数学 – 円の中にひそむ関数 - 三角関数 – (正弦・余弦の加法定理2)

学習環境

数学読本〈2〉簡単な関数/平面図形と式/指数関数・対数関数/三角関数 (松坂和夫(著)、岩波書店)の第8章(円の中にひそむ関数 - 三角関数)、8.2(加法定理)、正弦・余弦の加法定理、問16.を取り組んでみる。

問16.

$\begin{array}{l} \sin \frac{π}{12} \\ = \sin (\frac{π}{3} - \frac{π}{4}) \\ = \sin \frac{π}{3} \cos \frac{π}{4} - \cos \frac{π}{3} \sin \frac{π}{4} \\ = \frac{\sqrt{3}}{2} \cdot \frac{1}{\sqrt{2}} - \frac{1}{2} \frac{1}{\sqrt{2}} \\ = \frac{\sqrt{3} - 1}{2 \sqrt{2}} \\ = \frac{\sqrt{6} - \sqrt{2}}{4} \end{array}$
$\begin{array}{l} \cos \frac{π}{12} \\ = \cos (\frac{π}{3} - \frac{π}{4}) \\ = \cos \frac{π}{3} \cos \frac{π}{4} + \sin \frac{π}{3} \sin \frac{π}{4} \\ = \frac{1}{2} \cdot \frac{1}{\sqrt{2}} + \frac{\sqrt{3}}{2} \cdot \frac{1}{\sqrt{2}} \\ = \frac{\sqrt{2} + \sqrt{6}}{4} \end{array}$
$\begin{array}{l} \sin (\frac{π}{3} + \frac{π}{4}) \\ = \sin \frac{π}{3} \cos \frac{π}{4} + \cos \frac{π}{3} \sin \frac{π}{4} \\ = \frac{\sqrt{3}}{2} \cdot \frac{1}{\sqrt{2}} + \frac{1}{2} \cdot \frac{1}{\sqrt{2}} \\ = \frac{\sqrt{6} + \sqrt{2}}{4} \end{array}$
$\begin{array}{l} \cos (\frac{π}{3} + \frac{π}{4}) \\ = \cos \frac{π}{3} \cos \frac{π}{4} - \sin \frac{π}{3} \sin \frac{π}{4} \\ = \frac{1}{2} \cdot \frac{1}{\sqrt{2}} - \frac{\sqrt{3}}{2} \cdot \frac{1}{\sqrt{2}} \\ = \frac{\sqrt{2} - \sqrt{6}}{4} \end{array}$
$\sin \frac{2 π}{3} = \frac{\sqrt{3}}{2}$
$\cos \frac{2 π}{3} = - \frac{1}{2}$
$\sin 15^{°} = \frac{\sqrt{6} - \sqrt{2}}{4}$
$- \cos 15^{°} = - \frac{\sqrt{6} + \sqrt{2}}{4}$

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