Chapter 3 - Trigonometric Functions

LHS = $\frac{(\tan A + \sec A) - (\sec^2 A - \tan^2 A)}{\tan A - \sec A + 1}$
= $\frac{(\tan A + \sec A) - (\sec A - \tan A)(\sec A + \tan A)}{\tan A - \sec A + 1}$
= $\frac{(\tan A + \sec A)(1 - (\sec A - \tan A))}{\tan A - \sec A + 1}$
= $\frac{(\tan A + \sec A)(1 - \sec A + \tan A)}{\tan A - \sec A + 1}$
= $\tan A + \sec A = \frac{\sin A}{\cos A} + \frac{1}{\cos A} = \frac{1 + \sin A}{\cos A}$ = RHS.

$\tan 15^\circ = \tan(45^\circ - 30^\circ) = \frac{\tan 45^\circ - \tan 30^\circ}{1 + \tan 45^\circ \tan 30^\circ}$
= $\frac{1 - 1/\sqrt{3}}{1 + 1/\sqrt{3}} = \frac{\sqrt{3}-1}{\sqrt{3}+1}$
Rationalizing the denominator:
= $\frac{(\sqrt{3}-1)^2}{3-1} = \frac{3+1-2\sqrt{3}}{2} = \frac{4-2\sqrt{3}}{2} = 2 - \sqrt{3}$.

We know $2\cos^2 \theta = 1 + \cos 2\theta$.
LHS = $\frac{1}{2} [ (1+\cos 2x) + (1+\cos(2x + \frac{2\pi}{3})) + (1+\cos(2x - \frac{2\pi}{3})) ]$
= $\frac{1}{2} [ 3 + \cos 2x + \cos(2x + \frac{2\pi}{3}) + \cos(2x - \frac{2\pi}{3}) ]$
= $\frac{1}{2} [ 3 + \cos 2x + 2\cos 2x \cos \frac{2\pi}{3} ]$
= $\frac{1}{2} [ 3 + \cos 2x + 2\cos 2x (-\frac{1}{2}) ]$
= $\frac{1}{2} [ 3 + \cos 2x - \cos 2x ] = \frac{3}{2}$ = RHS.

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

LHS = $\frac{\sin x \cos y + \cos x \sin y}{\sin x \cos y - \cos x \sin y}$
Divide numerator and denominator by $\cos x \cos y$:
= $\frac{\frac{\sin x \cos y}{\cos x \cos y} + \frac{\cos x \sin y}{\cos x \cos y}}{\frac{\sin x \cos y}{\cos x \cos y} - \frac{\cos x \sin y}{\cos x \cos y}}$
= $\frac{\tan x + \tan y}{\tan x - \tan y}$ = RHS.

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

$2\sin x \cos x + \cos x = 0$
$\cos x (2\sin x + 1) = 0$
Case 1: $\cos x = 0 \implies x = (2n+1)\frac{\pi}{2}$
Case 2: $\sin x = -\frac{1}{2} = \sin(\pi + \frac{\pi}{6}) = \sin \frac{7\pi}{6}$
$\implies x = n\pi + (-1)^n \frac{7\pi}{6}$.

LHS = $(\sin 3x - \sin x) + \sin 2x$
= $2\cos \frac{3x+x}{2} \sin \frac{3x-x}{2} + 2\sin x \cos x$
= $2\cos 2x \sin x + 2\sin x \cos x$
= $2\sin x (\cos 2x + \cos x)$
= $2\sin x (2\cos \frac{2x+x}{2} \cos \frac{2x-x}{2})$
= $4\sin x \cos \frac{3x}{2} \cos \frac{x}{2}$ = RHS.

Let $x = \frac{\pi}{8}$. Then $2x = \frac{\pi}{4}$.
$\tan 2x = \frac{2\tan x}{1 - \tan^2 x} = 1$
$2\tan x = 1 - \tan^2 x \implies \tan^2 x + 2\tan x - 1 = 0$
$\tan x = \frac{-2 \pm \sqrt{4 - 4(1)(-1)}}{2} = \frac{-2 \pm \sqrt{8}}{2} = -1 \pm \sqrt{2}$
Since $\frac{\pi}{8}$ is in the first quadrant, $\tan \frac{\pi}{8} > 0$.
$\therefore \tan \frac{\pi}{8} = \sqrt{2} - 1$.

$\tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta}$
= $\frac{\frac{m}{m+1} + \frac{1}{2m+1}}{1 - \frac{m}{m+1} \cdot \frac{1}{2m+1}}$
= $\frac{m(2m+1) + (m+1)}{(m+1)(2m+1) - m}$
= $\frac{2m^2 + m + m + 1}{2m^2 + m + 2m + 1 - m}$
= $\frac{2m^2 + 2m + 1}{2m^2 + 2m + 1} = 1$
$\therefore \alpha + \beta = \frac{\pi}{4}$.

$\cos 15^\circ = \cos(45^\circ - 30^\circ) = \cos 45^\circ \cos 30^\circ + \sin 45^\circ \sin 30^\circ$
= $\frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} \cdot \frac{1}{2}$
= $\frac{\sqrt{3} + 1}{2\sqrt{2}} = \frac{\sqrt{6} + \sqrt{2}}{4}$.

Using $2\cos A \cos B = \cos(A+B) + \cos(A-B)$:
$2\cos \frac{\pi}{13} \cos \frac{9\pi}{13} = \cos \frac{10\pi}{13} + \cos \frac{8\pi}{13}$.
LHS = $\cos \frac{10\pi}{13} + \cos \frac{8\pi}{13} + \cos \frac{3\pi}{13} + \cos \frac{5\pi}{13}$.
$\cos \frac{10\pi}{13} = \cos(\pi - \frac{3\pi}{13}) = -\cos \frac{3\pi}{13}$.
$\cos \frac{8\pi}{13} = \cos(\pi - \frac{5\pi}{13}) = -\cos \frac{5\pi}{13}$.
LHS = $-\cos \frac{3\pi}{13} - \cos \frac{5\pi}{13} + \cos \frac{3\pi}{13} + \cos \frac{5\pi}{13} = 0$ = RHS.

LHS = $\sin 3x \sin x + \sin^2 x + \cos 3x \cos x - \cos^2 x$
= $(\cos 3x \cos x + \sin 3x \sin x) - (\cos^2 x - \sin^2 x)$
= $\cos(3x - x) - \cos 2x$
= $\cos 2x - \cos 2x = 0$ = RHS.

Using $\cos^2 A - \cos^2 B = \sin(B+A)\sin(B-A)$:
LHS = $\sin(6x+2x)\sin(6x-2x)$
= $\sin 8x \sin 4x$ = RHS.

Let $x = 18^\circ$. Then $5x = 90^\circ \implies 2x = 90^\circ - 3x$.
$\sin 2x = \cos 3x$
$2\sin x \cos x = 4\cos^3 x - 3\cos x$
Divide by $\cos x$ (since $\cos 18^\circ \neq 0$):
$2\sin x = 4\cos^2 x - 3 = 4(1-\sin^2 x) - 3$
$4\sin^2 x + 2\sin x - 1 = 0$
$\sin x = \frac{-2 \pm \sqrt{4 - 4(4)(-1)}}{8} = \frac{-2 \pm \sqrt{20}}{8} = \frac{-1 \pm \sqrt{5}}{4}$.
Since $18^\circ$ is in Q1, $\sin 18^\circ > 0$.
$\therefore \sin 18^\circ = \frac{\sqrt{5}-1}{4}$.

Expression = $\frac{\sqrt{3}}{\sin 20^\circ} - \frac{1}{\cos 20^\circ} = \frac{\sqrt{3}\cos 20^\circ - \sin 20^\circ}{\sin 20^\circ \cos 20^\circ}$
= $\frac{2(\frac{\sqrt{3}}{2}\cos 20^\circ - \frac{1}{2}\sin 20^\circ)}{\frac{1}{2}(2\sin 20^\circ \cos 20^\circ)}$
= $\frac{2(\sin 60^\circ \cos 20^\circ - \cos 60^\circ \sin 20^\circ)}{\frac{1}{2}\sin 40^\circ}$
= $\frac{2\sin(60^\circ - 20^\circ)}{\frac{1}{2}\sin 40^\circ} = \frac{4\sin 40^\circ}{\sin 40^\circ} = 4$.

$\tan x + \tan 2x = \sqrt{3} - \sqrt{3} \tan x \tan 2x$
$\tan x + \tan 2x = \sqrt{3}(1 - \tan x \tan 2x)$
$\frac{\tan x + \tan 2x}{1 - \tan x \tan 2x} = \sqrt{3}$
$\tan(x + 2x) = \sqrt{3}$
$\tan 3x = \tan \frac{\pi}{3}$
$3x = n\pi + \frac{\pi}{3}$
$x = \frac{n\pi}{3} + \frac{\pi}{9}$.

$(\sin 5x + \sin x) + \sin 3x = 0$
$2\sin 3x \cos 2x + \sin 3x = 0$
$\sin 3x (2\cos 2x + 1) = 0$
Case 1: $\sin 3x = 0 \implies 3x = n\pi \implies x = \frac{n\pi}{3}$.
Case 2: $\cos 2x = -\frac{1}{2} = \cos \frac{2\pi}{3} \implies 2x = 2n\pi \pm \frac{2\pi}{3} \implies x = n\pi \pm \frac{\pi}{3}$.

$\tan 3x = \tan(\frac{\pi}{2} - x)$
$3x = n\pi + \frac{\pi}{2} - x$
$4x = n\pi + \frac{\pi}{2} = (2n+1)\frac{\pi}{2}$
$x = (2n+1)\frac{\pi}{8}$.