We are asked to solve several trigonometric equations and prove some trigonometric identities. (a) Solve $\cos(2\theta) = \sin(\theta)$. (b) Solve $2\cos(6y) + 11\cos(6y)\sin(3y) = 0$. (c) Solve $4\sin^2(3t) - 3\sin(3t) = 1$. (d) Prove $\tan(\alpha + \beta)$. (e) Find $\tan(2\alpha)$.

1. Problem Description

We are asked to solve several trigonometric equations and prove some trigonometric identities.

(a) Solve $\cos(2\theta) = \sin(\theta)$.

(b) Solve $2\cos(6y) + 11\cos(6y)\sin(3y) = 0$.

(c) Solve $4\sin^2(3t) - 3\sin(3t) = 1$.

(d) Prove $\tan(\alpha + \beta)$.

(e) Find $\tan(2\alpha)$.

2. Solution Steps

(a) $\cos(2\theta) = \sin(\theta)$.

We know $\cos(2\theta) = 1 - 2\sin^2(\theta)$, so

$1 - 2\sin^2(\theta) = \sin(\theta)$

$2\sin^2(\theta) + \sin(\theta) - 1 = 0$

$(2\sin(\theta) - 1)(\sin(\theta) + 1) = 0$

So $\sin(\theta) = \frac{1}{2}$ or $\sin(\theta) = -1$.

If $\sin(\theta) = \frac{1}{2}$, then $\theta = \frac{\pi}{6} + 2\pi k$ or $\theta = \frac{5\pi}{6} + 2\pi k$ for integer $k$.

If $\sin(\theta) = -1$, then $\theta = \frac{3\pi}{2} + 2\pi k$ for integer $k$.

(b) $2\cos(6y) + 11\cos(6y)\sin(3y) = 0$

$\cos(6y)(2 + 11\sin(3y)) = 0$

So $\cos(6y) = 0$ or $2 + 11\sin(3y) = 0$.

If $\cos(6y) = 0$, then $6y = \frac{\pi}{2} + \pi k$ for integer $k$, so $y = \frac{\pi}{12} + \frac{\pi}{6}k$.

If $2 + 11\sin(3y) = 0$, then $\sin(3y) = -\frac{2}{11}$. Let $3y = \arcsin(-\frac{2}{11})$.

$3y = \arcsin(-\frac{2}{11}) + 2\pi k$ or $3y = \pi - \arcsin(-\frac{2}{11}) + 2\pi k = \pi + \arcsin(\frac{2}{11}) + 2\pi k$.

So $y = \frac{1}{3}\arcsin(-\frac{2}{11}) + \frac{2\pi}{3}k$ or $y = \frac{\pi}{3} + \frac{1}{3}\arcsin(\frac{2}{11}) + \frac{2\pi}{3}k$.

(c) $4\sin^2(3t) - 3\sin(3t) = 1$

$4\sin^2(3t) - 3\sin(3t) - 1 = 0$

$(4\sin(3t) + 1)(\sin(3t) - 1) = 0$

So $\sin(3t) = -\frac{1}{4}$ or $\sin(3t) = 1$.

If $\sin(3t) = -\frac{1}{4}$, then $3t = \arcsin(-\frac{1}{4}) + 2\pi k$ or $3t = \pi - \arcsin(-\frac{1}{4}) + 2\pi k = \pi + \arcsin(\frac{1}{4}) + 2\pi k$.

So $t = \frac{1}{3}\arcsin(-\frac{1}{4}) + \frac{2\pi}{3}k$ or $t = \frac{\pi}{3} + \frac{1}{3}\arcsin(\frac{1}{4}) + \frac{2\pi}{3}k$.

If $\sin(3t) = 1$, then $3t = \frac{\pi}{2} + 2\pi k$, so $t = \frac{\pi}{6} + \frac{2\pi}{3}k$.

(d) $\tan(\alpha + \beta) = \frac{\tan(\alpha) + \tan(\beta)}{1 - \tan(\alpha)\tan(\beta)}$

Proof:

$\tan(\alpha + \beta) = \frac{\sin(\alpha + \beta)}{\cos(\alpha + \beta)} = \frac{\sin\alpha\cos\beta + \cos\alpha\sin\beta}{\cos\alpha\cos\beta - \sin\alpha\sin\beta}$.

Divide both numerator and denominator by $\cos\alpha\cos\beta$:

$\tan(\alpha + \beta) = \frac{\frac{\sin\alpha\cos\beta}{\cos\alpha\cos\beta} + \frac{\cos\alpha\sin\beta}{\cos\alpha\cos\beta}}{\frac{\cos\alpha\cos\beta}{\cos\alpha\cos\beta} - \frac{\sin\alpha\sin\beta}{\cos\alpha\cos\beta}} = \frac{\frac{\sin\alpha}{\cos\alpha} + \frac{\sin\beta}{\cos\beta}}{1 - \frac{\sin\alpha}{\cos\alpha}\frac{\sin\beta}{\cos\beta}} = \frac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta}$.

(e) $\tan(2\alpha) = \tan(\alpha + \alpha) = \frac{\tan\alpha + \tan\alpha}{1 - \tan\alpha\tan\alpha} = \frac{2\tan\alpha}{1 - \tan^2\alpha}$

3. Final Answer

(a) $\theta = \frac{\pi}{6} + 2\pi k$, $\theta = \frac{5\pi}{6} + 2\pi k$, $\theta = \frac{3\pi}{2} + 2\pi k$ for integer $k$.

(b) $y = \frac{\pi}{12} + \frac{\pi}{6}k$, $y = \frac{1}{3}\arcsin(-\frac{2}{11}) + \frac{2\pi}{3}k$, $y = \frac{\pi}{3} + \frac{1}{3}\arcsin(\frac{2}{11}) + \frac{2\pi}{3}k$ for integer $k$.

(c) $t = \frac{1}{3}\arcsin(-\frac{1}{4}) + \frac{2\pi}{3}k$, $t = \frac{\pi}{3} + \frac{1}{3}\arcsin(\frac{1}{4}) + \frac{2\pi}{3}k$, $t = \frac{\pi}{6} + \frac{2\pi}{3}k$ for integer $k$.

(d) $\tan(\alpha + \beta) = \frac{\tan(\alpha) + \tan(\beta)}{1 - \tan(\alpha)\tan(\beta)}$

(e) $\tan(2\alpha) = \frac{2\tan\alpha}{1 - \tan^2\alpha}$