1.2 Trigonometric Functions

1.

(a)

(b)

(c)

$$1 = \cos^{2}{\frac{\alpha}{2}} + \sin^{2}{\frac{\alpha}{2}}$$

$$\cos{\alpha} = \cos{\frac{\alpha}{2} + \frac{\alpha}{2}} = \cos^{2}\frac{\alpha}{2} - \sin^{2}\frac{\alpha}{2}$$

Thus，

$$\cos^{2}{\frac{\alpha}{2}} = 1 + \cos{\alpha}$$

(d)

$$\sin{(\alpha + \beta)} = \sin{\alpha}\cos{\beta} + \cos{\alpha}\sin{\beta}$$

$$\sin{(\alpha - \beta)} = \sin{\alpha}\cos{\beta} - \cos{\alpha}\sin{\beta}$$

Thus，

$$\sin{\alpha}\cos{\beta} = \frac{1}{2}[\sin{(\alpha + \beta)} + \sin{(\alpha - \beta)}]$$

(e)

$$\sin{(A + B)} = \sin{A}\cos{B} + \cos{A}\sin{B}$$

$$\sin{(A - B)} = \sin{A}\cos{B} - \cos{A}\sin{B}$$

Let $A + B = \alpha, A- B = \beta$. Then

$$A = \frac{\alpha + \beta}{2}, B = \frac{\alpha - \beta}{2}$$

，

$$\sin{\alpha} + \sin{\beta} = 2[\sin{\frac{\alpha + \beta}{2}} \cos{\frac{\alpha - \beta}{2}}]$$

2.

(a) $${-\frac{\sqrt{2}}{2}}$$ (b) $${\frac{\sqrt{2} + \sqrt{6}}{4}}$$ (c) $${\frac{\sqrt{2 + \sqrt{2}}}{2}}$$

3.

(a) $${-\frac{\pi}{6}}$$ (b) $\pi$ (c) $${\frac{\pi}{4}}$$ (d) $${\frac{\pi}{3}}$$

4.

Let $${u = \sin^{-1}{x}, v = \cos^{-1}{x}}$$. Then $${x = \sin{u} (\frac{-\pi}{2} \leq u \leq \frac{\pi}{2}), x = \cos{v} (0 \leq v \leq \pi)}$$．
From this, $x = \sin{u} = \cos{v} = \sin(\frac{\pi}{2} - v)$．Thus, $${u + v = \frac{\pi}{2}}$$.

5.

(a) $${y = \sin^{-1}{(-x)} \Leftrightarrow -x = \sin{y} (\frac{-\pi}{2} \leq y \leq \frac{\pi}{2})}$$ implies $${x = - \sin{y} \Leftrightarrow x = \sin{(-y)} \Leftrightarrow}$$
$${-y = \sin^{-1}{x} \Leftrightarrow y = - \sin^{-1}{x}}$$

(b) $${y = \cos^{-1}{(-x)} \Leftrightarrow - x = \cos{y} (0 \leq y \leq \pi)}$$ implies $${x = - \cos{y} \Leftrightarrow x = \cos{(\pi - y)}}$$
$${\Leftrightarrow \cos^{-1}{x} = \pi - y \Leftrightarrow y = \pi - \cos^{-1}{x}}$$