1.2 Answer

1.1

1.

(a)

(b)

(c)

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

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

Then,，

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

(d)

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

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

Then，

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

(e)

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

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

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

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

Thus,

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

1.2

2.

(a)

$$\sin{\frac{5\pi}{4}} = -\frac{\sqrt{2}}{2}$$

or

$$\sin{\frac{5\pi}{4}} = \sin{(\pi + \frac{\pi}{4})} = - \sin{\frac{\pi}{4}} = -\frac{\sqrt{2}}{2}$$

(b)

$$\sin{\frac{7\pi}{12}} = \sin{(\frac{3\pi}{12} + \frac{4\pi}{12})}$$

$$= \sin{(\frac{\pi}{4} + \frac{\pi}{3})} = \sin{\frac{\pi}{4}}\cos{\frac{\pi}{3}} + \cos{\frac{\pi}{4}}\sin{\frac{\pi}{3}}$$

$$= \frac{\sqrt{2}}{2} \cdot \frac{1}{2} + \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{2} + \sqrt{6}}{4}$$

(c) $${\cos^{2}{\alpha} = \frac{1 + \cos{2\alpha}}{2}}$$ より

$$\cos{\frac{\pi}{8}} = \sqrt{\frac{1 + \cos{\frac{\pi}{4}}}{2}} = \frac{\sqrt{2 + \sqrt{2}}}{2}$$

3.

(a) Let $y = \sin^{-1}{(-\frac{1}{2})}$. Then

$$-\frac{1}{2} = \sin{y}, \ -\frac{\pi}{2} \leq y \leq \frac{\pi}{2}$$

Therefore, $${\sin^{-1}{(-\frac{1}{2})} = -\frac{\pi}{6}}$$

(b)Let $y = \cos^{-1}{(-1)}$. Then

$$-1 = \cos{y}, \ 0 \leq y \leq \pi$$

Thus, $${\cos^{-1}{(-1)} = \pi}$$

(c) Let $y = \tan^{-1}{(-1)}$. Then

$$-1 = \tan{y}, \ -\frac{\pi}{2} \leq y \leq \frac{\pi}{2}$$

Thus, $${\tan^{-1}{(-1)} = -\frac{\pi}{4}}$$

(d) Let $y = \tan^{-1}{(\sqrt{3})}$. Then

$$\sqrt{3} = \tan{y}, \ -\frac{\pi}{2} \leq y \leq \frac{\pi}{2}$$

Thus, $${\tan^{-1}{(\sqrt{3})} = \frac{\pi}{3}}$$

4.

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

Thus, $x = \sin{u} = \cos{v} = \sin{(\frac{\pi}{2} - v)}$．Therefore, $${u = \frac{\pi}{2} - v}$$ となり $${u + v = \frac{\pi}{2}}$$.

5.

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

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