準備
準備の準備
$$\begin{eqnarray}
\cos{\left(\alpha+\beta\right)}&=&\href{https://shikitenkai.blogspot.com/2019/06/blog-post.html}{
\cos{\left(\alpha\right)}\cos{\left(\beta\right)}-\sin{\left(\alpha\right)}\sin{\left(\beta\right)}
}
\\\cos{\left(\alpha-\beta\right)}&=&\cos{\left(\alpha\right)}\cos{\left(\beta\right)}+\sin{\left(\alpha\right)}\sin{\left(\beta\right)}
\\\sin{\left(\alpha+\beta\right)}&=&\href{https://shikitenkai.blogspot.com/2019/06/blog-post.html}{
\sin{\left(\alpha\right)}\cos{\left(\beta\right)}+\cos{\left(\alpha\right)}\sin{\left(\beta\right)}
}
\\\sin{\left(\alpha-\beta\right)}&=&\sin{\left(\alpha\right)}\cos{\left(\beta\right)}-\cos{\left(\alpha\right)}\sin{\left(\beta\right)}
\end{eqnarray}$$
準備1
$$\begin{eqnarray}
\\\cos{\left(\alpha-\beta\right)}-\cos{\left(\alpha+\beta\right)}&=&\cos{\left(\alpha\right)}\cos{\left(\beta\right)}+\sin{\left(\alpha\right)}\sin{\left(\beta\right)}
-\left\{\cos{\left(\alpha\right)}\cos{\left(\beta\right)}-\sin{\left(\alpha\right)}\sin{\left(\beta\right)}\right\}
\;\cdots\;準備の準備
\\&=&\cancel{\cos{\left(\alpha\right)}\cos{\left(\beta\right)}}+\sin{\left(\alpha\right)}\sin{\left(\beta\right)}
\cancel{-\cos{\left(\alpha\right)}\cos{\left(\beta\right)}}+\sin{\left(\alpha\right)}\sin{\left(\beta\right)}
\\&=&2\sin{\left(\alpha\right)}\sin{\left(\beta\right)}
\\\sin{\left(\alpha\right)}\sin{\left(\beta\right)}&=&\frac{\cos{\left(\alpha-\beta\right)}-\cos{\left(\alpha+\beta\right)}}{2}
\end{eqnarray}$$
準備2
$$\begin{eqnarray}
\\\cos{\left(\alpha-\beta\right)}+\cos{\left(\alpha+\beta\right)}&=&\cos{\left(\alpha\right)}\cos{\left(\beta\right)}+\sin{\left(\alpha\right)}\sin{\left(\beta\right)}
+\left\{\cos{\left(\alpha\right)}\cos{\left(\beta\right)}-\sin{\left(\alpha\right)}\sin{\left(\beta\right)}\right\}
\;\cdots\;準備の準備
\\&=&\cos{\left(\alpha\right)}\cos{\left(\beta\right)}+\cancel{\sin{\left(\alpha\right)}\sin{\left(\beta\right)}}
+\cos{\left(\alpha\right)}\cos{\left(\beta\right)}-\cancel{\sin{\left(\alpha\right)}\sin{\left(\beta\right)}}
\\&=&2\cos{\left(\alpha\right)}\cos{\left(\beta\right)}
\\\cos{\left(\alpha\right)}\cos{\left(\beta\right)}&=&\frac{\cos{\left(\alpha-\beta\right)}+\cos{\left(\alpha+\beta\right)}}{2}
\end{eqnarray}$$
準備3
$$\begin{eqnarray}
\cos{\left(2x\right)}&=&\cos{\left(x+x\right)}
\\&=&\cos(x)\cos(x)-\sin(x)\sin(x)\;\cdots\;準備の準備
\\&=&\cos^2{\left(x\right)}-\sin^2(x)
\\&=&(1-\sin^2(x))-\sin^2(x)
\\&=&1-2\sin^2(x)
\\\cos{\left(2x\right)}-1&=&-2\sin^2(x)
\\1-\cos{\left(2x\right)}&=&2\sin^2(x)
\\\sin^2(x)&=&\frac{1-\cos(2x)}{2}=\frac{1}{2}-\frac{\cos(2x)}{2}
\end{eqnarray}$$
準備4
$$\begin{eqnarray}
\cos{\left(2x\right)}&=&\cos{\left(x+x\right)}
\\&=&\cos(x)\cos(x)-\sin(x)\sin(x)\;\cdots\;準備の準備
\\&=&\cos^2{\left(x\right)}-\sin^2(x)
\\&=&\cos^2{\left(x\right)}-\left(1-\cos^2{\left(x\right)}\right)
\\&=&2\cos^2(x)-1
\\\cos{\left(2x\right)}+1&=&2\cos^2(x)
\\\cos^2(x)&=&\frac{1+\cos(2x)}{2}=\frac{1}{2}+\frac{\cos(2x)}{2}
\end{eqnarray}$$
準備5
$$\begin{eqnarray}
\\\sin{\left(\alpha-\beta\right)}+\sin{\left(\alpha+\beta\right)}&=&
\sin{\left(\alpha\right)}\cos{\left(\beta\right)}-\cos{\left(\alpha\right)}\sin{\left(\beta\right)}
+\left\{\sin{\left(\alpha\right)}\cos{\left(\beta\right)}+\cos{\left(\alpha\right)}\sin{\left(\beta\right)}\right\}
\;\cdots\;準備の準備
\\&=&\sin{\left(\alpha\right)}\cos{\left(\beta\right)}\cancel{-\cos{\left(\alpha\right)}\sin{\left(\beta\right)}}
+\sin{\left(\alpha\right)}\cos{\left(\beta\right)}\cancel{+\cos{\left(\alpha\right)}\sin{\left(\beta\right)}}
\\&=&2\sin{\left(\alpha\right)}\cos{\left(\beta\right)}
\\\sin{\left(\alpha\right)}\cos{\left(\beta\right)}&=&\frac{\sin{\left(\alpha-\beta\right)}+\sin{\left(\alpha+\beta\right)}}{2}
\end{eqnarray}$$
準備6
$$\begin{eqnarray}
\sin{\left(2x\right)}&=&\sin{\left(x+x\right)}
\\&=&\sin{\left(x\right)}\cos{\left(x\right)}+\cos{\left(x\right)}\sin{\left(x\right)}
\;\cdots\;準備の準備
\\&=&2\sin{\left(x\right)}\cos{\left(x\right)}
\\\sin{\left(x\right)}\cos{\left(x\right)}&=&\frac{\sin{\left(2x\right)}}{2}
\end{eqnarray}$$
\(\sin{\left(m\;x\right)}\sin{\left(n\;x\right)}\)
\(m,n\in\mathbb{N},m\ne n\)
$$\begin{eqnarray}
&&\int_{0}^{2\pi}\sin{\left(m\;x\right)}\sin{\left(n\;x\right)}\mathrm{d}x\;\cdots\;m,n\in\mathbb{N},m\ne n
\\&=&\int_{0}^{2\pi}\frac{1}{2}\left[
\cos{\left(mx-nx\right)}
-\cos{\left(mx+nx\right)}
\right]\mathrm{d}x\;\cdots\;準備1
\\&=&\int_{0}^{2\pi}\frac{1}{2}\left[
\cos{\left(\left(m-n\right)\;x\right)}
-\cos{\left(\left(m+n\right)\;x\right)}
\right]\mathrm{d}x
\\&=&\frac{1}{2}\left[
\int_{0}^{2\pi}\cos{\left(\left(m-n\right)\;x\right)}\mathrm{d}x
-\int_{0}^{2\pi}\cos{\left(\left(m+n\right)\;x\right)}\mathrm{d}x
\right]
\\&=&\frac{1}{2}\left[
\left[\frac{1}{m-n}\sin{\left(\left(m-n\right)\;x\right)}\right]_{0}^{2\pi}
-\left[\frac{1}{m+n}\sin{\left(\left(m+n\right)\;x\right)}\right]_{0}^{2\pi}
\right]
\\&=&\frac{1}{2}\left[
\frac{1}{m-n} \left[\sin{\left(\left(m-n\right)\;2\pi\right)} -
\sin{\left(\left(m-n\right)\;0\right)}\right]
-\frac{1}{m+n} \left[\sin{\left(\left(m+n\right)\;2\pi\right)} -
\sin{\left(\left(m+n\right)\;0\right)}\right]
\right]
\\&=&\frac{1}{2}\left[
\frac{1}{m-n}\left[0-0\right]
-\frac{1}{m+n}\left[0-0\right]
\right]
\\&=&\frac{1}{2}\left[
\frac{0}{m-n}
-\frac{0}{m+n}
\right]
\\&=&\frac{1}{2}\;0
\\&=&0
\end{eqnarray}$$
\(m,n\in\mathbb{N},m= n\)
$$\begin{eqnarray}
&&\int_{0}^{2\pi}\sin{\left(m\;x\right)}\sin{\left(n\;x\right)}\mathrm{d}x\;\cdots\;m,n\in\mathbb{N},m=n
\\&=&\int_{0}^{2\pi}\sin{\left(m\;x\right)}\sin{\left(m\;x\right)}\mathrm{d}x
\\&=&\int_{0}^{2\pi}\sin^2{\left(m\;x\right)}\mathrm{d}x
\\&=&\frac{1}{m}\int_{0}^{2m\pi}\sin^2{\left(u\right)}\mathrm{d}u\;\cdots\;u=mx,\frac{\mathrm{d}u}{\mathrm{d}x}=m,\mathrm{d}x=\frac{1}{m}\mathrm{d}u
\\&=&\frac{1}{m}\int_{0}^{2m\pi}\left[\frac{1}{2}-\frac{1}{2}\cos{\left(2u\right)}\right]\mathrm{d}u\;\cdots\;準備3
\\&=&\frac{1}{2m}\left[\int_{0}^{2m\pi}\mathrm{d}u-\int_{0}^{2m\pi}\cos{\left(2u\right)}\mathrm{d}u\right]
\\&=&\frac{1}{2m}\left[\left[u\right]_{0}^{2m\pi}-\left[\frac{1}{2}\sin{\left(2u\right)}\right]_0^{2m\pi}\right]
\\&=&\frac{1}{2m}\left[\left[2m\pi-0\right]-\left[\frac{1}{2}\sin{\left(2\cdot2m\pi\right)}-\frac{1}{2}\sin{\left(2\cdot0\right)}\right]\right]
\\&=&\frac{1}{2m}\left[2m\pi-\left[0-0\right]\right]
\\&=&\frac{1}{\cancel{2m}}\cancel{2m}\pi
\\&=&\pi
\end{eqnarray}$$
\(\cos{\left(m\;x\right)}\cos{\left(n\;x\right)}\)
\(m,n\in\mathbb{N},m\ne n\)
$$\begin{eqnarray}
&&\int_{0}^{2\pi}\cos{\left(m\;x\right)}\cos{\left(n\;x\right)}\mathrm{d}x\;\cdots\;m,n\in\mathbb{N},m\ne n
\\&=&\int_{0}^{2\pi}\frac{1}{2}\left[
\cos{\left(mx-nx\right)}
+\cos{\left(mx+nx\right)}
\right]\mathrm{d}x\;\cdots\;準備2
\\&=&\int_{0}^{2\pi}\frac{1}{2}\left[
\cos{\left(\left(m-n\right)\;x\right)}
+\cos{\left(\left(m+n\right)\;x\right)}
\right]\mathrm{d}x
\\&=&\frac{1}{2}\left[
\int_{0}^{2\pi}\cos{\left(\left(m-n\right)\;x\right)}\mathrm{d}x
+\int_{0}^{2\pi}\cos{\left(\left(m+n\right)\;x\right)}\mathrm{d}x
\right]
\\&=&\frac{1}{2}\left[
\left[\frac{1}{m-n}\sin{\left(\left(m-n\right)\;x\right)}\right]_{0}^{2\pi}
+\left[\frac{1}{m+n}\sin{\left(\left(m+n\right)\;x\right)}\right]_{0}^{2\pi}
\right]
\\&=&\frac{1}{2}\left[
\frac{1}{m-n} \left[\sin{\left(\left(m-n\right)\;2\pi\right)} -
\sin{\left(\left(m-n\right)\;0\right)}\right]
+\frac{1}{m+n} \left[\sin{\left(\left(m+n\right)\;2\pi\right)} -
\sin{\left(\left(m+n\right)\;0\right)}\right]
\right]
\\&=&\frac{1}{2}\left[
\frac{1}{m-n}\left[0-0\right]
+\frac{1}{m+n}\left[0-0\right]
\right]
\\&=&\frac{1}{2}\left[
\frac{0}{m-n}
+\frac{0}{m+n}
\right]
\\&=&\frac{1}{2}\;0
\\&=&0
\end{eqnarray}$$
\(m,n\in\mathbb{N},m= n\)
$$\begin{eqnarray}
&&\int_{0}^{2\pi}\cos{\left(m\;x\right)}\cos{\left(n\;x\right)}\mathrm{d}x\;\cdots\;m,n\in\mathbb{N},m=n
\\&=&\int_{0}^{2\pi}\cos{\left(m\;x\right)}\cos{\left(m\;x\right)}\mathrm{d}x
\\&=&\int_{0}^{2\pi}\cos^2{\left(m\;x\right)}\mathrm{d}x
\\&=&\frac{1}{m}\int_{0}^{2m\pi}\cos^2{\left(u\right)}\mathrm{d}u\;\cdots\;u=mx,\frac{\mathrm{d}u}{\mathrm{d}x}=m,\mathrm{d}x=\frac{1}{m}\mathrm{d}u
\\&=&\frac{1}{m}\int_{0}^{2m\pi}\left[\frac{1}{2}+\frac{1}{2}\cos{\left(2u\right)}\right]\mathrm{d}u\;\cdots\;準備4
\\&=&\frac{1}{2m}\left[\int_{0}^{2m\pi}\mathrm{d}u+\int_{0}^{2m\pi}\cos{\left(2u\right)}\mathrm{d}u\right]
\\&=&\frac{1}{2m}\left[\left[u\right]_{0}^{2m\pi}+\left[\frac{1}{2}\sin{\left(2u\right)}\right]_0^{2m\pi}\right]
\\&=&\frac{1}{2m}\left[\left[2m\pi-0\right]+\left[\frac{1}{2}\sin{\left(2\cdot2m\pi\right)}-\frac{1}{2}\sin{\left(2\cdot0\right)}\right]\right]
\\&=&\frac{1}{2m}\left[2m\pi+\left[0-0\right]\right]
\\&=&\frac{1}{\cancel{2m}}\cancel{2m}\pi
\\&=&\pi
\end{eqnarray}$$
\(\sin{\left(m\;x\right)}\cos{\left(n\;x\right)}\)
\(m,n\in\mathbb{N},m\ne n\)
$$\begin{eqnarray}
&&\int_{0}^{2\pi}\sin{\left(m\;x\right)}\cos{\left(n\;x\right)}\mathrm{d}x\;\cdots\;m,n\in\mathbb{N},m\ne n
\\&=&\int_{0}^{2\pi}\frac{1}{2}\left[
\sin{\left(mx-nx\right)}
+\sin{\left(mx+nx\right)}
\right]\mathrm{d}x\;\cdots\;準備5
\\&=&\int_{0}^{2\pi}\frac{1}{2}\left[
\sin{\left((m-n)x\right)}
+\sin{\left((m+n)x\right)}
\right]\mathrm{d}x
\\&=&\frac{1}{2}\left[
\int_{0}^{2\pi}\sin{\left((m-n)x\right)}\mathrm{d}x
+\int_{0}^{2\pi}\sin{\left((m+n)x\right)}\mathrm{d}x
\right]
\\&=&\frac{1}{2}\left[
\left[
\frac{-1}{m-n}
\cos{\left((m-n)x\right)}
\right]_{0}^{2\pi}
+\left[
\frac{-1}{m+n}
\cos{\left((m+n)x\right)}
\right]_{0}^{2\pi}
\right]
\\&=&\frac{1}{2}\left[
\frac{-1}{m-n}\left[
\cos{\left((m-n)2\pi\right)}-\cos{\left((m-n)0\right)}
\right]
+\frac{-1}{m+n}\left[
\cos{\left((m-n)2\pi\right)}-\cos{\left((m-n)0\right)}
\right]
\right]
\\&=&\frac{1}{2}\left[
\frac{-1}{m-n}\left[1-1\right]
+\frac{-1}{m+n}\left[1-1\right]
\right]
\\&=&\frac{1}{2}\left[
\frac{-1}{m-n}0
+\frac{-1}{m+n}0
\right]
\\&=&\frac{1}{2}\left[
0
+0
\right]
\\&=&\frac{1}{2}0
\\&=&0
\end{eqnarray}$$
\(m,n\in\mathbb{N},m= n\)
$$\begin{eqnarray}
&&\int_{0}^{2\pi}\sin{\left(m\;x\right)}\cos{\left(n\;x\right)}\mathrm{d}x\;\cdots\;m,n\in\mathbb{N},m= n
\\&=&\int_{0}^{2\pi}\sin{\left(m\;x\right)}\cos{\left(m\;x\right)}\mathrm{d}x
\\&=&\int_{0}^{2\pi}\frac{1}{2}\sin{\left(2m\;x\right)}\mathrm{d}x\;\cdots\;準備6
\\&=&\frac{1}{2}\int_{0}^{4m\pi}\frac{1}{2m}\sin{\left(u\right)}\mathrm{d}u\;\cdots\;u=2mx,\;\frac{\mathrm{d}u}{\mathrm{d}x}=2m,\;\mathrm{d}x=\frac{1}{2m}\mathrm{d}u
\\&=&\frac{1}{4m}\left[-\cos{\left(u\right)}\right]_{0}^{4m\pi}
\\&=&\frac{1}{4m}\left[-\cos{\left(4m\pi\right)}-\left(-\cos{\left(0\right)}\right)\right]
\\&=&\frac{1}{4m}\left[-1-\left(-1\right)\right]
\\&=&\frac{1}{4m}\left[-1+1\right]
\\&=&\frac{1}{4m}0
\\&=&0
\end{eqnarray}$$
まとめ
$$\begin{eqnarray}
m,n\in\mathbb{N}
\\\int_{0}^{2\pi}\sin{\left(m\;x\right)}\sin{\left(n\;x\right)}\mathrm{d}x&=&
\begin{cases}
\pi\;\cdots\;m=n
\\0\;\cdots\;m\ne n
\end{cases}
\\\int_{0}^{2\pi}\cos{\left(m\;x\right)}\cos{\left(n\;x\right)}\mathrm{d}x&=&
\begin{cases}
\pi\;\cdots\;m=n
\\0\;\cdots\;m\ne n
\end{cases}
\\\int_{0}^{2\pi}\sin{\left(m\;x\right)}\cos{\left(n\;x\right)}\mathrm{d}x&=&0
\end{eqnarray}$$
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