6.3 三角関数の公式
三角関数の積和公式
三角関数の積和公式（積$\to$和・差）を求めるため，
\begin{eqnarray}
&&e^{j(\alpha+\beta)} + e^{j(\alpha-\beta)}
\nonumber \\
&=& e^{j\alpha} (e^{j\beta} + e^{-j\beta})
\nonumber \\
&=& (\cos \alpha + j \sin \alpha) 2 \cos \beta
\nonumber \\
&=& 2 \cos \alpha \cos \beta + j2\sin \alpha \cos \beta
\nonumber \\
&=& \big\{ \cos (\alpha+\beta) + \cos (\alpha-\beta) \big\}
+ j \big\{ \sin (\alpha+\beta) + \sin (\alpha-\beta) \big\}
\end{eqnarray}
同様にして，
\begin{eqnarray}
&&e^{j(\alpha+\beta)} - e^{j(\alpha-\beta)}
\nonumber \\
&=& e^{j\alpha} (e^{j\beta} - e^{-j\beta})
\nonumber \\
&=& (\cos \alpha + j \sin \alpha) j2 \sin \beta
\nonumber \\
&=& j2 \cos \alpha \sin \beta - 2\sin \alpha \sin \beta
\nonumber \\
&=& \big\{ \cos (\alpha+\beta) - \cos (\alpha-\beta) \big\}
+ j \big\{ \sin (\alpha+\beta) - \sin (\alpha-\beta) \big\}
\end{eqnarray}
これより，
\begin{align}
&2\cos \alpha \cos \beta = \cos (\alpha+\beta) + \cos (\alpha-\beta)
\\
&2\sin \alpha \cos \beta = \sin (\alpha+\beta) + \sin (\alpha-\beta)
\\
&2\cos \alpha \sin \beta = \sin (\alpha+\beta) - \sin (\alpha-\beta)
\\
&-2\sin \alpha \sin \beta = \cos (\alpha+\beta) - \cos (\alpha-\beta)
\end{align}
三角関数の和積公式
三角関数の和積公式（和・差$\to$積）は，
\begin{eqnarray}
A &=& \alpha + \beta
\\
B &=& \alpha - \beta
\end{eqnarray}
とおいて，
\begin{eqnarray}
\alpha &=& \frac{A+B}{2}
\\
\beta &=& \frac{A-B}{2}
\end{eqnarray}
より，
\begin{eqnarray}
e^{jA} + e^{jB}
&=& e^{j\left( \frac{A+B}{2} + \frac{A-B}{2} \right)}
+ e^{j\left( \frac{A+B}{2} - \frac{A-B}{2} \right)}
\nonumber \\
&=& e^{j\frac{A+B}{2}} \left( e^{j\frac{A+B}{2}} + e^{j\frac{A-B}{2}} \right)
\nonumber \\
&=& \left\{ \cos \left( \frac{A+B}{2} \right) +j\sin \left( \frac{A+B}{2} \right) \right\}
2 \cos \left( \frac{A-B}{2} \right)
\nonumber \\
&=& \cos A + \cos B + j(\sin A + \sin B)
\\
e^{jA} - e^{jB}
&=& e^{j\left( \frac{A+B}{2} + \frac{A-B}{2} \right)}
- e^{j\left( \frac{A+B}{2} - \frac{A-B}{2} \right)}
\nonumber \\
&=& e^{j\frac{A+B}{2}} \left( e^{j\frac{A+B}{2}} - e^{j\frac{A-B}{2}} \right)
\nonumber \\
&=& \left\{ \cos \left( \frac{A+B}{2} \right) +j\sin \left( \frac{A+B}{2} \right) \right\}
j2 \sin \left( \frac{A-B}{2} \right)
\nonumber \\
&=& \cos A - \cos B + j(\sin A - \sin B)
\end{eqnarray}
これより，
\begin{gather}
\cos A + \cos B = 2\cos \left( \frac{A+B}{2} \right) \cos \left( \frac{A-B}{2} \right)
\\
\sin A + \sin B = 2\sin \left( \frac{A+B}{2} \right) \cos \left( \frac{A-B}{2} \right)
\\
\cos A - \cos B = -2\sin \left( \frac{A+B}{2} \right) \sin \left( \frac{A-B}{2} \right)
\\
\sin A - \sin B = 2\cos \left( \frac{A+B}{2} \right) \sin \left( \frac{A-B}{2} \right)
\end{gather}