三角関数の公式_オイラーの公式を使って

和積公式（積→和）

\( \Large \displaystyle \begin{eqnarray} \boldsymbol{ \LARGE {sin \ x \cdot sin \ y }}
&=& \frac{e^{ix} -e^{-ix}}{2i} \frac{e^{iy} -e^{-iy}}{2i}\\
&=& -\frac{1}{4} \left[ e^{i(x+y)} - e^{i(x-y)} - e^{-i(x-y)} + e^{-i(x+y)} \right] \\
&=& -\frac{1}{4} \left[ \left\{ e^{i(x+y)} + e^{-i(x+y)} \right\} - \left\{ e^{i(x-y)} + e^{-i(x-y)} \right\} \right] \\
&=& -\frac{1}{4} \left[ 2 \ cos (x+y) -2 \ cos (x-y) \right] \\
&=& \boldsymbol{ \LARGE {\frac{1}{2} \left[ \ cos (x-y) - \ cos (x+y) \right]}}
\end{eqnarray} \)

\( \Large \displaystyle \begin{eqnarray} \boldsymbol{ \LARGE {cos \ x \cdot cos \ y }}
&=& \frac{e^{ix} +e^{-ix}}{2} \frac{e^{iy} +e^{-iy}}{2}\\
&=& \frac{1}{4} \left[ e^{i(x+y)} + e^{i(x-y)} + e^{-i(x-y)} + e^{-i(x+y)} \right] \\
&=& \frac{1}{4} \left[ \left\{ e^{i(x+y)} + e^{-i(x+y)} \right\} + \left\{ e^{i(x-y)} + e^{-i(x-y)} \right\} \right] \\
&=& \frac{1}{4} \left[ 2 \ cos (x+y) +2 \ cos (x-y) \right] \\
&=& \boldsymbol{ \LARGE {\frac{1}{2} \left[ \ cos (x+y) + \ cos (x-y) \right]}}
\end{eqnarray} \)

\( \Large \displaystyle \begin{eqnarray} \boldsymbol{ \LARGE {sin \ x \cdot cos \ y }}
&=& \frac{e^{ix} -e^{-ix}}{2i} \frac{e^{iy} + e^{-iy}}{2}\\
&=& \frac{1}{4i} \left[ e^{i(x+y)} + e^{i(x-y)} - e^{-i(x-y)} - e^{-i(x+y)} \right] \\
&=& \frac{1}{4i} \left[ \left\{ e^{i(x+y)} - e^{-i(x+y)} \right\} - \left\{ e^{i(x-y)} - e^{-i(x-y)} \right\} \right] \\
&=& \frac{1}{2} \left[ \frac{ e^{i(x+y)} - e^{-i(x+y)} }{2i} - \frac{ e^{i(x-y)} - e^{-i(x-y)} }{2i} \right] \\
&=& \boldsymbol{ \LARGE {\frac{1}{2} \left[ \ sin (x+y) + \ sin (x-y) \right]}}
\end{eqnarray} \)

和積公式（和→積）

\( \Large \displaystyle \begin{eqnarray} \boldsymbol{ \LARGE {sin \ x + sin \ y}}
&=& Im \left[ e^{ix}+e^{iy} \right] \\
&=& Im \left[ e^{ \frac{i(x+y)}{2}} \left( e^{\frac{i(x-y)}{2}} + e^{\frac{-i(x-y)}{2}} \right) \right] \\
&=& Im \left[ 2 \ e^{ \frac{i(x+y)}{2}} cos {\frac{x-y}{2}} \right] \\
&=& Im \left[ 2\left( cos \frac{x+y}{2} +i \ sin \frac{x+y}{2} \right)cos {\frac{x-y}{2}} \right] \\
&=& \boldsymbol{ \LARGE {2 \ sin \frac{x+y}{2} \ cos \frac{x-y}{2}}}
\end{eqnarray} \)

\( \Large \displaystyle \begin{eqnarray} \boldsymbol{ \LARGE {sin \ x - sin \ y}}
&=& Im \left[ e^{ix}-e^{iy} \right] \\
&=& Im \left[ e^{ \frac{i(x+y)}{2}} \left( e^{\frac{i(x-y)}{2}} - e^{\frac{-i(x-y)}{2}} \right) \right] \\
&=& Im \left[ 2i \ e^{ \frac{i(x+y)}{2}} sin {\frac{x-y}{2}} \right] \\
&=& Im \left[ 2i \left( \ cos \frac{x+y}{2} +i \ sin \frac{x+y}{2} \right)sin {\frac{x-y}{2}} \right] \\
&=& \boldsymbol{ \LARGE {2 \ cos \frac{x+y}{2} \ sin \frac{x-y}{2}}}
\end{eqnarray} \)

\( \Large \displaystyle \begin{eqnarray} \boldsymbol{ \LARGE {cos \ x + cos \ y}}
&=& Re \left[ e^{ix}+e^{iy} \right] \\
&=& Re \left[ e^{ \frac{i(x+y)}{2}} \left( e^{\frac{i(x-y)}{2}} + e^{\frac{-i(x-y)}{2}} \right) \right] \\
&=& Re \left[ 2 \ e^{ \frac{i(x+y)}{2}} cos {\frac{x-y}{2}} \right] \\
&=& Re \left[ 2\left( cos \frac{x+y}{2} +i \ sin \frac{x+y}{2} \right)cos {\frac{x-y}{2}} \right] \\
&=& \boldsymbol{ \LARGE {2 \ cos \frac{x+y}{2} \ cos \frac{x-y}{2}}}
\end{eqnarray} \)

\( \Large \displaystyle \begin{eqnarray} \boldsymbol{ \LARGE {cos \ x - cos \ y}}
&=& Re \left[ e^{ix}-e^{iy} \right] \\
&=& Re \left[ e^{ \frac{i(x+y)}{2}} \left( e^{\frac{i(x-y)}{2}} - e^{\frac{-i(x-y)}{2}} \right) \right] \\
&=& Re \left[ 2i \ e^{ \frac{i(x+y)}{2}} sin {\frac{x-y}{2}} \right] \\
&=& Re \left[ 2i \left( \ cos \frac{x+y}{2} +i \ sin \frac{x+y}{2} \right)sin {\frac{x-y}{2}} \right] \\
&=& \boldsymbol{ \LARGE {-2 \ sin \frac{x+y}{2} \ sin \frac{x-y}{2}}}
\end{eqnarray} \)

次は，べき乗，