Trigonometric Functions

We use the conversion formula: Radian measure = $\dfrac{\pi}{180} \times$ Degree measure.

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

(ii) −47°30′
First convert minutes to degrees: $30' = \dfrac{30}{60}^\circ = \dfrac{1}{2}^\circ$

So $-47^\circ 30' = -47\dfrac{1}{2}^\circ = -\dfrac{95}{2}^\circ$
$$-\frac{95}{2}^\circ = \frac{\pi}{180} \times \left(-\frac{95}{2}\right) = -\frac{95\pi}{360} = -\frac{19\pi}{72} \text{ radian}$$

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

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

We use the conversion formula: Degree measure = $\dfrac{180}{\pi} \times$ Radian measure.

(i) $\dfrac{11}{16}$ radian
$$\frac{11}{16} \times \frac{180}{\pi} = \frac{11}{16} \times \frac{180 \times 7}{22} = \frac{11 \times 180 \times 7}{16 \times 22} = \frac{11 \times 1260}{352} = \frac{13860}{352} = 39\frac{3}{8}^\circ$$
$$= 39^\circ + \frac{3}{8} \times 60' = 39^\circ 22' 30''$$

(ii) $-4$ radian
$$-4 \times \frac{180}{\pi} = -4 \times \frac{180 \times 7}{22} = -\frac{5040}{22} = -229\frac{1}{11}^\circ$$
$$= -\left(229^\circ + \frac{1}{11} \times 60'\right) = -\left(229^\circ 5' \frac{5}{11}''\right) \approx -229^\circ 5' 27''$$

(iii) $\dfrac{5\pi}{3}$ radian
$$\frac{5\pi}{3} \times \frac{180}{\pi} = \frac{5 \times 180}{3} = 300^\circ$$

(iv) $\dfrac{7\pi}{6}$ radian
$$\frac{7\pi}{6} \times \frac{180}{\pi} = \frac{7 \times 180}{6} = 210^\circ$$

Given: The wheel makes 360 revolutions per minute.

Step 1: Find revolutions per second.
$$\text{Revolutions per second} = \frac{360}{60} = 6$$

Step 2: Each complete revolution = $2\pi$ radians.

Step 3: Radians turned in one second:
$$= 6 \times 2\pi = 12\pi \text{ radians}$$

Answer: The wheel turns through $12\pi$ radians in one second.

Given: Radius $r = 100$ cm, Arc length $l = 22$ cm.

Formula: $\theta = \dfrac{l}{r}$

Step 1: Find $\theta$ in radians.
$$\theta = \frac{22}{100} = \frac{11}{50} \text{ radian}$$

Step 2: Convert to degrees.
$$\theta = \frac{11}{50} \times \frac{180}{\pi} = \frac{11}{50} \times \frac{180 \times 7}{22} = \frac{11 \times 1260}{1100} = \frac{13860}{1100} = \frac{63}{5}^\circ = 12\frac{3}{5}^\circ$$
$$= 12^\circ + \frac{3}{5} \times 60' = 12^\circ 36'$$

Answer: The required angle is $12^\circ 36'$.

Given: Diameter = 40 cm, so radius $r = 20$ cm. Length of chord = 20 cm.

Step 1: Since the chord length equals the radius (both = 20 cm), the triangle formed by the two radii and the chord is equilateral.

Step 2: Therefore, the angle subtended at the centre by the chord:
$$\theta = 60^\circ = \frac{\pi}{3} \text{ radian}$$

Step 3: Length of minor arc:
$$l = r\theta = 20 \times \frac{\pi}{3} = \frac{20\pi}{3} \text{ cm}$$

Answer: The length of the minor arc is $\dfrac{20\pi}{3}$ cm.

Given: Let $r_1$ and $r_2$ be the radii of the two circles. The same arc length $l$ subtends angles $\theta_1 = 60°$ and $\theta_2 = 75°$ at the respective centres.

Convert to radians:
$$\theta_1 = 60^\circ = \frac{\pi}{3} \text{ radian}, \quad \theta_2 = 75^\circ = \frac{5\pi}{12} \text{ radian}$$

Using $l = r\theta$:
Since arc lengths are equal:
$$r_1 \theta_1 = r_2 \theta_2$$
$$r_1 \times \frac{\pi}{3} = r_2 \times \frac{5\pi}{12}$$
$$\frac{r_1}{r_2} = \frac{5\pi/12}{\pi/3} = \frac{5\pi}{12} \times \frac{3}{\pi} = \frac{5}{4}$$

Answer: $r_1 : r_2 = 5 : 4$.

Given: Length of pendulum = radius $r = 75$ cm.

Formula: $\theta = \dfrac{l}{r}$

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

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

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

Given: $\cos x = -\dfrac{1}{2}$, $x$ in third quadrant.

Step 1: Find $\sin x$.
$$\sin^2 x = 1 - \cos^2 x = 1 - \frac{1}{4} = \frac{3}{4}$$
$$\sin x = \pm\frac{\sqrt{3}}{2}$$
Since $x$ is in the third quadrant, \sin x < 0, so $\sin x = -\dfrac{\sqrt{3}}{2}$.

Step 2: Find remaining functions.
$$\tan x = \frac{\sin x}{\cos x} = \frac{-\sqrt{3}/2}{-1/2} = \sqrt{3}$$

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

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

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

Given: $\sin x = \dfrac{3}{5}$, $x$ in second quadrant.

Step 1: Find $\cos x$.
$$\cos^2 x = 1 - \sin^2 x = 1 - \frac{9}{25} = \frac{16}{25}$$
$$\cos x = \pm\frac{4}{5}$$
Since $x$ is in the second quadrant, \cos x < 0, so $\cos x = -\dfrac{4}{5}$.

Step 2: Find remaining functions.
$$\tan x = \frac{\sin x}{\cos x} = \frac{3/5}{-4/5} = -\frac{3}{4}$$

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

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

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

Given: $\cot x = \dfrac{3}{4}$, $x$ in third quadrant.

Step 1: Find $\tan x$.
$$\tan x = \frac{1}{\cot x} = \frac{4}{3}$$
(In third quadrant, \tan x > 0, consistent.)

Step 2: Find $\sec x$.
$$\sec^2 x = 1 + \tan^2 x = 1 + \frac{16}{9} = \frac{25}{9}$$
$$\sec x = \pm\frac{5}{3}$$
In third quadrant, \cos x < 0, so \sec x < 0. Thus $\sec x = -\dfrac{5}{3}$.

Step 3: Find $\cos x$.
$$\cos x = \frac{1}{\sec x} = -\frac{3}{5}$$

Step 4: Find $\sin x$.
$$\sin x = \tan x \cdot \cos x = \frac{4}{3} \times \left(-\frac{3}{5}\right) = -\frac{4}{5}$$

Step 5: Find $\csc x$.
$$\csc x = \frac{1}{\sin x} = -\frac{5}{4}$$

Given: $\sec x = \dfrac{13}{5}$, $x$ in fourth quadrant.

Step 1: Find $\cos x$.
$$\cos x = \frac{1}{\sec x} = \frac{5}{13}$$

Step 2: Find $\sin x$.
$$\sin^2 x = 1 - \cos^2 x = 1 - \frac{25}{169} = \frac{144}{169}$$
$$\sin x = \pm\frac{12}{13}$$
In fourth quadrant, \sin x < 0, so $\sin x = -\dfrac{12}{13}$.

Step 3: Find remaining functions.
$$\tan x = \frac{\sin x}{\cos x} = \frac{-12/13}{5/13} = -\frac{12}{5}$$

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

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

Given: $\tan x = -\dfrac{5}{12}$, $x$ in second quadrant.

Step 1: Find $\sec x$.
$$\sec^2 x = 1 + \tan^2 x = 1 + \frac{25}{144} = \frac{169}{144}$$
$$\sec x = \pm\frac{13}{12}$$
In second quadrant, \cos x < 0, so \sec x < 0. Thus $\sec x = -\dfrac{13}{12}$.

Step 2: Find $\cos x$.
$$\cos x = \frac{1}{\sec x} = -\frac{12}{13}$$

Step 3: Find $\sin x$.
$$\sin x = \tan x \cdot \cos x = \left(-\frac{5}{12}\right)\left(-\frac{12}{13}\right) = \frac{5}{13}$$
(Positive, consistent with second quadrant.)

Step 4: Find remaining functions.
$$\csc x = \frac{1}{\sin x} = \frac{13}{5}$$

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

Given: $\sin 765°$

We know that $\sin$ has a period of $360°$.
$$\sin 765° = \sin(765° - 2 \times 360°) = \sin(765° - 720°) = \sin 45°$$
$$= \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$$

Given: $\csc(-1410°)$

Step 1: Use the identity $\csc(-\theta) = -\csc(\theta)$.
$$\csc(-1410°) = -\csc(1410°)$$

Step 2: Reduce $1410°$ using periodicity ($360°$).
$$1410° = 3 \times 360° + 330°$$
$$\csc(1410°) = \csc(330°)$$

Step 3: Evaluate $\csc(330°)$.
$$\sin(330°) = \sin(360° - 30°) = -\sin 30° = -\frac{1}{2}$$
$$\csc(330°) = -2$$

Step 4:
$$\csc(-1410°) = -(-2) = 2$$

Given: $\tan\dfrac{19\pi}{3}$

$\tan$ has a period of $\pi$.
$$\frac{19\pi}{3} = 6\pi + \frac{\pi}{3}$$
$$\tan\frac{19\pi}{3} = \tan\left(6\pi + \frac{\pi}{3}\right) = \tan\frac{\pi}{3} = \sqrt{3}$$

Given: $\sin\left(-\dfrac{11\pi}{3}\right)$

Step 1: Use $\sin(-x) = -\sin x$.
$$\sin\left(-\frac{11\pi}{3}\right) = -\sin\frac{11\pi}{3}$$

Step 2: Reduce using period $2\pi$.
$$\frac{11\pi}{3} = \frac{12\pi - \pi}{3} = 4\pi - \frac{\pi}{3}$$
$$\sin\frac{11\pi}{3} = \sin\left(4\pi - \frac{\pi}{3}\right) = \sin\left(-\frac{\pi}{3}\right) = -\sin\frac{\pi}{3} = -\frac{\sqrt{3}}{2}$$

Step 3:
$$\sin\left(-\frac{11\pi}{3}\right) = -\left(-\frac{\sqrt{3}}{2}\right) = \frac{\sqrt{3}}{2}$$

Given: $\cot\left(-\dfrac{15\pi}{4}\right)$

Step 1: Use $\cot(-x) = -\cot x$.
$$\cot\left(-\frac{15\pi}{4}\right) = -\cot\frac{15\pi}{4}$$

Step 2: Reduce using period $\pi$.
$$\frac{15\pi}{4} = 3\pi + \frac{3\pi}{4}$$
$$\cot\frac{15\pi}{4} = \cot\left(3\pi + \frac{3\pi}{4}\right) = \cot\frac{3\pi}{4}$$

Step 3: Evaluate $\cot\dfrac{3\pi}{4}$.
$$\cot\frac{3\pi}{4} = \cot\left(\pi - \frac{\pi}{4}\right) = -\cot\frac{\pi}{4} = -1$$

Step 4:
$$\cot\left(-\frac{15\pi}{4}\right) = -(-1) = 1$$

Known values: $\sin\dfrac{\pi}{6} = \dfrac{1}{2}$, $\cos\dfrac{\pi}{3} = \dfrac{1}{2}$, $\tan\dfrac{\pi}{4} = 1$.

L.H.S.
$$= \left(\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right)^2 - (1)^2$$
$$= \frac{1}{4} + \frac{1}{4} - 1 = \frac{1}{2} - 1 = -\frac{1}{2} = \text{R.H.S.}$$

Hence proved.

Known values:
$\sin\dfrac{\pi}{6} = \dfrac{1}{2}$, $\cos\dfrac{\pi}{3} = \dfrac{1}{2}$.

$\csc\dfrac{7\pi}{6} = \csc\left(\pi + \dfrac{\pi}{6}\right) = -\csc\dfrac{\pi}{6} = -2$, so $\csc^2\dfrac{7\pi}{6} = 4$.

L.H.S.
$$= 2\left(\frac{1}{2}\right)^2 + 4 \times \left(\frac{1}{2}\right)^2$$
$$= 2 \times \frac{1}{4} + 4 \times \frac{1}{4} = \frac{1}{2} + 1 = \frac{3}{2} = \text{R.H.S.}$$

Hence proved.

Known values:
$\cot\dfrac{\pi}{6} = \sqrt{3}$, $\tan\dfrac{\pi}{6} = \dfrac{1}{\sqrt{3}}$.

$\csc\dfrac{5\pi}{6} = \csc\left(\pi - \dfrac{\pi}{6}\right) = \csc\dfrac{\pi}{6} = 2$.

L.H.S.
$$= (\sqrt{3})^2 + 2 + 3\left(\frac{1}{\sqrt{3}}\right)^2$$
$$= 3 + 2 + 3 \times \frac{1}{3} = 3 + 2 + 1 = 6 = \text{R.H.S.}$$

Hence proved.

Known values:
$\sin\dfrac{3\pi}{4} = \sin\left(\pi - \dfrac{\pi}{4}\right) = \sin\dfrac{\pi}{4} = \dfrac{1}{\sqrt{2}}$.

$\cos\dfrac{\pi}{4} = \dfrac{1}{\sqrt{2}}$, $\sec\dfrac{\pi}{3} = 2$.

L.H.S.
$$= 2\left(\frac{1}{\sqrt{2}}\right)^2 + 2\left(\frac{1}{\sqrt{2}}\right)^2 + 2(2)^2$$
$$= 2 \times \frac{1}{2} + 2 \times \frac{1}{2} + 2 \times 4 = 1 + 1 + 8 = 10 = \text{R.H.S.}$$

Hence proved.

(i) $\sin 75°$

Write $75° = 45° + 30°$.
$$\sin 75° = \sin(45° + 30°) = \sin 45°\cos 30° + \cos 45°\sin 30°$$
$$= \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} \cdot \frac{1}{2} = \frac{\sqrt{3}}{2\sqrt{2}} + \frac{1}{2\sqrt{2}} = \frac{\sqrt{3}+1}{2\sqrt{2}} = \frac{\sqrt{6}+\sqrt{2}}{4}$$

(ii) $\tan 15°$

Write $15° = 45° - 30°$.
$$\tan 15° = \tan(45° - 30°) = \frac{\tan 45° - \tan 30°}{1 + \tan 45°\tan 30°}$$
$$= \frac{1 - \dfrac{1}{\sqrt{3}}}{1 + 1 \cdot \dfrac{1}{\sqrt{3}}} = \frac{\dfrac{\sqrt{3}-1}{\sqrt{3}}}{\dfrac{\sqrt{3}+1}{\sqrt{3}}} = \frac{\sqrt{3}-1}{\sqrt{3}+1}$$

Rationalising:
$$= \frac{(\sqrt{3}-1)^2}{(\sqrt{3}+1)(\sqrt{3}-1)} = \frac{3 - 2\sqrt{3} + 1}{3 - 1} = \frac{4 - 2\sqrt{3}}{2} = 2 - \sqrt{3}$$

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

Let $A = \dfrac{\pi}{4} - x$ and $B = \dfrac{\pi}{4} - y$.

L.H.S.
$$= \cos\left[\left(\frac{\pi}{4}-x\right)+\left(\frac{\pi}{4}-y\right)\right]$$
$$= \cos\left[\frac{\pi}{2} - (x+y)\right]$$
$$= \sin(x+y) = \text{R.H.S.}$$

Hence proved.

Using the addition formula for tan:
$$\tan\left(\frac{\pi}{4}+x\right) = \frac{1+\tan x}{1-\tan x}, \quad \tan\left(\frac{\pi}{4}-x\right) = \frac{1-\tan x}{1+\tan x}$$

L.H.S.
$$= \frac{\dfrac{1+\tan x}{1-\tan x}}{\dfrac{1-\tan x}{1+\tan x}} = \frac{1+\tan x}{1-\tan x} \times \frac{1+\tan x}{1-\tan x} = \left(\frac{1+\tan x}{1-\tan x}\right)^2 = \text{R.H.S.}$$

Hence proved.

Using standard identities:
- $\cos(\pi+x) = -\cos x$
- $\cos(-x) = \cos x$
- $\sin(\pi-x) = \sin x$
- $\cos\left(\dfrac{\pi}{2}+x\right) = -\sin x$

L.H.S.
$$= \frac{(-\cos x)(\cos x)}{(\sin x)(-\sin x)} = \frac{-\cos^2 x}{-\sin^2 x} = \frac{\cos^2 x}{\sin^2 x} = \cot^2 x = \text{R.H.S.}$$

Hence proved.

Using standard identities:
- $\cos\left(\dfrac{3\pi}{2}+x\right) = \sin x$
- $\cos(2\pi+x) = \cos x$
- $\cot\left(\dfrac{3\pi}{2}-x\right) = \tan x$
- $\cot(2\pi+x) = \cot x$

L.H.S.
$$= \sin x \cdot \cos x \cdot [\tan x + \cot x]$$
$$= \sin x \cos x \left[\frac{\sin x}{\cos x} + \frac{\cos x}{\sin x}\right]$$
$$= \sin x \cos x \cdot \frac{\sin^2 x + \cos^2 x}{\sin x \cos x}$$
$$= \sin^2 x + \cos^2 x = 1 = \text{R.H.S.}$$

Hence proved.

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

Let $A = (n+2)x$ and $B = (n+1)x$.

L.H.S.
$$= \cos[(n+2)x - (n+1)x]$$
$$= \cos[x] = \cos x = \text{R.H.S.}$$

Hence proved.

Using sum-to-product formula: $\cos A - \cos B = -2\sin\dfrac{A+B}{2}\sin\dfrac{A-B}{2}$

Let $A = \dfrac{3\pi}{4}+x$, $B = \dfrac{3\pi}{4}-x$.

$$\frac{A+B}{2} = \frac{3\pi}{4}, \quad \frac{A-B}{2} = x$$

L.H.S.
$$= -2\sin\frac{3\pi}{4}\sin x = -2 \times \frac{1}{\sqrt{2}} \times \sin x = -\sqrt{2}\sin x = \text{R.H.S.}$$

Hence proved.

Using the identity: $\sin^2 A - \sin^2 B = \sin(A+B)\sin(A-B)$

L.H.S.
$$= \sin^2 6x - \sin^2 4x = \sin(6x+4x)\sin(6x-4x)$$
$$= \sin 10x \cdot \sin 2x = \sin 2x \sin 10x = \text{R.H.S.}$$

Hence proved.

Using the identity: $\cos^2 A - \cos^2 B = -(\cos^2 B - \cos^2 A) = \sin(A+B)\sin(B-A)$

Alternatively, use $\cos^2 A - \cos^2 B = (\cos A + \cos B)(\cos A - \cos B)$.

Using sum-to-product:
$$\cos A + \cos B = 2\cos\frac{A+B}{2}\cos\frac{A-B}{2}$$
$$\cos A - \cos B = -2\sin\frac{A+B}{2}\sin\frac{A-B}{2}$$

With $A = 2x$, $B = 6x$:
$$\cos 2x + \cos 6x = 2\cos 4x \cos 2x$$
$$\cos 2x - \cos 6x = -2\sin 4x \sin(-2x) = 2\sin 4x \sin 2x$$

L.H.S.
$$= (\cos 2x + \cos 6x)(\cos 2x - \cos 6x)$$
$$= (2\cos 4x \cos 2x)(2\sin 4x \sin 2x)$$
$$= 4\sin 4x \cos 4x \cdot \sin 2x \cos 2x$$
$$= \sin 8x \cdot \sin 4x = \sin 4x \sin 8x = \text{R.H.S.}$$

Hence proved.

L.H.S. $= \sin 2x + 2\sin 4x + \sin 6x$

Step 1: Group $\sin 6x + \sin 2x$ using sum-to-product:
$$\sin 6x + \sin 2x = 2\sin\frac{6x+2x}{2}\cos\frac{6x-2x}{2} = 2\sin 4x \cos 2x$$

Step 2:
$$\text{L.H.S.} = 2\sin 4x \cos 2x + 2\sin 4x = 2\sin 4x(\cos 2x + 1)$$

Step 3: Use $1 + \cos 2x = 2\cos^2 x$:
$$= 2\sin 4x \cdot 2\cos^2 x = 4\cos^2 x \sin 4x = \text{R.H.S.}$$

Hence proved.

L.H.S. $= \cot 4x(\sin 5x + \sin 3x)$

Using sum-to-product: $\sin 5x + \sin 3x = 2\sin 4x \cos x$

$$\text{L.H.S.} = \frac{\cos 4x}{\sin 4x} \cdot 2\sin 4x \cos x = 2\cos 4x \cos x$$

R.H.S. $= \cot x(\sin 5x - \sin 3x)$

Using sum-to-product: $\sin 5x - \sin 3x = 2\cos 4x \sin x$

$$\text{R.H.S.} = \frac{\cos x}{\sin x} \cdot 2\cos 4x \sin x = 2\cos 4x \cos x$$

L.H.S. = R.H.S. Hence proved.

Numerator: Using $\cos A - \cos B = -2\sin\dfrac{A+B}{2}\sin\dfrac{A-B}{2}$:
$$\cos 9x - \cos 5x = -2\sin 7x \sin 2x$$

Denominator: Using $\sin A - \sin B = 2\cos\dfrac{A+B}{2}\sin\dfrac{A-B}{2}$:
$$\sin 17x - \sin 3x = 2\cos 10x \sin 7x$$

L.H.S.
$$= \frac{-2\sin 7x \sin 2x}{2\cos 10x \sin 7x} = \frac{-\sin 2x}{\cos 10x} = \text{R.H.S.}$$

Hence proved.

Numerator: $\sin 5x + \sin 3x = 2\sin 4x \cos x$

Denominator: $\cos 5x + \cos 3x = 2\cos 4x \cos x$

L.H.S.
$$= \frac{2\sin 4x \cos x}{2\cos 4x \cos x} = \frac{\sin 4x}{\cos 4x} = \tan 4x = \text{R.H.S.}$$

Hence proved.

Numerator: $\sin x - \sin y = 2\cos\dfrac{x+y}{2}\sin\dfrac{x-y}{2}$

Denominator: $\cos x + \cos y = 2\cos\dfrac{x+y}{2}\cos\dfrac{x-y}{2}$

L.H.S.
$$= \frac{2\cos\dfrac{x+y}{2}\sin\dfrac{x-y}{2}}{2\cos\dfrac{x+y}{2}\cos\dfrac{x-y}{2}} = \frac{\sin\dfrac{x-y}{2}}{\cos\dfrac{x-y}{2}} = \tan\frac{x-y}{2} = \text{R.H.S.}$$

Hence proved.

Numerator: $\sin x + \sin 3x = 2\sin 2x \cos x$

Denominator: $\cos x + \cos 3x = 2\cos 2x \cos x$

L.H.S.
$$= \frac{2\sin 2x \cos x}{2\cos 2x \cos x} = \frac{\sin 2x}{\cos 2x} = \tan 2x = \text{R.H.S.}$$

Hence proved.

Numerator: $\sin x - \sin 3x = 2\cos\dfrac{x+3x}{2}\sin\dfrac{x-3x}{2} = 2\cos 2x \sin(-x) = -2\cos 2x \sin x$

Denominator: $\sin^2 x - \cos^2 x = -(\cos^2 x - \sin^2 x) = -\cos 2x$

L.H.S.
$$= \frac{-2\cos 2x \sin x}{-\cos 2x} = 2\sin x = \text{R.H.S.}$$

Hence proved.

Numerator:
$$\cos 4x + \cos 2x + \cos 3x = 2\cos 3x \cos x + \cos 3x = \cos 3x(2\cos x + 1)$$

(Using $\cos 4x + \cos 2x = 2\cos 3x \cos x$)

Denominator:
$$\sin 4x + \sin 2x + \sin 3x = 2\sin 3x \cos x + \sin 3x = \sin 3x(2\cos x + 1)$$

(Using $\sin 4x + \sin 2x = 2\sin 3x \cos x$)

L.H.S.
$$= \frac{\cos 3x(2\cos x + 1)}{\sin 3x(2\cos x + 1)} = \frac{\cos 3x}{\sin 3x} = \cot 3x = \text{R.H.S.}$$

Hence proved.

Strategy: Use $\cot 3x = \cot(2x + x)$.

$$\cot(2x+x) = \frac{\cot 2x \cot x - 1}{\cot x + \cot 2x}$$

$$\cot 3x (\cot x + \cot 2x) = \cot 2x \cot x - 1$$

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

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

$$= \cot x \cot 2x - \cot 2x \cot 3x - \cot 3x \cot x = \text{R.H.S.}$$

Hence proved.