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1.2 Answer

1.1

1.

(a)

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(b)

\begin{figure}\begin{center} \includegraphics[width=8cm]{CALCFIG/CALCANS/1-2-1-b.eps} \end{center}\end{figure}

(c)

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

Then,，

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

(d)


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

Then，

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

(e)


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

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

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

Thus,

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

1.2

2.

(a)

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

or

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

(b)

$\displaystyle \sin{\frac{7\pi}{12}}$ $\displaystyle =$ $\displaystyle \sin{(\frac{3\pi}{12} + \frac{4\pi}{12})}$  
  $\displaystyle =$ $\displaystyle \sin{(\frac{\pi}{4} + \frac{\pi}{3})} = \sin{\frac{\pi}{4}}\cos{\frac{\pi}{3}} + \cos{\frac{\pi}{4}}\sin{\frac{\pi}{3}}$  
  $\displaystyle =$ $\displaystyle \frac{\sqrt{2}}{2} \cdot \frac{1}{2} + \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{2} + \sqrt{6}}{4}$  

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

$\displaystyle \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

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

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

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

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

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

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

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

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

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

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

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

4.

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

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

5.

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

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

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Next: Limit of function Up: FUNCTIONS Previous: elementary functions