Chapter 3 - Trigonometric Functions

We know that $\cos \frac{5\pi}{8} = \cos(\pi - \frac{3\pi}{8}) = -\cos \frac{3\pi}{8}$ and $\cos \frac{7\pi}{8} = \cos(\pi - \frac{\pi}{8}) = -\cos \frac{\pi}{8}$.
So, the expression becomes $2(\cos^4 \frac{\pi}{8} + \cos^4 \frac{3\pi}{8})$.
Also $\frac{3\pi}{8} = \frac{\pi}{2} - \frac{\pi}{8}$, so $\cos \frac{3\pi}{8} = \sin \frac{\pi}{8}$.
Expression $= 2(\cos^4 \frac{\pi}{8} + \sin^4 \frac{\pi}{8}) = 2((\cos^2 \frac{\pi}{8} + \sin^2 \frac{\pi}{8})^2 - 2\cos^2 \frac{\pi}{8} \sin^2 \frac{\pi}{8})$
$= 2(1 - \frac{1}{2} \sin^2 \frac{\pi}{4}) = 2(1 - \frac{1}{2} (\frac{1}{\sqrt{2}})^2) = 2(1 - \frac{1}{4}) = 2(\frac{3}{4}) = \frac{3}{2}$.

We know that $AM \ge GM$.
$\frac{\cos^2 x + \sec^2 x}{2} \ge \sqrt{\cos^2 x \cdot \sec^2 x}$
$\frac{f(x)}{2} \ge \sqrt{1}$
$f(x) \ge 2$.

$\tan(\theta + \phi) = \frac{\tan \theta + \tan \phi}{1 - \tan \theta \tan \phi} = \frac{\frac{1}{2} + \frac{1}{3}}{1 - \frac{1}{2} \cdot \frac{1}{3}} = \frac{\frac{5}{6}}{1 - \frac{1}{6}} = \frac{\frac{5}{6}}{\frac{5}{6}} = 1$.
Since $\tan(\theta + \phi) = 1$, $\theta + \phi = \frac{\pi}{4}$.

We know that the range of $\sec \theta$ is $(-\infty, -1] \cup [1, \infty)$.
Therefore, $\sec \theta$ cannot be $\frac{1}{2}$.

$\tan 89^\circ = \tan(90^\circ - 1^\circ) = \cot 1^\circ$.
Similarly, $\tan 88^\circ = \cot 2^\circ$, and so on.
The expression becomes $(\tan 1^\circ \cot 1^\circ) (\tan 2^\circ \cot 2^\circ) \dots \tan 45^\circ$.
$= 1 \cdot 1 \dots 1 = 1$.

We know that $\cos 2\theta = \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta}$.
Put $\theta = 15^\circ$.
Expression $= \cos(2 \times 15^\circ) = \cos 30^\circ = \frac{\sqrt{3}}{2}$.

The series contains $\cos 90^\circ$.
Since $\cos 90^\circ = 0$, the product of the entire series is 0.

$\tan 75^\circ = 2+\sqrt{3}$ and $\cot 75^\circ = 2-\sqrt{3}$.
$\tan 75^\circ - \cot 75^\circ = (2+\sqrt{3}) - (2-\sqrt{3}) = 2\sqrt{3}$.

$\sin \frac{5\pi}{8} = \sin \frac{3\pi}{8}$ and $\sin \frac{7\pi}{8} = \sin \frac{\pi}{8}$.
Expression $= 2(\sin^2 \frac{\pi}{8} + \sin^2 \frac{3\pi}{8})$.
Since $\frac{3\pi}{8} = \frac{\pi}{2} - \frac{\pi}{8}$, $\sin \frac{3\pi}{8} = \cos \frac{\pi}{8}$.
$= 2(\sin^2 \frac{\pi}{8} + \cos^2 \frac{\pi}{8}) = 2(1) = 2$.

$\cos 156^\circ = \cos(180^\circ - 24^\circ) = -\cos 24^\circ$.
$\cos 132^\circ = \cos(180^\circ - 48^\circ) = -\cos 48^\circ$.
Sum $= \cos 12^\circ + \cos 84^\circ - \cos 24^\circ - \cos 48^\circ$.
Using $\cos C + \cos D$ and $\cos C - \cos D$ formulas, we get $-1/2$.

$\tan(A+B) = 1$ (as $A+B=45^\circ$).
$\tan 2A = \frac{2(1/2)}{1-1/4} = \frac{1}{3/4} = 4/3$.
$\tan(2A+B) = \frac{\tan 2A + \tan B}{1 - \tan 2A \tan B} = \frac{4/3 + 1/3}{1 - 4/9} = \frac{5/3}{5/9} = 3$.

$\sin 50^\circ + \sin 10^\circ - \sin 70^\circ = 2 \sin 30^\circ \cos 20^\circ - \sin 70^\circ$
$= 2(1/2) \cos 20^\circ - \cos 20^\circ = 0$.

$\cos \frac{7\pi}{16} = \sin \frac{\pi}{16}$ and $\cos \frac{5\pi}{16} = \sin \frac{3\pi}{16}$.
Sum $= (\cos^2 \frac{\pi}{16} + \sin^2 \frac{\pi}{16}) + (\cos^2 \frac{3\pi}{16} + \sin^2 \frac{3\pi}{16}) = 1 + 1 = 2$.

Using $AM \ge GM$, $\frac{2^{\sin x} + 2^{\cos x}}{2} \ge \sqrt{2^{\sin x + \cos x}}$.
Min value occurs when $\sin x + \cos x$ is minimum ($-\sqrt{2}$).
Min value $= 2 \cdot 2^{-\sqrt{2}/2} = 2^{1 - 1/\sqrt{2}}$.

$\sin^2(\theta - \phi) - \sin^2(\theta + \phi) = \sin(\theta - \phi + \theta + \phi) \sin(\theta - \phi - \theta - \phi) = \sin 2\theta \sin(-2\phi) = -\sin 2\theta \sin 2\phi$.
Expression $= \cos 2\theta \cos 2\phi - \sin 2\theta \sin 2\phi = \cos(2\theta + 2\phi)$.

$\cot(\frac{\pi}{4} - \theta) = \tan(\frac{\pi}{2} - (\frac{\pi}{4} - \theta)) = \tan(\frac{\pi}{4} + \theta)$.
Product $= \cot(\frac{\pi}{4} + \theta) \tan(\frac{\pi}{4} + \theta) = 1$.

$\cos \frac{14\pi}{15} = \cos(\pi - \frac{\pi}{15}) = -\cos \frac{\pi}{15}$.
Using $\cos A \cos 2A \cos 4A \dots = \frac{\sin 2^n A}{2^n \sin A}$, the value simplifies to $1/16$.

Put $x=0$.
$3(0-1)^4 + 6(0+1)^2 + 4(0+1) = 3(1) + 6(1) + 4(1) = 13$.

$\tan(\alpha - \beta) = \frac{(x+1)-(x-1)}{1+(x+1)(x-1)} = \frac{2}{1+x^2-1} = \frac{2}{x^2}$.
$\cot(\alpha - \beta) = \frac{x^2}{2}$.
$2 \cot(\alpha - \beta) = x^2$.

Squaring both sides: $\sin^2 \theta + \cos^2 \theta + 2\sin \theta \cos \theta = 1$.
$1 + \sin 2\theta = 1 \implies \sin 2\theta = 0$.

$\tan(\alpha + \beta) = 1 \implies \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta} = 1$.
$\tan \alpha + \tan \beta = 1 - \tan \alpha \tan \beta$.
$(1 + \tan \alpha)(1 + \tan \beta) = 1 + \tan \alpha + \tan \beta + \tan \alpha \tan \beta = 1 + 1 = 2$.

$(\sin^2 A + \cos^2 A)^3 = 1^3 = 1$.
Expansion gives $\sin^6 A + \cos^6 A + 3\sin^2 A \cos^2 A (\sin^2 A + \cos^2 A) = 1$.

$AM \ge GM$.
$\frac{4 \tan^2 \theta + 9 \cot^2 \theta}{2} \ge \sqrt{36}$.
Min value $= 2 \times 6 = 12$.

$\frac{\sin x + 1}{\cos x} = 2 \cos x \implies 1 + \sin x = 2 \cos^2 x = 2(1 - \sin^2 x)$.
$2\sin^2 x + \sin x - 1 = 0$.
$\sin x = 1/2, -1$.
If $\sin x = -1$, $\cos x = 0$ (undefined). So $\sin x = 1/2$.
Two solutions in $[0, 2\pi]$: $\pi/6, 5\pi/6$.

$\sin x + 1/\sin x = 2 \implies (\sin x - 1)^2 = 0 \implies \sin x = 1$.
So $1^{19} + 1^{19} = 2$.

$\cos 10^\circ > \sin 10^\circ$ because in $[0, 45^\circ]$, cosine decreases and sine increases, and $\cos 45^\circ = \sin 45^\circ$.

1 radian $\approx 57^\circ$, 2 radians $\approx 114^\circ$.
$\tan 1 > 0$, $\tan 2 < 0$ (2nd quadrant).
Wait, $\tan 1$ is positive, $\tan 2$ is negative. So $\tan 1 > \tan 2$.
Let's recheck options. If options are about values, $\tan 1 \approx 1.5$, $\tan 2 \approx -2$. So $\tan 1 > \tan 2$ is correct.

$1 \text{ radian} \approx 57^\circ$.
Since sine is increasing in $[0, 90^\circ]$, $\sin 1^\circ < \sin 57^\circ$.

$1 \text{ radian} \approx 57^\circ$.
Cosine is decreasing in $[0, 90^\circ]$.
So $\cos 1^\circ > \cos 57^\circ$.

Using $\cos(X-Y) = \cos X \cos Y + \sin X \sin Y$.
Here $X = (n+1)A, Y = (n-1)A$.
Expression $= \cos((n+1)A - (n-1)A) = \cos(2A)$.