Chapter 3 — TRIGONOMETRIC FUNCTIONS

We start with the given expression: \[ 2\cos \dfrac{\pi}{13}\cos \dfrac{9\pi}{13} +\cos \dfrac{3\pi}{13} +\cos \dfrac{5\pi}{13} \]

Using product-to-sum identity: \[ 2\cos A \cos B = \cos(A+B) + \cos(A-B) \]

\[ \begin{aligned} 2\cos \dfrac{\pi}{13}\cos \dfrac{9\pi}{13} &= \cos \dfrac{10\pi}{13} + \cos \dfrac{-8\pi}{13}\\ &= \cos \dfrac{10\pi}{13} + \cos \dfrac{8\pi}{13} \end{aligned} \]

So the expression becomes: \[ \cos \dfrac{10\pi}{13} +\cos \dfrac{8\pi}{13} +\cos \dfrac{3\pi}{13} +\cos \dfrac{5\pi}{13} \]

Now group terms: \[ \left(\cos \dfrac{10\pi}{13} + \cos \dfrac{3\pi}{13}\right) + \left(\cos \dfrac{8\pi}{13} + \cos \dfrac{5\pi}{13}\right) \]

Apply sum-to-product identity: \[ \cos x + \cos y = 2\cos \dfrac{x+y}{2}\cos \dfrac{x-y}{2} \]

First pair: \[ \begin{aligned} \cos \dfrac{10\pi}{13} + \cos \dfrac{3\pi}{13} &= 2\cos \dfrac{13\pi}{26}\cos \dfrac{7\pi}{26}\\ &= 2\cos \dfrac{\pi}{2}\cos \dfrac{7\pi}{26}\\ &= 0 \end{aligned} \]

Second pair: \[ \begin{aligned} \cos \dfrac{8\pi}{13} + \cos \dfrac{5\pi}{13} &= 2\cos \dfrac{13\pi}{26}\cos \dfrac{3\pi}{26}\\ &= 2\cos \dfrac{\pi}{2}\cos \dfrac{3\pi}{26}\\ &= 0 \end{aligned} \]

Therefore, total sum: \[ 0 + 0 = 0 \]

Hence, proved.

We start with: \[ \left( \sin 3x+\sin x\right)\sin x +\left( \cos 3x-\cos x\right)\cos x \]

Apply identities: \[ \sin 3x+\sin x = 2\sin 2x \cos x \] \[ \cos 3x-\cos x = -2\sin 2x \sin x \]

Substituting: \[ (2\sin 2x \cos x)\sin x + (-2\sin 2x \sin x)\cos x \]

Rearranging: \[ 2\sin 2x \cos x \sin x - 2\sin 2x \sin x \cos x \]

Factor: \[ 2\sin 2x \sin x \cos x - 2\sin 2x \sin x \cos x = 0 \]

Hence, proved.

Consider the left-hand side: \[ \left( \cos x+\cos y\right)^{2} +\left( \sin x-\sin y\right)^{2} \]

Using identities: \[ \cos x+\cos y =2\cos \dfrac{x+y}{2}\cos \dfrac{x-y}{2} \] \[ \sin x-\sin y =2\cos \dfrac{x+y}{2}\sin \dfrac{x-y}{2} \]

Substitute: \[ \left(2\cos \dfrac{x+y}{2}\cos \dfrac{x-y}{2}\right)^2 + \left(2\cos \dfrac{x+y}{2}\sin \dfrac{x-y}{2}\right)^2 \]

\[ = 4\cos^2\dfrac{x+y}{2}\cos^2\dfrac{x-y}{2} + 4\cos^2\dfrac{x+y}{2}\sin^2\dfrac{x-y}{2} \]

Factor out: \[ 4\cos^2\dfrac{x+y}{2} \left( \cos^2\dfrac{x-y}{2} + \sin^2\dfrac{x-y}{2} \right) \]

Using \(\sin^2\theta + \cos^2\theta = 1\): \[ =4\cos^2\dfrac{x+y}{2} \]

Hence, proved.

Consider the left-hand side: \[ \left( \cos x-\cos y\right)^{2} +\left( \sin x-\sin y\right)^{2} \]

Using identities: \[ \cos x-\cos y =-2\sin \dfrac{x+y}{2}\sin \dfrac{x-y}{2} \] \[ \sin x-\sin y =2\cos \dfrac{x+y}{2}\sin \dfrac{x-y}{2} \]

Substitute: \[ \left(-2\sin \dfrac{x+y}{2}\sin \dfrac{x-y}{2}\right)^2 + \left(2\cos \dfrac{x+y}{2}\sin \dfrac{x-y}{2}\right)^2 \]

\[ = 4\sin^2\dfrac{x-y}{2}\sin^2\dfrac{x+y}{2} + 4\sin^2\dfrac{x-y}{2}\cos^2\dfrac{x+y}{2} \]

Factor: \[ 4\sin^2\dfrac{x-y}{2} \left( \sin^2\dfrac{x+y}{2} + \cos^2\dfrac{x+y}{2} \right) \]

Using \(\sin^2\theta + \cos^2\theta = 1\): \[ =4\sin^2\dfrac{x-y}{2} \]

Hence, proved.

Consider the left-hand side: \[ \sin x+\sin 3x+\sin 5x+\sin 7x \]

Group terms: \[ (\sin x+\sin 7x)+(\sin 3x+\sin 5x) \]

Apply identity: \[ \sin x+\sin 7x =2\sin 4x\cos 3x \] \[ \sin 3x+\sin 5x =2\sin 4x\cos x \]

Substituting: \[ =2\sin 4x\cos 3x + 2\sin 4x\cos x \]

Factor: \[ =2\sin 4x(\cos 3x+\cos x) \]

Apply cosine identity: \[ \cos 3x+\cos x =2\cos 2x\cos x \]

Therefore: \[ =2\sin 4x \cdot 2\cos 2x\cos x =4\cos x\cos 2x\sin 4x \]

Hence, proved.

Consider the given expression: \[ \dfrac{(\sin 7x+\sin 5x)+(\sin 9x+\sin 3x)} {(\cos 7x+\cos 5x)+(\cos 9x+\cos 3x)} \]

Simplify numerator: \[ \sin 7x+\sin 5x = 2\sin 6x \cos x \] \[ \sin 9x+\sin 3x = 2\sin 6x \cos 3x \]

\[ \Rightarrow 2\sin 6x(\cos x+\cos 3x) \]

Apply identity: \[ \cos x+\cos 3x = 2\cos 2x\cos x \]

\[ \Rightarrow \text{Numerator} = 4\sin 6x\cos x\cos 2x \]

Similarly for denominator: \[ \cos 7x+\cos 5x = 2\cos 6x \cos x \] \[ \cos 9x+\cos 3x = 2\cos 6x \cos 3x \]

\[ \Rightarrow 2\cos 6x(\cos x+\cos 3x) \]

\[ \Rightarrow \text{Denominator} = 4\cos 6x\cos x\cos 2x \]

Therefore: \[ \dfrac{4\sin 6x\cos x\cos 2x} {4\cos 6x\cos x\cos 2x} = \tan 6x \]

Hence, proved.

Consider: \[ \sin 3x+\sin 2x-\sin x \]

Rearranging: \[ (\sin 3x - \sin x) + \sin 2x \]

Apply identity: \[ \sin 3x - \sin x = 2\cos 2x \sin x \]

So expression becomes: \[ 2\cos 2x \sin x + \sin 2x \]

Using \(\sin 2x = 2\sin x \cos x\): \[ =2\sin x \cos 2x + 2\sin x \cos x \]

Factor: \[ =2\sin x (\cos 2x + \cos x) \]

Apply identity: \[ \cos 2x + \cos x = 2\cos \dfrac{3x}{2}\cos \dfrac{x}{2} \]

Therefore: \[ \begin{aligned} &=2\sin x \cdot 2\cos \dfrac{3x}{2}\cos \dfrac{x}{2}\\ &=4\sin x\cos \dfrac{x}{2}\cos \dfrac{3x}{2} \end{aligned} \]

Hence, proved.

Given: \[ \tan x = -\dfrac{4}{3}, \quad x \text{ in quadrant II} \]

Using identity: \[ \begin{aligned} \sec^2 x &= 1 + \tan^2 x\\ &= 1 + \dfrac{16}{9}\\ &= \dfrac{25}{9} \end{aligned} \]

\[ \begin{aligned} \Rightarrow \sec x &= \dfrac{5}{3}\\ \Rightarrow \cos x &= \pm \dfrac{3}{5} \end{aligned} \]

Since \(x\) is in quadrant II: \[ \cos x = -\dfrac{3}{5} \]

Now using: \[ \begin{aligned} \cos^2 \dfrac{x}{2} = \dfrac{1+\cos x}{2} &= \dfrac{1 - \frac{3}{5}}{2}\\ &= \dfrac{2/5}{2}\\ &= \dfrac{1}{5} \end{aligned} \]

Since \(\dfrac{x}{2}\) lies in quadrant I: \[ \cos \dfrac{x}{2} = \dfrac{1}{\sqrt{5}} \]

\[\begin{aligned} \sin^2 \dfrac{x}{2} &= \dfrac{1 - \cos x}{2}\\ &= \dfrac{1 + \frac{3}{5}}{2}\\ &= \dfrac{8/5}{2}\\ &= \dfrac{4}{5} \end{aligned} \]

\[ \Rightarrow \sin \dfrac{x}{2} = \dfrac{2}{\sqrt{5}} \]

Finally: \[ \begin{aligned} \tan \dfrac{x}{2} &= \dfrac{\sin \frac{x}{2}}{\cos \frac{x}{2}}\\ &= \dfrac{2/\sqrt{5}}{1/\sqrt{5}} = 2 \end{aligned} \]

Hence, \[ \begin{aligned} \sin \dfrac{x}{2}&=\dfrac{2}{\sqrt{5}},\\ \cos \dfrac{x}{2}&=\dfrac{1}{\sqrt{5}},\\ \tan \dfrac{x}{2}&=2 \end{aligned} \]

Given: \[ \cos x = -\dfrac{1}{3}, \quad x \text{ in quadrant III} \]

Using half-angle identity: \[ \begin{aligned} \cos^2 \dfrac{x}{2} &= \dfrac{1+\cos x}{2}\\ &= \dfrac{1 - \frac{1}{3}}{2}\\ &= \dfrac{2/3}{2}\\ &= \dfrac{1}{3} \end{aligned} \]

Since \(x\) is in quadrant III, \(\dfrac{x}{2}\) lies in quadrant II, hence: \[ \cos \dfrac{x}{2} = -\dfrac{1}{\sqrt{3}} \]

Now: \[ \begin{aligned} \sin^2 \dfrac{x}{2} &= \dfrac{1-\cos x}{2}\\ &= \dfrac{1 + \frac{1}{3}}{2}\\ &= \dfrac{4/3}{2}\\ &= \dfrac{2}{3} \end{aligned} \]

Since \(\dfrac{x}{2}\) lies in quadrant II: \[ \sin \dfrac{x}{2} = \sqrt{\dfrac{2}{3}} \]

Therefore: \[ \begin{aligned} \tan \dfrac{x}{2} &= \dfrac{\sin \frac{x}{2}}{\cos \frac{x}{2}}\\ &= \dfrac{\sqrt{\frac{2}{3}}}{-\sqrt{\frac{1}{3}}}\\ &= -\sqrt{2} \end{aligned} \]

Hence, \[ \sin \dfrac{x}{2}=\sqrt{\dfrac{2}{3}},\quad \cos \dfrac{x}{2}=-\dfrac{1}{\sqrt{3}},\quad \tan \dfrac{x}{2}=-\sqrt{2} \]

Given: \[ \sin x = \dfrac{1}{4}, \quad x \text{ in quadrant II} \]

Using: \[ \begin{aligned} \cos^2 x &= 1 - \sin^2 x\\ &= 1 - \dfrac{1}{16}\\ &= \dfrac{15}{16} \end{aligned} \]

Since \(x\) lies in quadrant II: \[ \cos x = -\dfrac{\sqrt{15}}{4} \]

Now: \[ \begin{aligned} \cos^2 \dfrac{x}{2} &= \dfrac{1+\cos x}{2}\\ &= \dfrac{1 - \frac{\sqrt{15}}{4}}{2}\\ &= \dfrac{4-\sqrt{15}}{8} \end{aligned} \]

Since \(\dfrac{x}{2}\) lies in quadrant I: \[ \cos \dfrac{x}{2} = \sqrt{\dfrac{4-\sqrt{15}}{8}} \]

\[ \begin{aligned} \sin^2 \dfrac{x}{2} &= \dfrac{1-\cos x}{2}\\ &= \dfrac{1 + \frac{\sqrt{15}}{4}}{2}\\ &= \dfrac{4+\sqrt{15}}{8} \end{aligned} \]

\[ \Rightarrow \sin \dfrac{x}{2} = \sqrt{\dfrac{4+\sqrt{15}}{8}} \]

Using: \[ \begin{aligned} \tan \dfrac{x}{2} &= \dfrac{\sin x}{1+\cos x}\\ &= \dfrac{\frac{1}{4}}{1 - \frac{\sqrt{15}}{4}}\\ &= \dfrac{1}{4-\sqrt{15}} \end{aligned} \]

Rationalising: \[ \begin{aligned} \tan \dfrac{x}{2} &= \dfrac{1}{4-\sqrt{15}} \cdot \dfrac{4+\sqrt{15}}{4+\sqrt{15}}\\ &= 4+\sqrt{15} \end{aligned} \]

Hence, \[ \begin{aligned} \sin \dfrac{x}{2}&=\sqrt{\dfrac{4+\sqrt{15}}{8}},\\ \cos \dfrac{x}{2}&=\sqrt{\dfrac{4-\sqrt{15}}{8}},\\ \tan \dfrac{x}{2}&=4+\sqrt{15} \end{aligned} \]