NCERT Solutions For Class 11 Maths: Trigonometric Functions

Summary of the Chapter

This chapter provides a comprehensive introduction to trigonometric functions, moving from the right-angled triangle definitions to a more general circular function approach.

Key concepts covered:

• Angles: Measurement of angles in degrees and radians. The relationship is $2\pi \text{ radians} = 360^\circ$, or $\pi \text{ radians} = 180^\circ$.
• Radian Measure: The angle subtended at the center of a circle by an arc equal in length to the radius. The formula relating arc length ($l$), radius ($r$), and central angle ($\theta$ in radians) is $l = r\theta$.
• Trigonometric Functions: Defined using a unit circle (a circle with radius 1). For a point $P(x, y)$ on the unit circle corresponding to an angle $\theta$, $\cos \theta = x$ and $\sin \theta = y$.
• Six Functions: $\tan \theta = y/x$, $\csc \theta = 1/y$, $\sec \theta = 1/x$, and $\cot \theta = x/y$.
• Signs in Quadrants (ASTC Rule):
  • Quadrant I: All functions are positive.
  • Quadrant II: $\sin$ and $\csc$ are positive.
  • Quadrant III: $\tan$ and $\cot$ are positive.
  • Quadrant IV: $\cos$ and $\sec$ are positive.
• Domain and Range: The domain and range for all six trigonometric functions are established. For example, the domain of $\sin \theta$ is all real numbers ($R$) and its range is $[-1, 1]$.
• Fundamental Identities:
  • $\sin^2 x + \cos^2 x = 1$
  • $1 + \tan^2 x = \sec^2 x$
  • $1 + \cot^2 x = \csc^2 x$
• Periodicity: Trigonometric functions are periodic. $\sin$ and $\cos$ have a period of $2\pi$, while $\tan$ has a period of $\pi$.
• Sum and Difference Formulas: Formulas for $\cos(x \pm y)$, $\sin(x \pm y)$, and $\tan(x \pm y)$ are introduced, which are fundamental for further derivations.
• Double-Angle and Half-Angle Formulas: Formulas like $\cos 2x = \cos^2 x - \sin^2 x$, $\sin 2x = 2\sin x \cos x$, and others are derived.
• Trigonometric Equations: The concepts of principal solutions (solutions in the interval $[0, 2\pi)$) and general solutions (expressions that give all possible solutions) are introduced.

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NCERT Textbook Questions and Answers

Exercise 3.1

Question 1: Find the radian measures corresponding to the following degree measures:
(i) $25^\circ$
(ii) $-47^\circ 30'$
(iii) $240^\circ$
(iv) $520^\circ$

Answer-
We know that $180^\circ = \pi$ radians. Therefore, $1^\circ = \frac{\pi}{180}$ radians.

(i) $25^\circ = 25 \times \frac{\pi}{180} \text{ radians} = \frac{5\pi}{36} \text{ radians}$

(ii) $-47^\circ 30' = -47 \frac{30}{60}^\circ = -47.5^\circ = -\frac{95}{2}^\circ$
$-\frac{95}{2}^\circ = -\frac{95}{2} \times \frac{\pi}{180} \text{ radians} = -\frac{19}{2} \times \frac{\pi}{36} \text{ radians} = -\frac{19\pi}{72} \text{ radians}$

(iii) $240^\circ = 240 \times \frac{\pi}{180} \text{ radians} = \frac{4\pi}{3} \text{ radians}$

(iv) $520^\circ = 520 \times \frac{\pi}{180} \text{ radians} = \frac{26\pi}{9} \text{ radians}$

Question 2: Find the degree measures corresponding to the following radian measures (Use $\pi = 22/7$).
(i) $11/16$
(ii) $-4$
(iii) $5\pi/3$
(iv) $7\pi/6$

Answer-
We know that $\pi \text{ radians} = 180^\circ$. Therefore, $1 \text{ radian} = \frac{180}{\pi}^\circ$.

(i) $\frac{11}{16} \text{ radians} = \frac{11}{16} \times \frac{180}{\pi}^\circ = \frac{11}{16} \times \frac{180}{22/7}^\circ = \frac{11}{16} \times \frac{180 \times 7}{22}^\circ$
$= \frac{1}{16} \times \frac{180 \times 7}{2}^\circ = \frac{1}{16} \times 90 \times 7^\circ = \frac{45 \times 7}{8}^\circ = \frac{315}{8}^\circ$
$\frac{315}{8}^\circ = 39 \frac{3}{8}^\circ = 39^\circ + \frac{3}{8} \times 60' = 39^\circ + \frac{45}{2}' = 39^\circ + 22 \frac{1}{2}'$
$= 39^\circ 22' + \frac{1}{2} \times 60'' = 39^\circ 22' 30''$

(ii) $-4 \text{ radians} = -4 \times \frac{180}{\pi}^\circ = -4 \times \frac{180}{22/7}^\circ = \frac{-4 \times 180 \times 7}{22}^\circ = \frac{-2 \times 180 \times 7}{11}^\circ$
$= \frac{-2520}{11}^\circ = -229 \frac{1}{11}^\circ = -229^\circ - \frac{1}{11} \times 60' = -229^\circ - 5 \frac{5}{11}'$
$= -229^\circ 5' - \frac{5}{11} \times 60'' = -229^\circ 5' - \frac{300}{11}'' \approx -229^\circ 5' 27''$
(Note: $229^\circ 5' 27''$ is also acceptable if taken as positive conversion and adding negative sign)

(iii) $\frac{5\pi}{3} \text{ radians} = \frac{5\pi}{3} \times \frac{180}{\pi}^\circ = 5 \times 60^\circ = 300^\circ$

(iv) $\frac{7\pi}{6} \text{ radians} = \frac{7\pi}{6} \times \frac{180}{\pi}^\circ = 7 \times 30^\circ = 210^\circ$

Question 3: A wheel makes 360 revolutions in one minute. Through how many radians does it turn in one second?

Answer-
Revolutions in 1 minute (60 seconds) = 360
Revolutions in 1 second = $\frac{360}{60} = 6$
Angle turned in 1 revolution = $2\pi$ radians
Angle turned in 6 revolutions = $6 \times 2\pi = 12\pi$ radians
The wheel turns $12\pi$ radians in one second.

Question 4: 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 (Use $\pi = 22/7$).

Answer-
Given: radius $r = 100$ cm, arc length $l = 22$ cm.
We use the formula $\theta = \frac{l}{r}$, where $\theta$ is in radians.
$\theta = \frac{22}{100} \text{ radians} = \frac{11}{50} \text{ radians}$
Now, convert $\theta$ to degrees:
$\theta \text{ in degrees} = \frac{11}{50} \times \frac{180}{\pi}^\circ = \frac{11}{50} \times \frac{180}{22/7}^\circ = \frac{11}{50} \times \frac{180 \times 7}{22}^\circ$
$= \frac{1}{50} \times \frac{180 \times 7}{2}^\circ = \frac{1}{50} \times 90 \times 7^\circ = \frac{9 \times 7}{5}^\circ = \frac{63}{5}^\circ$
$\frac{63}{5}^\circ = 12 \frac{3}{5}^\circ = 12^\circ + \frac{3}{5} \times 60' = 12^\circ + 3 \times 12' = 12^\circ 36'$

Question 5: 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.

Answer-
Diameter = 40 cm, so radius $r = 20$ cm.
Length of chord = 20 cm.
The triangle formed by the two radii and the chord has all three sides equal to 20 cm.
Therefore, it is an equilateral triangle.
The angle $\theta$ subtended by the chord at the center is $60^\circ$.
To use $l = r\theta$, we must convert $\theta$ to radians:
$\theta = 60^\circ = 60 \times \frac{\pi}{180} = \frac{\pi}{3}$ radians.
Length of the minor arc $l = r\theta = 20 \times \frac{\pi}{3} = \frac{20\pi}{3}$ cm.

Question 6: If in two circles, arcs of the same length subtend angles $60^\circ$ and $75^\circ$ at the centre, find the ratio of their radii.

Answer-
Let the radii be $r_1$ and $r_2$.
Let the arc length be $l$ for both.
Given $\theta_1 = 60^\circ$ and $\theta_2 = 75^\circ$.
We must convert angles to radians:
$\theta_1 = 60^\circ = 60 \times \frac{\pi}{180} = \frac{\pi}{3}$ radians
$\theta_2 = 75^\circ = 75 \times \frac{\pi}{180} = \frac{5\pi}{12}$ radians
We know $l = r_1 \theta_1$ and $l = r_2 \theta_2$.
Therefore, $r_1 \theta_1 = r_2 \theta_2$
$r_1 \left(\frac{\pi}{3}\right) = r_2 \left(\frac{5\pi}{12}\right)$
$\frac{r_1}{3} = \frac{5r_2}{12}$
$\frac{r_1}{r_2} = \frac{5}{12} \times 3 = \frac{15}{12} = \frac{5}{4}$
The ratio of their radii $r_1 : r_2 = 5 : 4$.

Question 7: Find the angle in radian through which a pendulum swings if its length is 75 cm and the tip describes an arc of length:
(i) 10 cm
(ii) 15 cm
(iii) 21 cm

Answer-
The length of the pendulum is the radius $r = 75$ cm. The arc length is $l$.
The angle in radians is $\theta = \frac{l}{r}$.

(i) $l = 10$ cm. $\theta = \frac{10}{75} = \frac{2}{15}$ radians.

(ii) $l = 15$ cm. $\theta = \frac{15}{75} = \frac{1}{5}$ radians.

(iii) $l = 21$ cm. $\theta = \frac{21}{75} = \frac{7}{25}$ radians.

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Exercise 3.2

Question 1: Find the values of other five trigonometric functions if $\cos x = -1/2$, $x$ lies in the third quadrant.

Answer-
Given $\cos x = -1/2$ and $x$ is in Quadrant 3.
In Q3, $\sin x$ is negative and $\tan x$ is positive.

1. $\sec x = \frac{1}{\cos x} = \frac{1}{-1/2} = -2$

2. $\sin^2 x + \cos^2 x = 1$
   $\sin^2 x = 1 - \cos^2 x = 1 - (-1/2)^2 = 1 - 1/4 = 3/4$
   $\sin x = -\frac{\sqrt{3}}{2}$ (since $x$ is in Q3)

3. $\csc x = \frac{1}{\sin x} = \frac{1}{-\sqrt{3}/2} = -\frac{2}{\sqrt{3}}$

4. $\tan x = \frac{\sin x}{\cos x} = \frac{-\sqrt{3}/2}{-1/2} = \sqrt{3}$

5. $\cot x = \frac{1}{\tan x} = \frac{1}{\sqrt{3}}$

Question 2: Find the values of other five trigonometric functions if $\sin x = 3/5$, $x$ lies in second quadrant.

Answer-
Given $\sin x = 3/5$ and $x$ is in Quadrant 2.
In Q2, $\cos x$ is negative and $\tan x$ is negative.

1. $\csc x = \frac{1}{\sin x} = \frac{1}{3/5} = \frac{5}{3}$

2. $\sin^2 x + \cos^2 x = 1$
   $\cos^2 x = 1 - \sin^2 x = 1 - (3/5)^2 = 1 - 9/25 = 16/25$
   $\cos x = -\frac{4}{5}$ (since $x$ is in Q2)

3. $\sec x = \frac{1}{\cos x} = \frac{1}{-4/5} = -\frac{5}{4}$

4. $\tan x = \frac{\sin x}{\cos x} = \frac{3/5}{-4/5} = -\frac{3}{4}$

5. $\cot x = \frac{1}{\tan x} = \frac{1}{-3/4} = -\frac{4}{3}$

Question 3: Find the values of other five trigonometric functions if $\cot x = 3/4$, $x$ lies in third quadrant.

Answer-
Given $\cot x = 3/4$ and $x$ is in Quadrant 3.
In Q3, $\sin x$ is negative, $\cos x$ is negative, and $\tan x$ is positive.

1. $\tan x = \frac{1}{\cot x} = \frac{1}{3/4} = \frac{4}{3}$

2. $1 + \cot^2 x = \csc^2 x$
   $\csc^2 x = 1 + (3/4)^2 = 1 + 9/16 = 25/16$
   $\csc x = -\frac{5}{4}$ (since $x$ is in Q3)

3. $\sin x = \frac{1}{\csc x} = \frac{1}{-5/4} = -\frac{4}{5}$

4. $1 + \tan^2 x = \sec^2 x$
   $\sec^2 x = 1 + (4/3)^2 = 1 + 16/9 = 25/9$
   $\sec x = -\frac{5}{3}$ (since $x$ is in Q3)

5. $\cos x = \frac{1}{\sec x} = \frac{1}{-5/3} = -\frac{3}{5}$

Question 4: Find the values of other five trigonometric functions if $\sec x = 13/5$, $x$ lies in fourth quadrant.

Answer-
Given $\sec x = 13/5$ and $x$ is in Quadrant 4.
In Q4, $\sin x$ is negative, $\cos x$ is positive, and $\tan x$ is negative.

1. $\cos x = \frac{1}{\sec x} = \frac{1}{13/5} = \frac{5}{13}$

2. $\sin^2 x + \cos^2 x = 1$
   $\sin^2 x = 1 - \cos^2 x = 1 - (5/13)^2 = 1 - 25/169 = 144/169$
   $\sin x = -\frac{12}{13}$ (since $x$ is in Q4)

3. $\csc x = \frac{1}{\sin x} = \frac{1}{-12/13} = -\frac{13}{12}$

4. $\tan x = \frac{\sin x}{\cos x} = \frac{-12/13}{5/13} = -\frac{12}{5}$

5. $\cot x = \frac{1}{\tan x} = \frac{1}{-12/5} = -\frac{5}{12}$

Question 5: Find the values of other five trigonometric functions if $\tan x = -5/12$, $x$ lies in second quadrant.

Answer-
Given $\tan x = -5/12$ and $x$ is in Quadrant 2.
In Q2, $\sin x$ is positive, $\cos x$ is negative.

1. $\cot x = \frac{1}{\tan x} = \frac{1}{-5/12} = -\frac{12}{5}$

2. $1 + \tan^2 x = \sec^2 x$
   $\sec^2 x = 1 + (-5/12)^2 = 1 + 25/144 = 169/144$
   $\sec x = -\frac{13}{12}$ (since $x$ is in Q2)

3. $\cos x = \frac{1}{\sec x} = \frac{1}{-13/12} = -\frac{12}{13}$

4. We know $\tan x = \frac{\sin x}{\cos x}$, so $\sin x = \tan x \times \cos x$
   $\sin x = \left(-\frac{5}{12}\right) \times \left(-\frac{12}{13}\right) = \frac{5}{13}$

5. $\csc x = \frac{1}{\sin x} = \frac{1}{5/13} = \frac{13}{5}$

Question 6: Find the value of the trigonometric function $\sin 765^\circ$

Answer-
We can write $765^\circ$ as $n \times 360^\circ + \theta$.
$765^\circ = 2 \times 360^\circ + 45^\circ = 720^\circ + 45^\circ$
$\sin(765^\circ) = \sin(2 \times 360^\circ + 45^\circ)$
Since $\sin(2n\pi + \theta) = \sin \theta$,
$\sin(765^\circ) = \sin(45^\circ) = \frac{1}{\sqrt{2}}$

Question 7: Find the value of the trigonometric function $\csc (-1410^\circ)$

Answer-
We know $\csc(-\theta) = -\csc(\theta)$.
So, $\csc(-1410^\circ) = -\csc(1410^\circ)$.
Now, we write $1410^\circ$ as $n \times 360^\circ + \theta$.
$1410^\circ = 3 \times 360^\circ + 330^\circ = 1080^\circ + 330^\circ$
$-\csc(1410^\circ) = -\csc(3 \times 360^\circ + 330^\circ) = -\csc(330^\circ)$
We can write $\csc(330^\circ)$ as $\csc(360^\circ - 30^\circ)$.
In Q4, $\csc$ is negative, so $\csc(360^\circ - 30^\circ) = -\csc(30^\circ) = -2$.
Therefore, $-\csc(330^\circ) = -(-2) = 2$.

Alternatively:
$1410^\circ = 4 \times 360^\circ - 30^\circ = 1440^\circ - 30^\circ$
$-\csc(1410^\circ) = -\csc(4 \times 360^\circ - 30^\circ) = -\csc(-30^\circ)$
$= -(-\csc(30^\circ)) = \csc(30^\circ) = 2$.
So, $\csc(-1410^\circ) = 2$.

Question 8: Find the value of the trigonometric function $\tan (19\pi/3)$

Answer-
We can write $\frac{19\pi}{3}$ as $6\pi + \frac{\pi}{3} = 3 \times 2\pi + \frac{\pi}{3}$.
$\tan(19\pi/3) = \tan(6\pi + \pi/3)$
Since $\tan(2n\pi + \theta) = \tan \theta$,
$\tan(19\pi/3) = \tan(\pi/3) = \sqrt{3}$

Question 9: Find the value of the trigonometric function $\sin (-11\pi/3)$

Answer-
We know $\sin(-\theta) = -\sin(\theta)$.
So, $\sin(-11\pi/3) = -\sin(11\pi/3)$.
We can write $\frac{11\pi}{3}$ as $4\pi - \frac{\pi}{3} = 2 \times 2\pi - \frac{\pi}{3}$.
$-\sin(11\pi/3) = -\sin(4\pi - \pi/3)$
Since $\sin(2n\pi + \theta) = \sin \theta$, $\sin(4\pi - \pi/3) = \sin(-\pi/3)$.
$\sin(-\pi/3) = -\sin(\pi/3) = -\frac{\sqrt{3}}{2}$.
So, $\sin(-11\pi/3) = -(-\frac{\sqrt{3}}{2}) = \frac{\sqrt{3}}{2}$.

Question 10: Find the value of the trigonometric function $\cot (-15\pi/4)$

Answer-
We know $\cot(-\theta) = -\cot(\theta)$.
So, $\cot(-15\pi/4) = -\cot(15\pi/4)$.
We can write $\frac{15\pi}{4}$ as $4\pi - \frac{\pi}{4} = 2 \times 2\pi - \frac{\pi}{4}$.
$-\cot(15\pi/4) = -\cot(4\pi - \pi/4)$
Since $\cot(2n\pi + \theta) = \cot \theta$, $\cot(4\pi - \pi/4) = \cot(-\pi/4)$.
$\cot(-\pi/4) = -\cot(\pi/4) = -1$.
So, $\cot(-15\pi/4) = -(-1) = 1$.

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Exercise 3.3

Question 1: Prove that $\sin^2 \frac{\pi}{6} + \cos^2 \frac{\pi}{3} - \tan^2 \frac{\pi}{4} = -\frac{1}{2}$

Answer-
We take the Left Hand Side (LHS):
LHS = $\sin^2 \frac{\pi}{6} + \cos^2 \frac{\pi}{3} - \tan^2 \frac{\pi}{4}$
We know the values:
$\sin(\pi/6) = 1/2$
$\cos(\pi/3) = 1/2$
$\tan(\pi/4) = 1$
Substitute these values into the LHS:
LHS = $(1/2)^2 + (1/2)^2 - (1)^2$
LHS = $1/4 + 1/4 - 1$
LHS = $1/2 - 1 = -1/2$
LHS = RHS (Right Hand Side)
Hence, proved.

Question 2: Prove that $2\sin^2 \frac{\pi}{6} + \csc^2 \frac{7\pi}{6} \cos^2 \frac{\pi}{3} = \frac{3}{2}$

Answer-
LHS = $2\sin^2 \frac{\pi}{6} + \csc^2 \frac{7\pi}{6} \cos^2 \frac{\pi}{3}$
First, find the value of $\csc(7\pi/6)$:
$\csc(7\pi/6) = \csc(\pi + \pi/6)$
Since $\csc$ is negative in Quadrant 3, $\csc(\pi + \theta) = -\csc(\theta)$.
$\csc(7\pi/6) = -\csc(\pi/6) = -2$
Now substitute all values into the LHS:
$\sin(\pi/6) = 1/2$
$\cos(\pi/3) = 1/2$
$\csc(7\pi/6) = -2$
LHS = $2(1/2)^2 + (-2)^2 (1/2)^2$
LHS = $2(1/4) + (4)(1/4)$
LHS = $1/2 + 1 = 3/2$
LHS = RHS
Hence, proved.

Question 3: Prove that $\cot^2 \frac{\pi}{6} + \csc \frac{5\pi}{6} + 3\tan^2 \frac{\pi}{6} = 6$

Answer-
LHS = $\cot^2 \frac{\pi}{6} + \csc \frac{5\pi}{6} + 3\tan^2 \frac{\pi}{6}$
First, find the value of $\csc(5\pi/6)$:
$\csc(5\pi/6) = \csc(\pi - \pi/6)$
Since $\csc$ is positive in Quadrant 2, $\csc(\pi - \theta) = \csc(\theta)$.
$\csc(5\pi/6) = \csc(\pi/6) = 2$
Now substitute all values into the LHS:
$\cot(\pi/6) = \sqrt{3}$
$\tan(\pi/6) = 1/\sqrt{3}$
$\csc(5\pi/6) = 2$
LHS = $(\sqrt{3})^2 + 2 + 3(1/\sqrt{3})^2$
LHS = $3 + 2 + 3(1/3)$
LHS = $3 + 2 + 1 = 6$
LHS = RHS
Hence, proved.

Question 4: Prove that $2\sin^2 \frac{3\pi}{4} + 2\cos^2 \frac{\pi}{4} + 2\sec^2 \frac{\pi}{3} = 10$

Answer-
LHS = $2\sin^2 \frac{3\pi}{4} + 2\cos^2 \frac{\pi}{4} + 2\sec^2 \frac{\pi}{3}$
First, find the value of $\sin(3\pi/4)$:
$\sin(3\pi/4) = \sin(\pi - \pi/4)$
Since $\sin$ is positive in Quadrant 2, $\sin(\pi - \theta) = \sin(\theta)$.
$\sin(3\pi/4) = \sin(\pi/4) = 1/\sqrt{2}$
Now substitute all values into the LHS:
$\sin(3\pi/4) = 1/\sqrt{2}$
$\cos(\pi/4) = 1/\sqrt{2}$
$\sec(\pi/3) = 2$
LHS = $2(1/\sqrt{2})^2 + 2(1/\sqrt{2})^2 + 2(2)^2$
LHS = $2(1/2) + 2(1/2) + 2(4)$
LHS = $1 + 1 + 8 = 10$
LHS = RHS
Hence, proved.

Question 5: Find the value of: (i) $\sin 75^\circ$ (ii) $\tan 15^\circ$

Answer-
(i) $\sin 75^\circ$
We can write $75^\circ = 45^\circ + 30^\circ$
$\sin(75^\circ) = \sin(45^\circ + 30^\circ)$
Using the formula $\sin(A + B) = \sin A \cos B + \cos A \sin B$:
$\sin(75^\circ) = \sin(45^\circ)\cos(30^\circ) + \cos(45^\circ)\sin(30^\circ)$
$\sin(75^\circ) = (1/\sqrt{2})(\sqrt{3}/2) + (1/\sqrt{2})(1/2)$
$\sin(75^\circ) = \frac{\sqrt{3}}{2\sqrt{2}} + \frac{1}{2\sqrt{2}} = \frac{\sqrt{3} + 1}{2\sqrt{2}}$
Rationalizing the denominator: $\frac{(\sqrt{3} + 1)}{2\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{6} + \sqrt{2}}{4}$
$\sin 75^\circ = \frac{\sqrt{6} + \sqrt{2}}{4}$

(ii) $\tan 15^\circ$
We can write $15^\circ = 45^\circ - 30^\circ$
$\tan(15^\circ) = \tan(45^\circ - 30^\circ)$
Using the formula $\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$:
$\tan(15^\circ) = \frac{\tan 45^\circ - \tan 30^\circ}{1 + \tan 45^\circ \tan 30^\circ}$
$\tan(15^\circ) = \frac{1 - 1/\sqrt{3}}{1 + 1 \cdot (1/\sqrt{3})} = \frac{( \sqrt{3} - 1 ) / \sqrt{3}}{( \sqrt{3} + 1 ) / \sqrt{3}}$
$\tan(15^\circ) = \frac{\sqrt{3} - 1}{\sqrt{3} + 1}$
Rationalizing the denominator:
$\tan(15^\circ) = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} \times \frac{\sqrt{3} - 1}{\sqrt{3} - 1} = \frac{(\sqrt{3} - 1)^2}{(\sqrt{3})^2 - 1^2}$
$\tan(15^\circ) = \frac{3 - 2\sqrt{3} + 1}{3 - 1} = \frac{4 - 2\sqrt{3}}{2} = 2 - \sqrt{3}$
$\tan 15^\circ = 2 - \sqrt{3}$

Question 6: Prove that $\cos(\frac{\pi}{4} - x)\cos(\frac{\pi}{4} - y) - \sin(\frac{\pi}{4} - x)\sin(\frac{\pi}{4} - y) = \sin(x + y)$

Answer-
LHS = $\cos(\frac{\pi}{4} - x)\cos(\frac{\pi}{4} - y) - \sin(\frac{\pi}{4} - x)\sin(\frac{\pi}{4} - y)$
This expression is in the form $\cos A \cos B - \sin A \sin B$, which is equal to $\cos(A + B)$.
Let $A = (\frac{\pi}{4} - x)$ and $B = (\frac{\pi}{4} - y)$.
LHS = $\cos(A + B) = \cos\left[ \left(\frac{\pi}{4} - x\right) + \left(\frac{\pi}{4} - y\right) \right]$
LHS = $\cos\left( \frac{\pi}{4} + \frac{\pi}{4} - x - y \right) = \cos\left( \frac{2\pi}{4} - (x + y) \right)$
LHS = $\cos\left( \frac{\pi}{2} - (x + y) \right)$
Using the identity $\cos(\pi/2 - \theta) = \sin \theta$:
LHS = $\sin(x + y)$
LHS = RHS
Hence, proved.

Question 7: Prove that $\frac{\tan(\frac{\pi}{4} + x)}{\tan(\frac{\pi}{4} - x)} = \left(\frac{1 + \tan x}{1 - \tan x}\right)^2$

Answer-
LHS = $\frac{\tan(\frac{\pi}{4} + x)}{\tan(\frac{\pi}{4} - x)}$
We use the sum and difference formulas for $\tan$:
Numerator: $\tan(\frac{\pi}{4} + x) = \frac{\tan(\pi/4) + \tan x}{1 - \tan(\pi/4) \tan x} = \frac{1 + \tan x}{1 - \tan x}$
Denominator: $\tan(\frac{\pi}{4} - x) = \frac{\tan(\pi/4) - \tan x}{1 + \tan(\pi/4) \tan x} = \frac{1 - \tan x}{1 + \tan x}$
Now, substitute these back into the LHS:
LHS = $\frac{ (1 + \tan x) / (1 - \tan x) }{ (1 - \tan x) / (1 + \tan x) }$
LHS = $\frac{1 + \tan x}{1 - \tan x} \times \frac{1 + \tan x}{1 - \tan x}$
LHS = $\frac{(1 + \tan x)^2}{(1 - \tan x)^2} = \left(\frac{1 + \tan x}{1 - \tan x}\right)^2$
LHS = RHS
Hence, proved.

Question 8: Prove that $\frac{\cos(\pi + x) \cos(-x)}{\sin(\pi - x) \cos(\frac{\pi}{2} + x)} = \cot^2 x$

Answer-
LHS = $\frac{\cos(\pi + x) \cos(-x)}{\sin(\pi - x) \cos(\frac{\pi}{2} + x)}$
We simplify each term using allied angle identities:
$\cos(\pi + x) = -\cos x$ (Q3, $\cos$ is negative)
$\cos(-x) = \cos x$ ($\cos$ is an even function)
$\sin(\pi - x) = \sin x$ (Q2, $\sin$ is positive)
$\cos(\frac{\pi}{2} + x) = -\sin x$ (Q2, $\cos$ is negative, and $\cos$ changes to $\sin$)
Substitute these into the LHS:
LHS = $\frac{(-\cos x)(\cos x)}{(\sin x)(-\sin x)}$
LHS = $\frac{-\cos^2 x}{-\sin^2 x} = \frac{\cos^2 x}{\sin^2 x}$
LHS = $\cot^2 x$
LHS = RHS
Hence, proved.

Question 9: Prove that $\cos(\frac{3\pi}{2} + x) \cos(2\pi + x) \left[ \cot(\frac{3\pi}{2} - x) + \cot(2\pi + x) \right] = 1$

Answer-
LHS = $\cos(\frac{3\pi}{2} + x) \cos(2\pi + x) \left[ \cot(\frac{3\pi}{2} - x) + \cot(2\pi + x) \right]$
Simplify each term:
$\cos(\frac{3\pi}{2} + x) = \sin x$ (Q4, $\cos$ is positive, $\cos$ changes to $\sin$)
$\cos(2\pi + x) = \cos x$ (Q1, period of $2\pi$)
$\cot(\frac{3\pi}{2} - x) = \tan x$ (Q3, $\cot$ is positive, $\cot$ changes to $\tan$)
$\cot(2\pi + x) = \cot x$ (Q1, period of $2\pi$)
Substitute these into the LHS:
LHS = $(\sin x)(\cos x) [\tan x + \cot x]$
Now, express $\tan x$ and $\cot x$ in terms of $\sin x$ and $\cos x$:
LHS = $(\sin x)(\cos x) \left[ \frac{\sin x}{\cos x} + \frac{\cos x}{\sin x} \right]$
LHS = $(\sin x)(\cos x) \left[ \frac{\sin^2 x + \cos^2 x}{\sin x \cos x} \right]$
Using the identity $\sin^2 x + \cos^2 x = 1$:
LHS = $(\sin x)(\cos x) \left[ \frac{1}{\sin x \cos x} \right]$
LHS = $1$
LHS = RHS
Hence, proved.

Question 10: Prove that $\sin(n + 1)x \sin(n + 2)x + \cos(n + 1)x \cos(n + 2)x = \cos x$

Answer-
LHS = $\sin(n + 1)x \sin(n + 2)x + \cos(n + 1)x \cos(n + 2)x$
Rearranging the terms:
LHS = $\cos(n + 2)x \cos(n + 1)x + \sin(n + 2)x \sin(n + 1)x$
This expression is in the form $\cos A \cos B + \sin A \sin B$, which is equal to $\cos(A - B)$.
Let $A = (n + 2)x$ and $B = (n + 1)x$.
LHS = $\cos(A - B) = \cos[ (n + 2)x - (n + 1)x ]$
LHS = $\cos( (n+2 - n - 1)x )$
LHS = $\cos( 1 \cdot x ) = \cos x$
LHS = RHS
Hence, proved.

Question 11: Prove that $\cos(\frac{3\pi}{4} + x) - \cos(\frac{3\pi}{4} - x) = -\sqrt{2} \sin x$

Answer-
LHS = $\cos(\frac{3\pi}{4} + x) - \cos(\frac{3\pi}{4} - x)$
We use the sum-to-product formula $\cos A - \cos B = -2 \sin\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right)$.
Let $A = \frac{3\pi}{4} + x$ and $B = \frac{3\pi}{4} - x$.
$\frac{A+B}{2} = \frac{(\frac{3\pi}{4} + x) + (\frac{3\pi}{4} - x)}{2} = \frac{2 \cdot (3\pi/4)}{2} = \frac{3\pi}{4}$
$\frac{A-B}{2} = \frac{(\frac{3\pi}{4} + x) - (\frac{3\pi}{4} - x)}{2} = \frac{2x}{2} = x$
Substitute these into the formula:
LHS = $-2 \sin\left(\frac{3\pi}{4}\right) \sin(x)$
We need to find $\sin(3\pi/4)$:
$\sin(3\pi/4) = \sin(\pi - \pi/4) = \sin(\pi/4) = \frac{1}{\sqrt{2}}$
LHS = $-2 \left(\frac{1}{\sqrt{2}}\right) \sin(x)$
LHS = $-\frac{2}{\sqrt{2}} \sin(x) = -\sqrt{2} \sin x$
LHS = RHS
Hence, proved.

Question 12: Prove that $\sin^2 6x - \sin^2 4x = \sin 2x \sin 10x$

Answer-
LHS = $\sin^2 6x - \sin^2 4x$
We use the identity $\sin^2 A - \sin^2 B = \sin(A + B) \sin(A - B)$.
Let $A = 6x$ and $B = 4x$.
LHS = $\sin(6x + 4x) \sin(6x - 4x)$
LHS = $\sin(10x) \sin(2x)$
LHS = $\sin 2x \sin 10x$
LHS = RHS
Hence, proved.

Question 13: Prove that $\cos^2 2x - \cos^2 6x = \sin 4x \sin 8x$

Answer-
LHS = $\cos^2 2x - \cos^2 6x$
We use the identity $\cos^2 A = 1 - \sin^2 A$.
LHS = $(1 - \sin^2 2x) - (1 - \sin^2 6x)$
LHS = $1 - \sin^2 2x - 1 + \sin^2 6x$
LHS = $\sin^2 6x - \sin^2 2x$
Now we use the identity $\sin^2 A - \sin^2 B = \sin(A + B) \sin(A - B)$.
Let $A = 6x$ and $B = 2x$.
LHS = $\sin(6x + 2x) \sin(6x - 2x)$
LHS = $\sin(8x) \sin(4x)$
LHS = $\sin 4x \sin 8x$
LHS = RHS
Hence, proved.

Question 14: Prove that $\sin 2x + 2\sin 4x + \sin 6x = 4\cos^2 x \sin 4x$

Answer-
LHS = $\sin 2x + 2\sin 4x + \sin 6x$
Group the outer terms:
LHS = $(\sin 6x + \sin 2x) + 2\sin 4x$
Use the sum-to-product formula $\sin A + \sin B = 2 \sin\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right)$.
LHS = $\left[ 2 \sin\left(\frac{6x+2x}{2}\right) \cos\left(\frac{6x-2x}{2}\right) \right] + 2\sin 4x$
LHS = $2 \sin(4x) \cos(2x) + 2\sin 4x$
Factor out $2\sin 4x$:
LHS = $2\sin 4x (\cos 2x + 1)$
Use the double-angle identity $\cos 2x = 2\cos^2 x - 1$.
$\cos 2x + 1 = (2\cos^2 x - 1) + 1 = 2\cos^2 x$
Substitute this back:
LHS = $2\sin 4x (2\cos^2 x)$
LHS = $4\cos^2 x \sin 4x$
LHS = RHS
Hence, proved.

Question 15: Prove that $\cot 4x (\sin 5x + \sin 3x) = \cot x (\sin 5x - \sin 3x)$

Answer-
We will simplify both LHS and RHS.
LHS = $\cot 4x (\sin 5x + \sin 3x)$
Use $\sin A + \sin B = 2 \sin\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right)$:
LHS = $\frac{\cos 4x}{\sin 4x} \left[ 2 \sin\left(\frac{5x+3x}{2}\right) \cos\left(\frac{5x-3x}{2}\right) \right]$
LHS = $\frac{\cos 4x}{\sin 4x} [ 2 \sin(4x) \cos(x) ]$
LHS = $2 \cos 4x \cos x$

RHS = $\cot x (\sin 5x - \sin 3x)$
Use $\sin A - \sin B = 2 \cos\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right)$:
RHS = $\frac{\cos x}{\sin x} \left[ 2 \cos\left(\frac{5x+3x}{2}\right) \sin\left(\frac{5x-3x}{2}\right) \right]$
RHS = $\frac{\cos x}{\sin x} [ 2 \cos(4x) \sin(x) ]$
RHS = $2 \cos x \cos 4x$

Since LHS = $2 \cos 4x \cos x$ and RHS = $2 \cos 4x \cos x$,
LHS = RHS
Hence, proved.

Question 16: Prove that $\frac{\cos 9x - \cos 5x}{\sin 17x - \sin 3x} = -\frac{\sin 2x}{\cos 10x}$

Answer-
LHS = $\frac{\cos 9x - \cos 5x}{\sin 17x - \sin 3x}$
Use sum-to-product formulas:
$\cos A - \cos B = -2 \sin\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right)$
$\sin A - \sin B = 2 \cos\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right)$
Numerator: $\cos 9x - \cos 5x = -2 \sin\left(\frac{9x+5x}{2}\right) \sin\left(\frac{9x-5x}{2}\right) = -2 \sin(7x) \sin(2x)$
Denominator: $\sin 17x - \sin 3x = 2 \cos\left(\frac{17x+3x}{2}\right) \sin\left(\frac{17x-3x}{2}\right) = 2 \cos(10x) \sin(7x)$
Substitute back into the LHS:
LHS = $\frac{-2 \sin(7x) \sin(2x)}{2 \cos(10x) \sin(7x)}$
Cancel the $2$ and $\sin(7x)$ terms:
LHS = $\frac{-\sin(2x)}{\cos(10x)}$
LHS = RHS
Hence, proved.

Question 17: Prove that $\frac{\sin 5x + \sin 3x}{\cos 5x + \cos 3x} = \tan 4x$

Answer-
LHS = $\frac{\sin 5x + \sin 3x}{\cos 5x + \cos 3x}$
Use sum-to-product formulas:
$\sin A + \sin B = 2 \sin\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right)$
$\cos A + \cos B = 2 \cos\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right)$
Numerator: $\sin 5x + \sin 3x = 2 \sin\left(\frac{5x+3x}{2}\right) \cos\left(\frac{5x-3x}{2}\right) = 2 \sin(4x) \cos(x)$
Denominator: $\cos 5x + \cos 3x = 2 \cos\left(\frac{5x+3x}{2}\right) \cos\left(\frac{5x-3x}{2}\right) = 2 \cos(4x) \cos(x)$
Substitute back into the LHS:
LHS = $\frac{2 \sin(4x) \cos(x)}{2 \cos(4x) \cos(x)}$
Cancel the $2$ and $\cos(x)$ terms:
LHS = $\frac{\sin(4x)}{\cos(4x)} = \tan 4x$
LHS = RHS
Hence, proved.

Question 18: Prove that $\frac{\sin x - \sin y}{\cos x + \cos y} = \tan\left(\frac{x - y}{2}\right)$

Answer-
LHS = $\frac{\sin x - \sin y}{\cos x + \cos y}$
Use sum-to-product formulas:
$\sin A - \sin B = 2 \cos\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right)$
$\cos A + \cos B = 2 \cos\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right)$
Numerator: $\sin x - \sin y = 2 \cos\left(\frac{x+y}{2}\right) \sin\left(\frac{x-y}{2}\right)$
Denominator: $\cos x + \cos y = 2 \cos\left(\frac{x+y}{2}\right) \cos\left(\frac{x-y}{2}\right)$
Substitute back into the LHS:
LHS = $\frac{2 \cos\left(\frac{x+y}{2}\right) \sin\left(\frac{x-y}{2}\right)}{2 \cos\left(\frac{x+y}{2}\right) \cos\left(\frac{x-y}{2}\right)}$
Cancel the $2$ and $\cos\left(\frac{x+y}{2}\right)$ terms:
LHS = $\frac{\sin\left(\frac{x-y}{2}\right)}{\cos\left(\frac{x-y}{2}\right)} = \tan\left(\frac{x - y}{2}\right)$
LHS = RHS
Hence, proved.

Question 19: Prove that $\frac{\sin x + \sin 3x}{\cos x + \cos 3x} = \tan 2x$

Answer-
LHS = $\frac{\sin x + \sin 3x}{\cos x + \cos 3x} = \frac{\sin 3x + \sin x}{\cos 3x + \cos x}$
Use sum-to-product formulas:
$\sin A + \sin B = 2 \sin\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right)$
$\cos A + \cos B = 2 \cos\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right)$
Numerator: $\sin 3x + \sin x = 2 \sin\left(\frac{3x+x}{2}\right) \cos\left(\frac{3x-x}{2}\right) = 2 \sin(2x) \cos(x)$
Denominator: $\cos 3x + \cos x = 2 \cos\left(\frac{3x+x}{2}\right) \cos\left(\frac{3x-x}{2}\right) = 2 \cos(2x) \cos(x)$
Substitute back into the LHS:
LHS = $\frac{2 \sin(2x) \cos(x)}{2 \cos(2x) \cos(x)}$
Cancel the $2$ and $\cos(x)$ terms:
LHS = $\frac{\sin(2x)}{\cos(2x)} = \tan 2x$
LHS = RHS
Hence, proved.

Question 20: Prove that $\frac{\sin x - \sin 3x}{\sin^2 x - \cos^2 x} = 2\sin x$

Answer-
LHS = $\frac{\sin x - \sin 3x}{\sin^2 x - \cos^2 x}$
Numerator: Use $\sin A - \sin B = 2 \cos\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right)$
$\sin x - \sin 3x = 2 \cos\left(\frac{x+3x}{2}\right) \sin\left(\frac{x-3x}{2}\right)$
$= 2 \cos(2x) \sin(-x)$
Since $\sin(-x) = -\sin x$:
$= -2 \cos(2x) \sin(x)$
Denominator: $\sin^2 x - \cos^2 x = -(\cos^2 x - \sin^2 x)$
Using the identity $\cos 2x = \cos^2 x - \sin^2 x$:
$= -\cos(2x)$
Substitute back into the LHS:
LHS = $\frac{-2 \cos(2x) \sin(x)}{-\cos(2x)}$
Cancel the $-\cos(2x)$ terms:
LHS = $2\sin x$
LHS = RHS
Hence, proved.

Question 21: Prove that $\frac{\cos 4x + \cos 3x + \cos 2x}{\sin 4x + \sin 3x + \sin 2x} = \cot 3x$

Answer-
LHS = $\frac{\cos 4x + \cos 3x + \cos 2x}{\sin 4x + \sin 3x + \sin 2x}$
Group the outer terms in the numerator and denominator:
LHS = $\frac{(\cos 4x + \cos 2x) + \cos 3x}{(\sin 4x + \sin 2x) + \sin 3x}$
Use sum-to-product formulas:
$\cos A + \cos B = 2 \cos\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right)$
$\sin A + \sin B = 2 \sin\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right)$
Numerator: $(\cos 4x + \cos 2x) + \cos 3x = \left[ 2 \cos\left(\frac{4x+2x}{2}\right) \cos\left(\frac{4x-2x}{2}\right) \right] + \cos 3x$
$= 2 \cos(3x) \cos(x) + \cos 3x$
Denominator: $(\sin 4x + \sin 2x) + \sin 3x = \left[ 2 \sin\left(\frac{4x+2x}{2}\right) \cos\left(\frac{4x-2x}{2}\right) \right] + \sin 3x$
$= 2 \sin(3x) \cos(x) + \sin 3x$
Substitute back into the LHS:
LHS = $\frac{2 \cos(3x) \cos(x) + \cos 3x}{2 \sin(3x) \cos(x) + \sin 3x}$
Factor out $\cos 3x$ from the numerator and $\sin 3x$ from the denominator:
LHS = $\frac{\cos 3x (2 \cos x + 1)}{\sin 3x (2 \cos x + 1)}$
Cancel the $(2 \cos x + 1)$ term:
LHS = $\frac{\cos 3x}{\sin 3x} = \cot 3x$
LHS = RHS
Hence, proved.

Question 22: Prove that $\cot x \cot 2x - \cot 2x \cot 3x - \cot 3x \cot x = 1$

Answer-
We start with the relationship $3x = 2x + x$.
Apply $\cot$ to both sides:
$\cot(3x) = \cot(2x + x)$
Use the formula $\cot(A + B) = \frac{\cot A \cot B - 1}{\cot B + \cot A}$:
$\cot(3x) = \frac{\cot 2x \cot x - 1}{\cot x + \cot 2x}$
Cross-multiply:
$\cot(3x) (\cot x + \cot 2x) = \cot 2x \cot x - 1$
$\cot 3x \cot x + \cot 3x \cot 2x = \cot 2x \cot x - 1$
Rearrange the terms to match the required expression:
$1 = \cot 2x \cot x - \cot 3x \cot x - \cot 3x \cot 2x$
$1 = \cot x \cot 2x - \cot 2x \cot 3x - \cot 3x \cot x$
Hence, proved.

Question 23: Prove that $\tan 4x = \frac{4\tan x (1 - \tan^2 x)}{1 - 6\tan^2 x + \tan^4 x}$

Answer-
LHS = $\tan 4x = \tan(2 \cdot 2x)$
Use the double-angle formula $\tan 2A = \frac{2\tan A}{1 - \tan^2 A}$.
First, let $A = 2x$:
$\tan 4x = \frac{2\tan 2x}{1 - \tan^2(2x)}$
Now, use the same formula again for $\tan 2x$:
$\tan 2x = \frac{2\tan x}{1 - \tan^2 x}$
Substitute this into the expression for $\tan 4x$:
Numerator: $2\tan 2x = 2 \left( \frac{2\tan x}{1 - \tan^2 x} \right) = \frac{4\tan x}{1 - \tan^2 x}$
Denominator: $1 - \tan^2(2x) = 1 - \left( \frac{2\tan x}{1 - \tan^2 x} \right)^2$
$= 1 - \frac{4\tan^2 x}{(1 - \tan^2 x)^2}$
$= \frac{(1 - \tan^2 x)^2 - 4\tan^2 x}{(1 - \tan^2 x)^2}$
$= \frac{(1 - 2\tan^2 x + \tan^4 x) - 4\tan^2 x}{(1 - \tan^2 x)^2}$
$= \frac{1 - 6\tan^2 x + \tan^4 x}{(1 - \tan^2 x)^2}$
Now, combine the numerator and denominator for $\tan 4x$:
LHS = $\frac{ \frac{4\tan x}{1 - \tan^2 x} }{ \frac{1 - 6\tan^2 x + \tan^4 x}{(1 - \tan^2 x)^2} }$
LHS = $\frac{4\tan x}{1 - \tan^2 x} \times \frac{(1 - \tan^2 x)^2}{1 - 6\tan^2 x + \tan^4 x}$
Cancel one $(1 - \tan^2 x)$ term:
LHS = $\frac{4\tan x (1 - \tan^2 x)}{1 - 6\tan^2 x + \tan^4 x}$
LHS = RHS
Hence, proved.

Question 24: Prove that $\cos 4x = 1 - 8\sin^2 x \cos^2 x$

Answer-
LHS = $\cos 4x = \cos(2 \cdot 2x)$
Use the double-angle formula $\cos 2A = 1 - 2\sin^2 A$.
Let $A = 2x$:
LHS = $1 - 2\sin^2(2x)$
LHS = $1 - 2(\sin 2x)^2$
Now use the double-angle formula $\sin 2x = 2\sin x \cos x$:
LHS = $1 - 2(2\sin x \cos x)^2$
LHS = $1 - 2(4\sin^2 x \cos^2 x)$
LHS = $1 - 8\sin^2 x \cos^2 x$
LHS = RHS
Hence, proved.

Question 25: Prove that $\cos 6x = 32\cos^6 x - 48\cos^4 x + 18\cos^2 x - 1$

Answer-
LHS = $\cos 6x = \cos(3 \cdot 2x)$
Use the triple-angle formula $\cos 3A = 4\cos^3 A - 3\cos A$.
Let $A = 2x$:
LHS = $4\cos^3(2x) - 3\cos(2x)$
Now use the double-angle formula $\cos 2x = 2\cos^2 x - 1$:
LHS = $4(2\cos^2 x - 1)^3 - 3(2\cos^2 x - 1)$
Expand $(2\cos^2 x - 1)^3$ using $(a-b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$:
$(2\cos^2 x)^3 - 3(2\cos^2 x)^2(1) + 3(2\cos^2 x)(1)^2 - (1)^3$
$= 8\cos^6 x - 3(4\cos^4 x) + 6\cos^2 x - 1$
$= 8\cos^6 x - 12\cos^4 x + 6\cos^2 x - 1$
Now substitute this back into the expression for LHS:
LHS = $4(8\cos^6 x - 12\cos^4 x + 6\cos^2 x - 1) - 3(2\cos^2 x - 1)$
LHS = $(32\cos^6 x - 48\cos^4 x + 24\cos^2 x - 4) - (6\cos^2 x - 3)$
LHS = $32\cos^6 x - 48\cos^4 x + 24\cos^2 x - 4 - 6\cos^2 x + 3$
LHS = $32\cos^6 x - 48\cos^4 x + 18\cos^2 x - 1$
LHS = RHS
Hence, proved.

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Miscellaneous Exercise on Chapter 3

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

Answer-
LHS = $2\cos \frac{\pi}{13} \cos \frac{9\pi}{13} + \cos \frac{3\pi}{13} + \cos \frac{5\pi}{13}$
Use the product-to-sum formula $2\cos A \cos B = \cos(A + B) + \cos(A - B)$ on the first term.
$2\cos \frac{\pi}{13} \cos \frac{9\pi}{13} = \cos\left(\frac{\pi}{13} + \frac{9\pi}{13}\right) + \cos\left(\frac{\pi}{13} - \frac{9\pi}{13}\right)$
$= \cos\left(\frac{10\pi}{13}\right) + \cos\left(-\frac{8\pi}{13}\right)$
Since $\cos(-\theta) = \cos \theta$:
$= \cos\left(\frac{10\pi}{13}\right) + \cos\left(\frac{8\pi}{13}\right)$
Substitute this back into the LHS:
LHS = $\cos\left(\frac{10\pi}{13}\right) + \cos\left(\frac{8\pi}{13}\right) + \cos \frac{3\pi}{13} + \cos \frac{5\pi}{13}$
We can write $10\pi/13$ as $\pi - 3\pi/13$ and $8\pi/13$ as $\pi - 5\pi/13$.
$\cos(10\pi/13) = \cos(\pi - 3\pi/13) = -\cos(3\pi/13)$
$\cos(8\pi/13) = \cos(\pi - 5\pi/13) = -\cos(5\pi/13)$
Substitute these into the LHS:
LHS = $(-\cos \frac{3\pi}{13}) + (-\cos \frac{5\pi}{13}) + \cos \frac{3\pi}{13} + \cos \frac{5\pi}{13}$
LHS = $0$
LHS = RHS
Hence, proved.

Question 2: Prove that $(\sin 3x + \sin x)\sin x + (\cos 3x - \cos x)\cos x = 0$

Answer-
LHS = $(\sin 3x + \sin x)\sin x + (\cos 3x - \cos x)\cos x$
Method 1: Expand the terms
LHS = $\sin 3x \sin x + \sin^2 x + \cos 3x \cos x - \cos^2 x$
LHS = $(\cos 3x \cos x + \sin 3x \sin x) - (\cos^2 x - \sin^2 x)$
Use $\cos(A - B) = \cos A \cos B + \sin A \sin B$ and $\cos 2x = \cos^2 x - \sin^2 x$:
LHS = $\cos(3x - x) - \cos(2x)$
LHS = $\cos(2x) - \cos(2x) = 0$
LHS = RHS
Method 2: Use sum-to-product formulas
$\sin 3x + \sin x = 2 \sin(2x) \cos(x)$
$\cos 3x - \cos x = -2 \sin(2x) \sin(x)$
LHS = $(2 \sin 2x \cos x)\sin x + (-2 \sin 2x \sin x)\cos x$
LHS = $2 \sin 2x \sin x \cos x - 2 \sin 2x \sin x \cos x$
LHS = $0$
LHS = RHS
Hence, proved.

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

Answer-
LHS = $(\cos x + \cos y)^2 + (\sin x - \sin y)^2$
Use sum-to-product formulas:
$\cos x + \cos y = 2 \cos\left(\frac{x+y}{2}\right) \cos\left(\frac{x-y}{2}\right)$
$\sin x - \sin y = 2 \cos\left(\frac{x+y}{2}\right) \sin\left(\frac{x-y}{2}\right)$
Substitute these into the LHS:
LHS = $\left[ 2 \cos\left(\frac{x+y}{2}\right) \cos\left(\frac{x-y}{2}\right) \right]^2 + \left[ 2 \cos\left(\frac{x+y}{2}\right) \sin\left(\frac{x-y}{2}\right) \right]^2$
LHS = $4 \cos^2\left(\frac{x+y}{2}\right) \cos^2\left(\frac{x-y}{2}\right) + 4 \cos^2\left(\frac{x+y}{2}\right) \sin^2\left(\frac{x-y}{2}\right)$
Factor out $4 \cos^2\left(\frac{x+y}{2}\right)$:
LHS = $4 \cos^2\left(\frac{x+y}{2}\right) \left[ \cos^2\left(\frac{x-y}{2}\right) + \sin^2\left(\frac{x-y}{2}\right) \right]$
Using the identity $\sin^2 \theta + \cos^2 \theta = 1$:
LHS = $4 \cos^2\left(\frac{x+y}{2}\right) [ 1 ]$
LHS = $4 \cos^2\left(\frac{x+y}{2}\right)$
LHS = RHS
Hence, proved.

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

Answer-
LHS = $(\cos x - \cos y)^2 + (\sin x - \sin y)^2$
Use sum-to-product formulas:
$\cos x - \cos y = -2 \sin\left(\frac{x+y}{2}\right) \sin\left(\frac{x-y}{2}\right)$
$\sin x - \sin y = 2 \cos\left(\frac{x+y}{2}\right) \sin\left(\frac{x-y}{2}\right)$
Substitute these into the LHS:
LHS = $\left[ -2 \sin\left(\frac{x+y}{2}\right) \sin\left(\frac{x-y}{2}\right) \right]^2 + \left[ 2 \cos\left(\frac{x+y}{2}\right) \sin\left(\frac{x-y}{2}\right) \right]^2$
LHS = $4 \sin^2\left(\frac{x+y}{2}\right) \sin^2\left(\frac{x-y}{2}\right) + 4 \cos^2\left(\frac{x+y}{2}\right) \sin^2\left(\frac{x-y}{2}\right)$
Factor out $4 \sin^2\left(\frac{x-y}{2}\right)$:
LHS = $4 \sin^2\left(\frac{x-y}{2}\right) \left[ \sin^2\left(\frac{x+y}{2}\right) + \cos^2\left(\frac{x+y}{2}\right) \right]$
Using the identity $\sin^2 \theta + \cos^2 \theta = 1$:
LHS = $4 \sin^2\left(\frac{x-y}{2}\right) [ 1 ]$
LHS = $4 \sin^2\left(\frac{x-y}{2}\right)$
LHS = RHS
Hence, proved.

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

Answer-
LHS = $\sin x + \sin 3x + \sin 5x + \sin 7x$
Group terms:
LHS = $(\sin 7x + \sin x) + (\sin 5x + \sin 3x)$
Use $\sin A + \sin B = 2 \sin\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right)$:
$(\sin 7x + \sin x) = 2 \sin\left(\frac{7x+x}{2}\right) \cos\left(\frac{7x-x}{2}\right) = 2 \sin(4x) \cos(3x)$
$(\sin 5x + \sin 3x) = 2 \sin\left(\frac{5x+3x}{2}\right) \cos\left(\frac{5x-3x}{2}\right) = 2 \sin(4x) \cos(x)$
Substitute back into the LHS:
LHS = $2 \sin(4x) \cos(3x) + 2 \sin(4x) \cos(x)$
Factor out $2 \sin(4x)$:
LHS = $2 \sin(4x) [\cos(3x) + \cos(x)]$
Now use $\cos A + \cos B = 2 \cos\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right)$:
$\cos(3x) + \cos(x) = 2 \cos\left(\frac{3x+x}{2}\right) \cos\left(\frac{3x-x}{2}\right) = 2 \cos(2x) \cos(x)$
Substitute this back:
LHS = $2 \sin(4x) [2 \cos(2x) \cos(x)]$
LHS = $4 \sin(4x) \cos(2x) \cos(x)$
LHS = $4\cos x \cos 2x \sin 4x$
LHS = RHS
Hence, proved.

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

Answer-
LHS = $\frac{(\sin 7x + \sin 5x) + (\sin 9x + \sin 3x)}{(\cos 7x + \cos 5x) + (\cos 9x + \cos 3x)}$
Apply sum-to-product formulas to all four pairs:
$\sin 7x + \sin 5x = 2 \sin(6x) \cos(x)$
$\sin 9x + \sin 3x = 2 \sin(6x) \cos(3x)$
$\cos 7x + \cos 5x = 2 \cos(6x) \cos(x)$
$\cos 9x + \cos 3x = 2 \cos(6x) \cos(3x)$
Substitute these into the LHS:
LHS = $\frac{2 \sin(6x) \cos(x) + 2 \sin(6x) \cos(3x)}{2 \cos(6x) \cos(x) + 2 \cos(6x) \cos(3x)}$
Factor out $2\sin(6x)$ from the numerator and $2\cos(6x)$ from the denominator:
LHS = $\frac{2 \sin(6x) (\cos x + \cos 3x)}{2 \cos(6x) (\cos x + \cos 3x)}$
Cancel the $2$ and $(\cos x + \cos 3x)$ terms:
LHS = $\frac{\sin(6x)}{\cos(6x)} = \tan 6x$
LHS = RHS
Hence, proved.

Question 7: Prove that $\sin 3x + \sin 2x - \sin x = 4\sin x \cos\left(\frac{x}{2}\right) \cos\left(\frac{3x}{2}\right)$

Answer-
LHS = $\sin 3x + \sin 2x - \sin x$
Group the outer terms:
LHS = $(\sin 3x - \sin x) + \sin 2x$
Use $\sin A - \sin B = 2 \cos\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right)$:
$\sin 3x - \sin x = 2 \cos\left(\frac{3x+x}{2}\right) \sin\left(\frac{3x-x}{2}\right) = 2 \cos(2x) \sin(x)$
Substitute back, and use $\sin 2x = 2\sin x \cos x$:
LHS = $2 \cos(2x) \sin(x) + 2\sin x \cos x$
Factor out $2\sin x$:
LHS = $2\sin x (\cos 2x + \cos x)$
Now use $\cos A + \cos B = 2 \cos\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right)$:
$\cos 2x + \cos x = 2 \cos\left(\frac{2x+x}{2}\right) \cos\left(\frac{2x-x}{2}\right) = 2 \cos\left(\frac{3x}{2}\right) \cos\left(\frac{x}{2}\right)$
Substitute this back:
LHS = $2\sin x \left[ 2 \cos\left(\frac{3x}{2}\right) \cos\left(\frac{x}{2}\right) \right]$
LHS = $4\sin x \cos\left(\frac{x}{2}\right) \cos\left(\frac{3x}{2}\right)$
LHS = RHS
Hence, proved.

Question 8: Find $\sin(x/2)$, $\cos(x/2)$ and $\tan(x/2)$ if $\tan x = -4/3$, $x$ lies in quadrant II.

Answer-
Given $\tan x = -4/3$ and $x$ is in Quadrant II.
This means $\frac{\pi}{2} < x < \pi$.
Dividing by 2, we get $\frac{\pi}{4} < \frac{x}{2} < \frac{\pi}{2}$.
This means $x/2$ lies in Quadrant I. In Q1, $\sin(x/2)$, $\cos(x/2)$, and $\tan(x/2)$ are all positive.

We know the formula $\tan x = \frac{2\tan(x/2)}{1 - \tan^2(x/2)}$.
Let $t = \tan(x/2)$.
$-4/3 = \frac{2t}{1 - t^2}$
$-4(1 - t^2) = 3(2t)$
$-4 + 4t^2 = 6t$
$4t^2 - 6t - 4 = 0$
$2t^2 - 3t - 2 = 0$
Factor the quadratic equation:
$(2t + 1)(t - 2) = 0$
So, $t = -1/2$ or $t = 2$.
Since $x/2$ is in Q1, $\tan(x/2)$ must be positive.
Therefore, $\tan(x/2) = 2$.

Now we find $\sin(x/2)$ and $\cos(x/2)$ using $\tan(x/2) = 2/1$ (Opposite/Adjacent).
Hypotenuse = $\sqrt{2^2 + 1^2} = \sqrt{5}$.
Since $x/2$ is in Q1:
$\sin(x/2) = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{2}{\sqrt{5}} = \frac{2\sqrt{5}}{5}$
$\cos(x/2) = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{1}{\sqrt{5}} = \frac{\sqrt{5}}{5}$

Final Answer:
$\sin(x/2) = \frac{2}{\sqrt{5}}$
$\cos(x/2) = \frac{1}{\sqrt{5}}$
$\tan(x/2) = 2$

Question 9: Find $\sin(x/2)$, $\cos(x/2)$ and $\tan(x/2)$ if $\cos x = -1/3$, $x$ lies in quadrant III.

Answer-
Given $\cos x = -1/3$ and $x$ is in Quadrant III.
This means $\pi < x < \frac{3\pi}{2}$.
Dividing by 2, we get $\frac{\pi}{2} < \frac{x}{2} < \frac{3\pi}{4}$.
This means $x/2$ lies in Quadrant II. In Q2, $\sin(x/2)$ is positive, $\cos(x/2)$ is negative, and $\tan(x/2)$ is negative.

We use the half-angle identities:

1. $\sin^2(x/2) = \frac{1 - \cos x}{2}$
   $\sin^2(x/2) = \frac{1 - (-1/3)}{2} = \frac{1 + 1/3}{2} = \frac{4/3}{2} = \frac{4}{6} = \frac{2}{3}$
   Since $x/2$ is in Q2, $\sin(x/2)$ is positive.
   $\sin(x/2) = \sqrt{\frac{2}{3}} = \frac{\sqrt{2}}{\sqrt{3}} = \frac{\sqrt{6}}{3}$

2. $\cos^2(x/2) = \frac{1 + \cos x}{2}$
   $\cos^2(x/2) = \frac{1 + (-1/3)}{2} = \frac{1 - 1/3}{2} = \frac{2/3}{2} = \frac{1}{3}$
   Since $x/2$ is in Q2, $\cos(x/2)$ is negative.
   $\cos(x/2) = -\sqrt{\frac{1}{3}} = -\frac{1}{\sqrt{3}} = -\frac{\sqrt{3}}{3}$

3. $\tan(x/2) = \frac{\sin(x/2)}{\cos(x/2)}$
   $\tan(x/2) = \frac{\sqrt{2/3}}{-\sqrt{1/3}} = -\sqrt{\frac{2/3}{1/3}} = -\sqrt{2}$

Final Answer:
$\sin(x/2) = \frac{\sqrt{6}}{3}$
$\cos(x/2) = -\frac{\sqrt{3}}{3}$
$\tan(x/2) = -\sqrt{2}$

Question 10: Find $\sin(x/2)$, $\cos(x/2)$ and $\tan(x/2)$ if $\sin x = 1/4$, $x$ lies in quadrant II.

Answer-
Given $\sin x = 1/4$ and $x$ is in Quadrant II.
This means $\frac{\pi}{2} < x < \pi$.
Dividing by 2, we get $\frac{\pi}{4} < \frac{x}{2} < \frac{\pi}{2}$.
This means $x/2$ lies in Quadrant I. In Q1, all trigonometric functions are positive.

First, we need to find $\cos x$. Since $x$ is in Q2, $\cos x$ is negative.
$\cos^2 x = 1 - \sin^2 x = 1 - (1/4)^2 = 1 - 1/16 = 15/16$
$\cos x = -\sqrt{\frac{15}{16}} = -\frac{\sqrt{15}}{4}$

Now we use the half-angle identities:

1. $\sin^2(x/2) = \frac{1 - \cos x}{2}$
   $\sin^2(x/2) = \frac{1 - (-\sqrt{15}/4)}{2} = \frac{1 + \sqrt{15}/4}{2} = \frac{(4 + \sqrt{15})/4}{2} = \frac{4 + \sqrt{15}}{8}$
   Since $x/2$ is in Q1, $\sin(x/2)$ is positive.
   $\sin(x/2) = \sqrt{\frac{4 + \sqrt{15}}{8}} = \frac{\sqrt{4 + \sqrt{15}}}{\sqrt{8}} = \frac{\sqrt{4 + \sqrt{15}}}{2\sqrt{2}}$
   (To simplify $\sqrt{8+2\sqrt{15}} = \sqrt{(\sqrt{5}+\sqrt{3})^2} = \sqrt{5}+\sqrt{3}$)
   $\sin(x/2) = \frac{\sqrt{8 + 2\sqrt{15}}}{4} = \frac{\sqrt{5} + \sqrt{3}}{4}$

2. $\cos^2(x/2) = \frac{1 + \cos x}{2}$
   $\cos^2(x/2) = \frac{1 + (-\sqrt{15}/4)}{2} = \frac{1 - \sqrt{15}/4}{2} = \frac{(4 - \sqrt{15})/4}{2} = \frac{4 - \sqrt{15}}{8}$
   Since $x/2$ is in Q1, $\cos(x/2)$ is positive.
   $\cos(x/2) = \sqrt{\frac{4 - \sqrt{15}}{8}} = \frac{\sqrt{4 - \sqrt{15}}}{\sqrt{8}} = \frac{\sqrt{4 - \sqrt{15}}}{2\sqrt{2}}$
   (To simplify $\sqrt{8-2\sqrt{15}} = \sqrt{(\sqrt{5}-\sqrt{3})^2} = \sqrt{5}-\sqrt{3}$)
   $\cos(x/2) = \frac{\sqrt{8 - 2\sqrt{15}}}{4} = \frac{\sqrt{5} - \sqrt{3}}{4}$

3. $\tan(x/2) = \frac{\sin(x/2)}{\cos(x/2)}$
   $\tan(x/2) = \frac{(\sqrt{5} + \sqrt{3})/4}{(\sqrt{5} - \sqrt{3})/4} = \frac{\sqrt{5} + \sqrt{3}}{\sqrt{5} - \sqrt{3}}$
   Rationalizing: $\frac{\sqrt{5} + \sqrt{3}}{\sqrt{5} - \sqrt{3}} \times \frac{\sqrt{5} + \sqrt{3}}{\sqrt{5} + \sqrt{3}} = \frac{(\sqrt{5} + \sqrt{3})^2}{5 - 3}$
   $= \frac{5 + 3 + 2\sqrt{15}}{2} = \frac{8 + 2\sqrt{15}}{2} = 4 + \sqrt{15}$

Final Answer:
$\sin(x/2) = \frac{\sqrt{5} + \sqrt{3}}{4}$
$\cos(x/2) = \frac{\sqrt{5} - \sqrt{3}}{4}$
$\tan(x/2) = 4 + \sqrt{15}$