3角関数

三角関数の合成

by nomura · 2019年11月23日

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三角関数の合成

(1)

\begin{aligned} a \sin θ + b \cos θ & = \sqrt{a^{2} + b^{2}} \sin (θ + α) \end{aligned}

\begin{aligned} α & = \arcsin \frac{b}{\sqrt{a^{2} + b^{2}}} \\ = \arccos \frac{a}{\sqrt{a^{2} + b^{2}}} \end{aligned}

(2)

\begin{aligned} a \sin θ + b \cos θ & = \sqrt{a^{2} + b^{2}} \cos (θ - β) \end{aligned}

\begin{aligned} β & = \arcsin \frac{a}{\sqrt{a^{2} + b^{2}}} \\ = \arccos \frac{b}{\sqrt{a^{2} + b^{2}}} \end{aligned}

(1)

\begin{aligned} α & = \arcsin \frac{b}{\sqrt{a^{2} + b^{2}}} \\ = \arccos \frac{a}{\sqrt{a^{2} + b^{2}}} \end{aligned}

とおくと、

\begin{aligned} a \sin θ + b \cos θ & = \sqrt{a^{2} + b^{2}} (\sin θ \frac{a}{\sqrt{a^{2} + b^{2}}} + \cos θ \frac{b}{\sqrt{a^{2} + b^{2}}}) \\ = \sqrt{a^{2} + b^{2}} (\sin θ \cos α + \cos θ \sin α) \\ = \sqrt{a^{2} + b^{2}} \sin (θ + α) \end{aligned}

(2)

(1)より、

\begin{aligned} a \sin θ + b \cos θ & = \sqrt{a^{2} + b^{2}} \sin (θ + α) \\ = \sqrt{a^{2} + b^{2}} \cos (θ + α - \frac{π}{2}) \\ = \sqrt{a^{2} + b^{2}} \cos (θ - β), β = \frac{π}{2} - α \end{aligned}

\begin{aligned} β & = \frac{π}{2} - α \\ = \frac{π}{2} - \arcsin \frac{b}{\sqrt{a^{2} + b^{2}}} \\ = \arccos \frac{b}{\sqrt{a^{2} + b^{2}}} \end{aligned}

同様に、

\begin{aligned} β & = \frac{π}{2} - α \\ = \frac{π}{2} - \arccos \frac{a}{\sqrt{a^{2} + b^{2}}} \\ = \arcsin \frac{a}{\sqrt{a^{2} + b^{2}}} \end{aligned}

これより、

\begin{aligned} β & = \arccos \frac{b}{\sqrt{a^{2} + b^{2}}} \\ = \arcsin \frac{a}{\sqrt{a^{2} + b^{2}}} \end{aligned}

(2)別解

\begin{aligned} β & = \arccos \frac{b}{\sqrt{a^{2} + b^{2}}} \\ = \arcsin \frac{a}{\sqrt{a^{2} + b^{2}}} \end{aligned}

とおくと、

\begin{aligned} a \sin θ + b \cos θ & = \sqrt{a^{2} + b^{2}} (\cos θ \frac{b}{\sqrt{a^{2} + b^{2}}} + \sin θ \frac{a}{\sqrt{a^{2} + b^{2}}}) \\ = \sqrt{a^{2} + b^{2}} (\cos θ \cos β + \sin θ \sin β) \\ = \sqrt{a^{2} + b^{2}} \sin (θ - β) \end{aligned}

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