Chapter 3 - Trigonometric Functions

LHS = $(\sin 7x + \sin x) + (\sin 5x + \sin 3x)$
= $2 \sin 4x \cos 3x + 2 \sin 4x \cos x$
= $2 \sin 4x (\cos 3x + \cos x)$
= $2 \sin 4x (2 \cos 2x \cos x)$
= $4 \cos x \cos 2x \sin 4x$ = RHS.

LHS = $\cos^2 x + \cos^2 y + 2\cos x \cos y + \sin^2 x + \sin^2 y - 2\sin x \sin y$
= $(\cos^2 x + \sin^2 x) + (\cos^2 y + \sin^2 y) + 2(\cos x \cos y - \sin x \sin y)$
= $1 + 1 + 2\cos(x+y)$
= $2 + 2\cos(x+y)$
= $2(1 + \cos(x+y))$
= $2(2\cos^2 \frac{x+y}{2})$
= $4\cos^2 \frac{x+y}{2}$ = RHS.

Expression = $(\tan 81^\circ + \tan 9^\circ) - (\tan 63^\circ + \tan 27^\circ)$
= $(\cot 9^\circ + \tan 9^\circ) - (\cot 27^\circ + \tan 27^\circ)$
= $\frac{\cos 9^\circ}{\sin 9^\circ} + \frac{\sin 9^\circ}{\cos 9^\circ} - (\frac{\cos 27^\circ}{\sin 27^\circ} + \frac{\sin 27^\circ}{\cos 27^\circ})$
= $\frac{1}{\sin 9^\circ \cos 9^\circ} - \frac{1}{\sin 27^\circ \cos 27^\circ}$
= $\frac{2}{\sin 18^\circ} - \frac{2}{\sin 54^\circ}$
= $2 [\frac{\sin 54^\circ - \sin 18^\circ}{\sin 18^\circ \sin 54^\circ}]$
= $2 [\frac{\cos 36^\circ - \sin 18^\circ}{\sin 18^\circ \cos 36^\circ}]$
= $2 [\frac{\frac{\sqrt{5}+1}{4} - \frac{\sqrt{5}-1}{4}}{\frac{\sqrt{5}-1}{4} \cdot \frac{\sqrt{5}+1}{4}}] = 2 [\frac{\frac{2}{4}}{\frac{5-1}{16}}] = 2 [\frac{1/2}{4/16}] = 2 [\frac{1/2}{1/4}] = 4$.

LHS = $\frac{\frac{1}{\cos 8x} - 1}{\frac{1}{\cos 4x} - 1} = \frac{1 - \cos 8x}{\cos 8x} \cdot \frac{\cos 4x}{1 - \cos 4x}$
= $\frac{2\sin^2 4x}{\cos 8x} \cdot \frac{\cos 4x}{2\sin^2 2x}$
= $\frac{\sin 4x (2\sin 4x \cos 4x)}{\cos 8x (2\sin^2 2x)}$
= $\frac{\sin 4x \sin 8x}{\cos 8x (2\sin^2 2x)} = \tan 8x \frac{\sin 4x}{2\sin^2 2x}$
= $\tan 8x \frac{2\sin 2x \cos 2x}{2\sin^2 2x} = \tan 8x \frac{\cos 2x}{\sin 2x} = \frac{\tan 8x}{\tan 2x}$ = RHS.

$\tan x + \tan 2x = -\tan 3x + \tan x \tan 2x \tan 3x$
$\tan x + \tan 2x = -\tan 3x (1 - \tan x \tan 2x)$
$\frac{\tan x + \tan 2x}{1 - \tan x \tan 2x} = -\tan 3x$
$\tan(x+2x) = -\tan 3x$
$\tan 3x = -\tan 3x \implies 2\tan 3x = 0 \implies \tan 3x = 0$
$3x = n\pi \implies x = \frac{n\pi}{3}$.

$a^2 + b^2 = (\sin \alpha + \sin \beta)^2 + (\cos \alpha + \cos \beta)^2$
= $\sin^2 \alpha + \sin^2 \beta + 2\sin \alpha \sin \beta + \cos^2 \alpha + \cos^2 \beta + 2\cos \alpha \cos \beta$
= $2 + 2(\cos \alpha \cos \beta + \sin \alpha \sin \beta) = 2(1 + \cos(\alpha - \beta)) = 4\cos^2 \frac{\alpha-\beta}{2}$.
Also, $\frac{a}{b} = \frac{2\sin \frac{\alpha+\beta}{2} \cos \frac{\alpha-\beta}{2}}{2\cos \frac{\alpha+\beta}{2} \cos \frac{\alpha-\beta}{2}} = \tan \frac{\alpha+\beta}{2}$.
$\sin(\alpha+\beta) = \frac{2\tan \frac{\alpha+\beta}{2}}{1+\tan^2 \frac{\alpha+\beta}{2}} = \frac{2(a/b)}{1+(a/b)^2} = \frac{2ab}{b^2+a^2}$.

$\frac{\sin(\theta + 2\alpha)}{\sin \theta} = \frac{1}{n}$.
Apply Componendo and Dividendo:
$\frac{\sin(\theta + 2\alpha) + \sin \theta}{\sin(\theta + 2\alpha) - \sin \theta} = \frac{1+n}{1-n}$
$\frac{2\sin(\theta+\alpha)\cos \alpha}{2\cos(\theta+\alpha)\sin \alpha} = \frac{1+n}{1-n}$
$\tan(\theta+\alpha) \cot \alpha = \frac{1+n}{1-n}$
$\tan(\theta+\alpha) = \frac{1+n}{1-n} \tan \alpha$.

$2\cos \theta \cos 3\theta \cos 2\theta = \frac{1}{2}$
$(\cos 4\theta + \cos 2\theta) \cos 2\theta = \frac{1}{2}$
$2\cos 4\theta \cos 2\theta + 2\cos^2 2\theta = 1$
$(\cos 6\theta + \cos 2\theta) + (1 + \cos 4\theta) = 1$
$\cos 6\theta + \cos 4\theta + \cos 2\theta = 0$
$\cos 6\theta + \cos 2\theta + \cos 4\theta = 0$
$2\cos 4\theta \cos 2\theta + \cos 4\theta = 0$
$\cos 4\theta (2\cos 2\theta + 1) = 0$
Case 1: $\cos 4\theta = 0 \implies 4\theta = (2n+1)\frac{\pi}{2} \implies \theta = (2n+1)\frac{\pi}{8}$.
Case 2: $\cos 2\theta = -1/2 \implies 2\theta = 2n\pi \pm \frac{2\pi}{3} \implies \theta = n\pi \pm \frac{\pi}{3}$.

$\tan x + \frac{\tan x + \sqrt{3}}{1 - \sqrt{3}\tan x} + \frac{\tan x - \sqrt{3}}{1 + \sqrt{3}\tan x} = 3$
$\tan x + \frac{(\tan x + \sqrt{3})(1 + \sqrt{3}\tan x) + (\tan x - \sqrt{3})(1 - \sqrt{3}\tan x)}{1 - 3\tan^2 x} = 3$
$\tan x + \frac{\tan x + \sqrt{3}\tan^2 x + \sqrt{3} + 3\tan x + \tan x - \sqrt{3}\tan^2 x - \sqrt{3} + 3\tan x}{1 - 3\tan^2 x} = 3$
$\tan x + \frac{8\tan x}{1 - 3\tan^2 x} = 3$
$\frac{\tan x - 3\tan^3 x + 8\tan x}{1 - 3\tan^2 x} = 3$
$\frac{9\tan x - 3\tan^3 x}{1 - 3\tan^2 x} = 3$
$3 \left( \frac{3\tan x - \tan^3 x}{1 - 3\tan^2 x} \right) = 3$
$3 \tan 3x = 3 \implies \tan 3x = 1$
$3x = n\pi + \frac{\pi}{4} \implies x = \frac{n\pi}{3} + \frac{\pi}{12}$.

$\sin \theta - \sin 2\theta = \cos 2\theta - \cos \theta$
$-2\cos \frac{3\theta}{2} \sin \frac{\theta}{2} = -2\sin \frac{3\theta}{2} \sin \frac{\theta}{2}$
$2\sin \frac{\theta}{2} (\sin \frac{3\theta}{2} - \cos \frac{3\theta}{2}) = 0$
Case 1: $\sin \frac{\theta}{2} = 0 \implies \frac{\theta}{2} = n\pi \implies \theta = 2n\pi$.
Case 2: $\tan \frac{3\theta}{2} = 1 \implies \frac{3\theta}{2} = n\pi + \frac{\pi}{4} \implies \theta = \frac{2n\pi}{3} + \frac{\pi}{6}$.

$2(\frac{1}{\sqrt{2}})^2 + 2(\frac{1}{\sqrt{2}})^2 + 2(2)^2 = 2(1/2) + 2(1/2) + 8 = 1 + 1 + 8 = 10$.

$\sin 18^\circ \sin(180^\circ + 54^\circ)$ - Wait, $13\pi/10 = 234^\circ = 180+54$.
$\sin 18^\circ (-\sin 54^\circ) = -(\frac{\sqrt{5}-1}{4})(\frac{\sqrt{5}+1}{4}) = -\frac{5-1}{16} = -\frac{4}{16} = -\frac{1}{4}$.

$\frac{4\pi}{9} = \frac{\pi}{2} - \frac{\pi}{18} \implies \sin \frac{4\pi}{9} = \cos \frac{\pi}{18}$.
$\frac{7\pi}{18} = \frac{\pi}{2} - \frac{\pi}{9} \implies \sin \frac{7\pi}{18} = \cos \frac{\pi}{9}$.
Sum $= (\sin^2 \frac{\pi}{18} + \cos^2 \frac{\pi}{18}) + (\sin^2 \frac{\pi}{9} + \cos^2 \frac{\pi}{9}) = 1 + 1 = 2$.

$\cos 160^\circ = -\cos 20^\circ$, $\cos 140^\circ = -\cos 40^\circ$, etc.
Pairs cancel out: $(20, 160), (40, 140), (60, 120), (80, 100)$.
Remaining terms: $\cos 180^\circ = -1$.

$\cos \frac{7\pi}{16} = \sin \frac{\pi}{16}$. Sum $= 1 + 1 = 2$.

Contains $\cos 90^\circ = 0$.

$\tan(A+B) = \frac{1/2+1/3}{1-1/6} = 1$.

$\sin^6 x + \cos^6 x = 1 - 3\sin^2 x \cos^2 x$.
$(a^2-1) = 2\sin x \cos x$.
Expression $= 1 - 3(\frac{a^2-1}{2})^2 = 1 - \frac{3}{4}(a^2-1)^2$.

Max value is $\sqrt{a^2+b^2}$, min is $-\sqrt{a^2+b^2}$. So $|c| \le \sqrt{a^2+b^2} \implies c^2 \le a^2+b^2$.

Using $\frac{\sin 2^n A}{2^n \sin A}$ with $A=\pi/7, n=3$.
$\frac{\sin(8\pi/7)}{8 \sin(\pi/7)} = \frac{-\sin(\pi/7)}{8 \sin(\pi/7)} = -\frac{1}{8}$.

$\sin 10^\circ \sin 50^\circ \sin 70^\circ = \sin 10^\circ \cos 40^\circ \cos 20^\circ$.
Multiply/divide by $2\cos 10^\circ$: $\frac{1}{2} \sin 20^\circ \cos 20^\circ \cos 40^\circ / \cos 10^\circ = \frac{1}{8} \sin 80^\circ / \cos 10^\circ = 1/8$.

$2 \cos \frac{\pi}{4} \cos x = 2 \frac{1}{\sqrt{2}} \cos x = \sqrt{2} \cos x$.

Max value = $\sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{4} = 2$.

In first quadrant, $\sin \theta < \cos \theta$ for $\theta < 45^\circ$.

$\sin 120^\circ = \frac{\sqrt{3}}{2}$. $\cos 150^\circ = -\frac{\sqrt{3}}{2}$.

$x + 1/x \ge 2$ or $\le -2$. $\sin \theta$ cannot exceed 1.

$\sin^2(x+\pi) = (-\sin x)^2 = \sin^2 x$.