Chapter 3: Trigonometric Functions

Start with the key conversion fact: $180^{\circ}=\pi$ radians, which immediately gives us $1^{\circ}=\frac{\pi}{180}$ radians.

(i) Multiply $25°$ by the conversion factor:

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

∴ Radian measure of $25^{\circ}$ is $\frac{5\pi}{36}$.

(ii) First convert the minutes: $30^{\prime}=\left(\frac{30}{60}\right)^{\circ}=\frac{1}{2}^{\circ}$, so the angle becomes $-47\frac{1}{2}^{\circ}$.

$-47^{\circ} 30^{\prime}=-\frac{95}{2} \times \frac{\pi}{180}$ radians $=-\frac{19\pi}{72}$ radians

∴ Radian measure of $-47^{\circ} 30^{\prime}$ is $-\frac{19\pi}{72}$.

(iii) Apply the same conversion:

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

∴ Radian measure of $240^{\circ}$ is $\frac{4\pi}{3}$.

(iv) Apply the same conversion:

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

∴ Radian measure of $520^{\circ}$ is $\frac{26\pi}{9}$.

The reverse conversion rule is: $\pi$ radians $=180^{\circ}$, so $1 \text{ radian }=\frac{180^{\circ}}{\pi}$.

(i) Multiply by the conversion factor and substitute $\pi = \frac{22}{7}$:

$\frac{11}{16} \text{ radians} = \frac{11}{16} \times \frac{180^{\circ}}{\pi} = \left(\frac{11 \times 45}{4} \times \frac{7}{22}\right)^{\circ} = \frac{315^{\circ}}{8}$

Now perform the long division: $\frac{315}{8} = 39^{\circ}$ remainder $3 \Rightarrow \frac{3 \times 60}{8} = 22^{\prime}$ remainder $4 \Rightarrow \frac{4 \times 60}{8} = 30^{\prime\prime}$

Ans. $39^{\circ} 22^{\prime} 30^{\prime\prime}$

(ii) Substitute and simplify:

$-4 \text{ radians} = -4 \times \frac{180^{\circ}}{\pi} = \left(-720 \times \frac{7}{22}\right)^{\circ} = -\frac{2520^{\circ}}{11}$

Performing long division: $\frac{2520}{11} = 229^{\circ}$ remainder $1 \Rightarrow \frac{1 \times 60}{11} = 5^{\prime}$ remainder $5 \Rightarrow \frac{5 \times 60}{11} \approx 27^{\prime\prime}$

Ans. $-229^{\circ} 5^{\prime} 27^{\prime\prime}$ approximately.

(iii) Since $\pi$ cancels directly:

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

(iv) Similarly, $\pi$ cancels directly:

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

Begin by scaling down the rate from minutes to seconds:

Number of revolutions in one minute $= 360$

$\Rightarrow$ Number of revolutions in one second $= \frac{360}{60} = 6$

Now use the fact that one full revolution corresponds to a complete circle:

Since 1 revolution $= 360^{\circ} = 2\pi$ radians,

$\therefore 6$ revolutions $= 6 \times 2\pi = 12\pi$ radians

$\Rightarrow$ Number of radians turned in one second $= 12\pi$.

Identify the given quantities first:

$l =$ length of arc $= 22$ cm, $r =$ radius of circle $= 100$ cm

Let $\theta$ be the central angle. Apply the arc-length formula:

$\theta = \frac{l}{r} \text{ radians} = \frac{22}{100} \text{ radians} = \frac{11}{50} \times \frac{180^{\circ}}{\pi}$

$= \left(\frac{11 \times 18}{5} \times \frac{7}{22}\right)^{\circ} = \frac{63^{\circ}}{5}$

Breaking this into degrees and minutes: $\frac{63}{5} = 12^{\circ}$ remainder $3 \Rightarrow \frac{3 \times 60}{5} = 36^{\prime}$

Ans. $12^{\circ} 36^{\prime}$

Extract the radius from the diameter:

Radius of circle $= \frac{1}{2} \times 40$ cm $= 20$ cm

Chord $AB = 20$ cm

∴ In triangle OAB, all three sides are equal: $OA = OB = AB = 20$ cm

$\Rightarrow$ Triangle OAB is equilateral $\Rightarrow$ Each angle $= 60^{\circ}$

Let arc $AB = l$ cm. Convert the central angle to radians:

$\theta = \angle AOB = 60^{\circ} = 60 \times \frac{\pi}{180} = \frac{\pi}{3} \text{ radians}, \quad r = 20 \text{ cm}$

Apply the arc-length formula:

$\therefore \quad l = r\theta = 20 \times \frac{\pi}{3} \text{ cm} = \frac{20\pi}{3} \text{ cm}$

Let the radii of the two circles be $r_1$ and $r_2$ respectively, and let the common arc length be $l$.

For the first circle:

$\theta = 60^{\circ} = 60 \times \frac{\pi}{180} = \frac{\pi}{3} \text{ radians}$

$\therefore \quad r_1 = \frac{l}{\pi/3} = \frac{3l}{\pi} \quad \cdots (i)$

For the second circle:

$\theta = 75^{\circ} = 75 \times \frac{\pi}{180} = \frac{5\pi}{12} \text{ radians}$

$\therefore \quad r_2 = \frac{l}{5\pi/12} = \frac{12l}{5\pi} \quad \cdots (ii)$

Divide equation $(i)$ by equation $(ii)$:

$\frac{r_1}{r_2} = \frac{3l}{\pi} \times \frac{5\pi}{12l} = \frac{5}{4}$

$\therefore \quad r_1 : r_2 = 5 : 4$

Think of the pendulum as the radius of a circle — the string length is the radius and the arc traced by the tip is the arc length.

Here, $r = 75$ cm.

(i) $l = 10$ cm

$\therefore \quad \theta = \frac{l}{r} = \frac{10}{75} = \frac{2}{15} \text{ radians}$

(ii) $l = 15$ cm

$\therefore \quad \theta = \frac{l}{r} = \frac{15}{75} = \frac{1}{5} \text{ radians}$

(iii) $l = 21$ cm

$\therefore \quad \theta = \frac{l}{r} = \frac{21}{75} = \frac{7}{25} \text{ radians}$

Start with the reciprocal relationship:

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

Use the Pythagorean identity to find $\sin x$:

$\sin^2 x = 1 – \cos^2 x = 1 – \frac{1}{4} = \frac{3}{4}$

$\Rightarrow \quad \sin x = \pm\frac{\sqrt{3}}{2}$

Since $x$ is in the third quadrant, $\sin x$ is negative there:

$\therefore \quad \sin x = -\frac{\sqrt{3}}{2}$

$\operatorname{cosec} x = \frac{1}{\sin x} = -\frac{2}{\sqrt{3}}$

Use the ratio definition to find $\tan x$ and its reciprocal:

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

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

Take the reciprocal first:

$\therefore \quad \operatorname{cosec} x = \frac{1}{\sin x} = \frac{5}{3}$

Apply the Pythagorean identity for $\cos x$:

$\cos^2 x = 1 – \sin^2 x = 1 – \frac{9}{25} = \frac{16}{25}$

$\Rightarrow \quad \cos x = \pm\frac{4}{5}$

Since $x$ is in the second quadrant, $\cos x$ is negative there:

$\therefore \quad \cos x = -\frac{4}{5}$

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

Complete the remaining two ratios:

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

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

The reciprocal of $\cot x$ gives $\tan x$ immediately:

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

Apply the cosec-cot identity:

$\operatorname{cosec}^2 x = 1 + \cot^2 x = 1 + \frac{9}{16} = \frac{25}{16}$

$\Rightarrow \quad \operatorname{cosec} x = \pm\frac{5}{4}$

Since $x$ is in the third quadrant, $\operatorname{cosec} x$ is negative there:

$\therefore \quad \operatorname{cosec} x = -\frac{5}{4}$

$\sin x = \frac{1}{\operatorname{cosec} x} = -\frac{4}{5}$

Express $\cos x$ using the product $\cot x \cdot \sin x$:

$\cos x = \cot x \cdot \sin x = \frac{3}{4}\left(-\frac{4}{5}\right) = -\frac{3}{5}$

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

Flip $\sec x$ to get $\cos x$:

$\therefore \quad \cos x = \frac{1}{\sec x} = \frac{5}{13}$

Use the Pythagorean identity:

$\sin^2 x = 1 – \cos^2 x = 1 – \frac{25}{169} = \frac{144}{169}$

$\Rightarrow \quad \sin x = \pm\frac{12}{13}$

Since $x$ is in the fourth quadrant, $\sin x$ is negative there:

$\therefore \quad \sin x = -\frac{12}{13}$

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

Complete the remaining two ratios:

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

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

Take the reciprocal to find $\cot x$:

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

Use the sec-tan identity:

$\sec^2 x = 1 + \tan^2 x = 1 + \frac{25}{144} = \frac{169}{144}$

$\Rightarrow \quad \sec x = \pm\frac{13}{12}$

Since $x$ is in the second quadrant, $\sec x$ is negative there:

$\therefore \quad \sec x = -\frac{13}{12}$

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

Express $\sin x$ using the $\tan x \cdot \cos x$ product:

$\sin x = \tan x \cdot \cos x = \left(-\frac{5}{12}\right)\left(-\frac{12}{13}\right) = \frac{5}{13}$

$\operatorname{cosec} x = \frac{1}{\sin x} = \frac{13}{5}$

Reduce the angle by stripping out full rotations ($360°$ each):

$\sin 765^{\circ} = \sin\left(2 \times 360^{\circ} + 45^{\circ}\right) = \sin 45^{\circ} = \frac{1}{\sqrt{2}}$

$\left[\because \sin\left(n \times 360^{\circ} + x\right) = \sin x,\ n \in \mathbb{Z}\right]$

Add a suitable multiple of $360°$ to bring the angle into a recognisable range:

$\operatorname{cosec}\left(-1410^{\circ}\right) = \operatorname{cosec}\left(4 \times 360^{\circ} – 1410^{\circ}\right) = \operatorname{cosec}\left(1440^{\circ} – 1410^{\circ}\right) = \operatorname{cosec} 30^{\circ} = 2$

$\left[\because \operatorname{cosec} x = \operatorname{cosec}\left(n \times 360^{\circ} + x\right),\ n \in \mathbb{Z}\right]$

Separate out complete multiples of $\pi$ using algebra:

$\tan\frac{19\pi}{3} = \tan\left(\frac{18\pi + \pi}{3}\right) = \tan\left(6\pi + \frac{\pi}{3}\right) = \tan\frac{\pi}{3} = \sqrt{3}$

$\left[\because \tan(n\pi + x) = \tan x,\ n \in \mathbb{Z}\right]$

Add a convenient multiple of $2\pi$ to make the angle positive:

$\sin\left(-\frac{11\pi}{3}\right) = \sin\left(4\pi – \frac{11\pi}{3}\right) \quad \left[\because \sin x = \sin(2n\pi + x),\ n \in \mathbb{Z}\right]$

$= \sin\frac{\pi}{3} = \frac{\sqrt{3}}{2}$

Shift by an appropriate multiple of $\pi$ to land in the standard range:

$\cot\left(-\frac{15\pi}{4}\right) = \cot\left(4\pi – \frac{15\pi}{4}\right) \quad \left[\because \cot x = \cot(n\pi + x),\ n \in \mathbb{Z}\right]$

$= \cot\frac{\pi}{4} = 1$

Plug in the standard values $\sin\frac{\pi}{6} = \frac{1}{2}$, $\cos\frac{\pi}{3} = \frac{1}{2}$, $\tan\frac{\pi}{4} = 1$:

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

First find $\operatorname{cosec}\frac{7\pi}{6}$ by expressing the angle as $\pi + \frac{\pi}{6}$:

$\operatorname{cosec}\frac{7\pi}{6} = \operatorname{cosec}\left(\pi + \frac{\pi}{6}\right) = -\operatorname{cosec}\frac{\pi}{6} = -2$

Now substitute all standard values ($\sin\frac{\pi}{6} = \frac{1}{2}$, $\cos\frac{\pi}{3} = \frac{1}{2}$):

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

First evaluate $\operatorname{cosec}\frac{5\pi}{6}$ using the supplementary angle identity:

$\operatorname{cosec}\frac{5\pi}{6} = \operatorname{cosec}\left(\pi – \frac{\pi}{6}\right) = \operatorname{cosec}\frac{\pi}{6} = 2$

Substitute all known values ($\cot\frac{\pi}{6} = \sqrt{3}$, $\tan\frac{\pi}{6} = \frac{1}{\sqrt{3}}$):

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

Simplify $\sin\frac{3\pi}{4}$ first by writing it as a supplementary angle:

$\sin\frac{3\pi}{4} = \sin\left(\pi – \frac{\pi}{4}\right) = \sin\frac{\pi}{4} = \frac{1}{\sqrt{2}}$

Now substitute all values ($\cos\frac{\pi}{4} = \frac{1}{\sqrt{2}}$, $\sec\frac{\pi}{3} = 2$):

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

(i) Split $75°$ as a sum of two standard angles, then apply the sine addition formula:

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

$\left[\because \sin(x + y) = \sin x\cos y + \cos x\sin y\right]$

$= \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} \cdot \frac{1}{2} = \frac{\sqrt{3}+1}{2\sqrt{2}}$

(ii) Express $15°$ as a difference of two standard angles and apply the tangent subtraction formula:

$\tan 15^{\circ} = \tan(45^{\circ} – 30^{\circ}) = \frac{\tan 45^{\circ} – \tan 30^{\circ}}{1 + \tan 45^{\circ}\tan 30^{\circ}}$

$\left[\because \tan(x – y) = \frac{\tan x – \tan y}{1 + \tan x\tan y}\right]$

$= \frac{1 – \frac{1}{\sqrt{3}}}{1 + \frac{1}{\sqrt{3}}} = \frac{\sqrt{3}-1}{\sqrt{3}+1}$

Rationalise by multiplying numerator and denominator by $(\sqrt{3}-1)$:

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

Let $\frac{\pi}{4} – x = A$ and $\frac{\pi}{4} – y = B$:

$\text{L.H.S.} = \cos A\cos B – \sin A\sin B = \cos(A + B)$

$= \cos\left(\frac{\pi}{4} – x + \frac{\pi}{4} – y\right) = \cos\left(\frac{\pi}{2} – (x+y)\right)$

$= \sin(x+y) = \text{R.H.S.} \quad \left[\because \cos\left(\frac{\pi}{2} – \theta\right) = \sin\theta\right]$

Expand each tangent using the addition and subtraction formulas, recalling $\tan\frac{\pi}{4} = 1$:

Reduce each factor using standard allied angle results:

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

Reduce each factor one by one using allied angle results:

$\text{L.H.S.} = \sin x \cdot \cos x \cdot [\tan x + \cot x]$

$= \sin x\cos x\left[\frac{\sin x}{\cos x} + \frac{\cos x}{\sin x}\right]$

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

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

$\text{L.H.S.} = \sin A\sin B + \cos A\cos B = \cos(A – B)$

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

Expand each cosine using the compound angle formulas, then combine like terms:

$\text{L.H.S.} = \cos\frac{3\pi}{4}\cos x – \sin\frac{3\pi}{4}\sin x – \left(\cos\frac{3\pi}{4}\cos x + \sin\frac{3\pi}{4}\sin x\right)$

$= -2\sin\frac{3\pi}{4}\sin x = -2\sin\left(\pi – \frac{\pi}{4}\right)\sin x = -2\sin\frac{\pi}{4}\sin x = -2\cdot\frac{1}{\sqrt{2}}\sin x = -\sqrt{2}\sin x = \text{R.H.S.}$

Directly apply the factorisation identity $\sin^2 A – \sin^2 B = \sin(A+B)\sin(A-B)$:

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

Convert to sines first using $\cos^2\theta = 1 – \sin^2\theta$:

$\text{L.H.S.} = (1 – \sin^2 2x) – (1 – \sin^2 6x) = \sin^2 6x – \sin^2 2x$

Now apply the factorisation identity $\sin^2 A – \sin^2 B = \sin(A+B)\sin(A-B)$:

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

Group the outer terms together and apply the sum-to-product formula:

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

$\left[\because \sin C + \sin D = 2\sin\frac{C+D}{2}\cos\frac{C-D}{2}\right]$

Factor out $2\sin 4x$, then use $\cos 2x = 2\cos^2 x – 1$:

$= 2\sin 4x(\cos 2x + 1) = 2\sin 4x(2\cos^2 x – 1 + 1) = 4\cos^2 x\sin 4x = \text{R.H.S.}$

L.H.S.:

$\cot 4x(\sin 5x + \sin 3x) = \frac{\cos 4x}{\sin 4x} \cdot 2\sin 4x\cos x = 2\cos 4x\cos x \quad \cdots (i)$

$\left[\because \sin C + \sin D = 2\sin\frac{C+D}{2}\cos\frac{C-D}{2}\right]$

R.H.S.:

$\cot x(\sin 5x – \sin 3x) = \frac{\cos x}{\sin x} \cdot 2\cos 4x\sin x = 2\cos x\cos 4x \quad \cdots (ii)$

$\left[\because \sin C – \sin D = 2\cos\frac{C+D}{2}\sin\frac{C-D}{2}\right]$

From $(i)$ and $(ii)$, L.H.S. = R.H.S.

Apply the C–D formulas to both numerator and denominator:

Apply the sum-to-product formulas to both numerator and denominator:

Convert the numerator and denominator using C–D formulas, then cancel the common factor:

Apply the sum-to-product formulas on both top and bottom:

Flip the sign in both numerator and denominator to make the algebra cleaner:

$\text{L.H.S.} = \frac{\sin 3x – \sin x}{\cos^2 x – \sin^2 x}$

Apply the C–D formula on the numerator and recall $\cos^2 x – \sin^2 x = \cos 2x$:

Group the first and third terms in both numerator and denominator:

Factor $\cos 3x$ from the numerator and $\sin 3x$ from the denominator:

Write $3x = 2x + x$ and apply the cotangent addition formula:

$\cot 3x = \frac{\cot 2x\cot x – 1}{\cot x + \cot 2x}$

Cross-multiply:

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

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

Rearrange to get the required expression:

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

Apply the double angle formula $\tan 2A = \frac{2\tan A}{1 – \tan^2 A}$ twice — first for $\tan 2x$, then for $\tan 4x = \tan(2 \cdot 2x)$:

Apply the double angle formula $\cos 2A = 1 – 2\sin^2 A$ with $A = 2x$:

$\cos 4x = 1 – 2\sin^2 2x = 1 – 2(2\sin x\cos x)^2 = 1 – 8\sin^2 x\cos^2 x$

Apply $\cos 2A = 2\cos^2 A – 1$ with $A = 3x$, then expand $\cos 3x = 4\cos^3 x – 3\cos x$:

$\cos 6x = 2\cos^2 3x – 1 = 2(4\cos^3 x – 3\cos x)^2 – 1$

$= 2(16\cos^6 x – 24\cos^4 x + 9\cos^2 x) – 1$

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

Convert the product using $2\cos A\cos B = \cos(A+B) + \cos(A-B)$:

$\text{L.H.S.} = \cos\frac{10\pi}{13} + \cos\frac{8\pi}{13} + \cos\frac{3\pi}{13} + \cos\frac{5\pi}{13}$

Rewrite using the supplementary identity $\cos(\pi – \theta) = -\cos\theta$:

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

$= -\cos\frac{3\pi}{13} – \cos\frac{5\pi}{13} + \cos\frac{3\pi}{13} + \cos\frac{5\pi}{13} = 0 = \text{R.H.S.}$

Apply sum-to-product and difference-to-product formulas on the bracketed expressions:

$\text{L.H.S.} = 2\sin 2x\cos x\cdot\sin x – 2\sin 2x\sin x\cdot\cos x = 0 = \text{R.H.S.}$

Apply C–D formulas to write both brackets in product form:

$\text{L.H.S.} = \left(2\cos\frac{x+y}{2}\cos\frac{x-y}{2}\right)^2 + \left(2\cos\frac{x+y}{2}\sin\frac{x-y}{2}\right)^2$

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

Apply C–D formulas to convert each bracket:

$\text{L.H.S.} = \left(-2\sin\frac{x+y}{2}\sin\frac{x-y}{2}\right)^2 + \left(2\cos\frac{x+y}{2}\sin\frac{x-y}{2}\right)^2$

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

Pair up the terms symmetrically:

$\text{L.H.S.} = (\sin 7x + \sin x) + (\sin 5x + \sin 3x) = 2\sin 4x\cos 3x + 2\sin 4x\cos x$

Factor out $2\sin 4x$:

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

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

Apply C–D formulas separately to each pair in numerator and denominator:

$\text{L.H.S.} = \frac{2\sin 6x\cos x + 2\sin 6x\cos 3x}{2\cos 6x\cos x + 2\cos 6x\cos 3x}$

Factor and cancel the common bracket:

Group $\sin 3x$ with $-\sin x$ first, apply the C–D formula, then factor:

$\text{L.H.S.} = (\sin 3x – \sin x) + \sin 2x = 2\cos 2x\sin x + 2\sin x\cos x = 2\sin x(\cos 2x + \cos x)$

Apply the C–D formula on $(\cos 2x + \cos x)$:

$= 2\sin x \cdot 2\cos\frac{3x}{2}\cos\frac{x}{2} = 4\sin x\cos\frac{x}{2}\cos\frac{3x}{2} = \text{R.H.S.}$

Since $x$ is in quadrant II: $\frac{\pi}{2} < x < \pi$, so $\frac{\pi}{4} < \frac{x}{2} < \frac{\pi}{2}$ — all ratios of $\frac{x}{2}$ are positive.

Find $\cos x$ using the sec-tan identity:

$\sec^2 x = 1 + \frac{16}{9} = \frac{25}{9} \Rightarrow \sec x = -\frac{5}{3} \Rightarrow \cos x = -\frac{3}{5}$

Apply the half-angle formulas:

$\sin\frac{x}{2} = \sqrt{\frac{1-\cos x}{2}} = \sqrt{\frac{1+\frac{3}{5}}{2}} = \frac{2}{\sqrt{5}}$

$\cos\frac{x}{2} = \sqrt{\frac{1+\cos x}{2}} = \sqrt{\frac{1-\frac{3}{5}}{2}} = \frac{1}{\sqrt{5}}$

$\therefore \quad \tan\frac{x}{2} = \frac{2/\sqrt{5}}{1/\sqrt{5}} = 2$

Since $x$ is in quadrant III: $\pi < x < \frac{3\pi}{2}$, so $\frac{\pi}{2} < \frac{x}{2} < \frac{3\pi}{4}$ — $\sin\frac{x}{2} > 0$, $\cos\frac{x}{2} < 0$, $\tan\frac{x}{2} < 0$.

$\sin\frac{x}{2} = \sqrt{\frac{1-\cos x}{2}} = \sqrt{\frac{1+\frac{1}{3}}{2}} = \sqrt{\frac{2}{3}}$

$\cos\frac{x}{2} = -\sqrt{\frac{1+\cos x}{2}} = -\sqrt{\frac{1-\frac{1}{3}}{2}} = -\frac{1}{\sqrt{3}}$

$\therefore \quad \tan\frac{x}{2} = \frac{\sqrt{2/3}}{-1/\sqrt{3}} = -\sqrt{2}$

Since $x$ is in quadrant II: $\frac{\pi}{2} < x < \pi$, so $\frac{x}{2}$ is in the first quadrant — all ratios are positive.

Find $\cos x$ (negative in quadrant II):

$\cos^2 x = 1 – \frac{1}{16} = \frac{15}{16} \Rightarrow \cos x = -\frac{\sqrt{15}}{4}$

Apply half-angle formulas and simplify the nested square roots:

Rationalise to find $\tan\frac{x}{2}$: