac{2 \sin 6x (\cos x + \cos 3x)}{2 \cos 6x (\cos x + \cos 3x)}$$

Cancel $2$ and $(\cos x + \cos 3x)$ (assuming the expression is non-zero):

$$\text{LHS} = \frac{\sin 6x}{\cos 6x} = \tan 6x = \text{RHS}$$

7. Prove: $\sin 3x + \sin 2x – \sin x = 4 \sin x \cos \frac{x}{2} \cos \frac{3x}{2}$

Group $\sin 3x$ and $-\sin x$:

$$\text{LHS} = (\sin 3x – \sin x) + \sin 2x$$

Use the $\sin C – \sin D$ Sum-to-Product Formula: $\sin C – \sin D = 2 \cos \left(\frac{C+D}{2}\right) \sin \left(\frac{C-D}{2}\right)$.

$$\sin 3x – \sin x = 2 \cos \left(\frac{4x}{2}\right) \sin \left(\frac{2x}{2}\right) = 2 \cos 2x \sin x$$

Substitute back into LHS:

$$\text{LHS} = 2 \cos 2x \sin x + \sin 2x$$

Use the Double Angle Identity $\sin 2x = 2 \sin x \cos x$:

$$\text{LHS} = 2 \cos 2x \sin x + 2 \sin x \cos x$$

Factor out $2 \sin x$:

$$\text{LHS} = 2 \sin x (\cos 2x + \cos x)$$

Apply the $\cos C + \cos D$ Sum-to-Product Formula to the term in brackets:

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

Substitute back:

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

$$\text{LHS} = \mathbf{4 \sin x \cos \frac{x}{2} \cos \frac{3x}{2}} = \text{RHS}$$

Half-Angle Values (Exercises 8-10)

The Half-Angle Formulas are:

$$\sin \frac{x}{2} = \pm \sqrt{\frac{1 – \cos x}{2}}$$

$$\cos \frac{x}{2} = \pm \sqrt{\frac{1 + \cos x}{2}}$$

$$\tan \frac{x}{2} = \frac{\sin x}{1 + \cos x} = \frac{1 – \cos x}{\sin x}$$

The sign ($\pm$) depends on the quadrant of $\mathbf{x/2}$.

8. $\tan x = -\frac{4}{3}$, $x$ in quadrant II ($\frac{\pi}{2} < x < \pi$)

1. Determine the quadrant of $x/2$:$$\frac{\pi}{4} < \frac{x}{2} < \frac{\pi}{2}$$$x/2$ is in Quadrant I. All trig functions are positive.
2. Find $\cos x$:Since $\tan x = -4/3$, we use $1 + \tan^2 x = \sec^2 x$.$$\sec^2 x = 1 + \left(-\frac{4}{3}\right)^2 = 1 + \frac{16}{9} = \frac{25}{9}$$$\sec x = \pm 5/3$. In Q.II, $\cos x$ is negative.$$\cos x = -\frac{3}{5}$$
3. Find $\sin x$:In Q.II, $\sin x$ is positive. $\sin x = \sqrt{1 – (-3/5)^2} = \sqrt{1 – 9/25} = \sqrt{16/25} = 4/5$.
4. Calculate Half-Angles (Positive since $x/2$ is in Q.I):$$\sin \frac{x}{2} = \sqrt{\frac{1 – \cos x}{2}} = \sqrt{\frac{1 – (-3/5)}{2}} = \sqrt{\frac{8/5}{2}} = \sqrt{\frac{4}{5}} = \mathbf{\frac{2}{\sqrt{5}}}$$$$\cos \frac{x}{2} = \sqrt{\frac{1 + \cos x}{2}} = \sqrt{\frac{1 + (-3/5)}{2}} = \sqrt{\frac{2/5}{2}} = \sqrt{\frac{1}{5}} = \mathbf{\frac{1}{\sqrt{5}}}$$$$\tan \frac{x}{2} = \frac{\sin (x/2)}{\cos (x/2)} = \frac{2/\sqrt{5}}{1/\sqrt{5}} = \mathbf{2}$$

9. $\cos x = -\frac{1}{3}$, $x$ in quadrant III ($\pi < x < \frac{3\pi}{2}$)

1. Determine the quadrant of $x/2$:$$\frac{\pi}{2} < \frac{x}{2} < \frac{3\pi}{4}$$$x/2$ is in Quadrant II. $\sin(x/2)$ is positive, $\cos(x/2)$ and $\tan(x/2)$ are negative.
2. Find $\sin x$:In Q.III, $\sin x$ is negative.$$\sin x = -\sqrt{1 – \left(-\frac{1}{3}\right)^2} = -\sqrt{1 – \frac{1}{9}} = -\sqrt{\frac{8}{9}} = -\frac{2\sqrt{2}}{3}$$
3. Calculate Half-Angles:$$\sin \frac{x}{2} = +\sqrt{\frac{1 – \cos x}{2}} = \sqrt{\frac{1 – (-1/3)}{2}} = \sqrt{\frac{4/3}{2}} = \sqrt{\frac{2}{3}} = \mathbf{\frac{\sqrt{6}}{3}}$$$$\cos \frac{x}{2} = -\sqrt{\frac{1 + \cos x}{2}} = -\sqrt{\frac{1 + (-1/3)}{2}} = -\sqrt{\frac{2/3}{2}} = -\sqrt{\frac{1}{3}} = \mathbf{-\frac{\sqrt{3}}{3}}$$$$\tan \frac{x}{2} = \frac{\sin (x/2)}{\cos (x/2)} = \frac{\sqrt{2/3}}{-\sqrt{1/3}} = -\sqrt{2} = \mathbf{-\sqrt{2}}$$

10. $\sin x = \frac{1}{4}$, $x$ in quadrant II ($\frac{\pi}{2} < x < \pi$)

1. Determine the quadrant of $x/2$:$$\frac{\pi}{4} < \frac{x}{2} < \frac{\pi}{2}$$$x/2$ is in Quadrant I. All trig functions are positive.
2. Find $\cos x$:In Q.II, $\cos x$ is negative.$$\cos x = -\sqrt{1 – \left(\frac{1}{4}\right)^2} = -\sqrt{1 – \frac{1}{16}} = -\sqrt{\frac{15}{16}} = -\frac{\sqrt{15}}{4}$$
3. Calculate Half-Angles (Positive since $x/2$ is in Q.I):$$\sin \frac{x}{2} = \sqrt{\frac{1 – \cos x}{2}} = \sqrt{\frac{1 – (-\sqrt{15}/4)}{2}} = \sqrt{\frac{4 + \sqrt{15}}{8}} = \mathbf{\frac{\sqrt{4 + \sqrt{15}}}{2\sqrt{2}}}$$$$\cos \frac{x}{2} = \sqrt{\frac{1 + \cos x}{2}} = \sqrt{\frac{1 + (-\sqrt{15}/4)}{2}} = \sqrt{\frac{4 – \sqrt{15}}{8}} = \mathbf{\frac{\sqrt{4 – \sqrt{15}}}{2\sqrt{2}}}$$$$\tan \frac{x}{2} = \frac{\sin x}{1 + \cos x} = \frac{1/4}{1 + (-\sqrt{15}/4)} = \frac{1/4}{(4 – \sqrt{15})/4} = \frac{1}{4 – \sqrt{15}}$$Rationalize:$$\tan \frac{x}{2} = \frac{1}{4 – \sqrt{15}} \times \frac{4 + \sqrt{15}}{4 + \sqrt{15}} = \frac{4 + \sqrt{15}}{16 – 15} = \mathbf{4 + \sqrt{15}}$$