lim x→π/2 cot(x) を求める
\(\lim_{x\rightarrow \frac{\pi}{2}} \cot{\left(x\right)}\)を求める
高階の微分を求めておく
準備として\(\csc{\left(x\right)}\)の微分を求める． $$\begin{eqnarray} \frac{\mathrm{d}}{\mathrm{d}x}\csc{\left(x\right)} &=&\frac{\mathrm{d}}{\mathrm{d}x}\frac{1}{\sin{\left(x\right)}} \\&=&\sin^{-1}{\left(x\right)} \\&=&-\sin^{-2}{\left(x\right)}\left(\frac{\mathrm{d}}{\mathrm{d}x}\sin{\left(x\right)}\right) \;\cdots\;u=\sin{\left(x\right)},f=u^{-1},\frac{\mathrm{d}f}{\mathrm{d}x}=\frac{\mathrm{d}f}{\mathrm{d}u}\frac{\mathrm{d}u}{\mathrm{d}x}=-u^{-2}\frac{\mathrm{d}u}{\mathrm{d}x} \\&=&-\sin^{-2}{\left(x\right)}\left(\cos{\left(x\right)}\right) \\&=&-\frac{\cos{\left(x\right)}}{\sin^2{\left(x\right)} } \\&=&-\frac{1}{\sin{\left(x\right)}}\frac{\cos{\left(x\right)}}{\sin{\left(x\right)}} \\&=&-\csc{\left(x\right)}\cot{\left(x\right)} \end{eqnarray}$$ 一階から順に求めておく． $$\begin{eqnarray} \frac{\mathrm{d}}{\mathrm{d}x}\cot{\left(x\right)} &=&\frac{\mathrm{d}}{\mathrm{d}x}\frac{\cos{\left(x\right)}}{\sin{\left(x\right)}} \\&=&\frac{ (\frac{\mathrm{d}}{\mathrm{d}x}\cos{\left(x\right)})\sin{\left(x\right)} -\cos{\left(x\right)}(\frac{\mathrm{d}}{\mathrm{d}x}\sin{\left(x\right)}) }{\sin^2{\left(x\right)}} \;\cdots\;\left(\frac{u}{v}\right)^\prime=\frac{u^\prime v-uv^\prime}{v^2} \\&&\;\cdots\;\left(\frac{u}{v}\right)^\prime=\left(uv^{-1}\right)^\prime =u\left(v^{-1}\right)^\prime+\left(u^\prime\right)v^{-1} =u\left(-v^{-2}v^\prime\right)+ \left(u^\prime\right)v^{-1}\cdot vv^{-1} =v^{-2}\left(u^\prime v-uv^\prime \right) =\frac{u^\prime v-uv^\prime}{v^2} \\&=&\frac{ (-\sin{\left(x\right)})\sin{\left(x\right)} -\cos{\left(x\right)(\cos{\left(x\right)})} }{\sin^2{\left(x\right)}} \\&=&-\frac{ \sin^2{\left(x\right)}+\cos^2{\left(x\right)} }{\sin^2{\left(x\right)}} \\&=&-\frac{1}{\sin^2{\left(x\right)}} \\&=&-\csc^2{\left(x\right)} \\\; \\\frac{\mathrm{d}^2}{\mathrm{d}x^2}\cot{\left(x\right)} &=&\frac{\mathrm{d}}{\mathrm{d}x}\left(-\csc^2{\left(x\right)}\right) \\&=&-2\csc{\left(x\right)}\frac{\mathrm{d}}{\mathrm{d}x}\csc{\left(x\right)} \;\cdots\;u=\csc{\left(x\right)},f=-u^2,\frac{\mathrm{d}f}{\mathrm{d}x}=\frac{\mathrm{d}f}{\mathrm{d}u}\frac{\mathrm{d}u}{\mathrm{d}x}=-2u\frac{\mathrm{d}u}{\mathrm{d}x} \\&=&-2\csc{\left(x\right)}\left(-\csc{\left(x\right)}\cot{\left(x\right)}\right)\;\cdots\;準備\;参照 \\&=&2\csc^2{\left(x\right)}\cot{\left(x\right)} \;\cdots\;\frac{\mathrm{d}}{\mathrm{d}x}\csc^2{\left(x\right)}=-2\csc^2{\left(x\right)}\cot{\left(x\right)}でもある(後で使う)． \\\; \\\frac{\mathrm{d}^3}{\mathrm{d}x^3}\cot{\left(x\right)} &=&\frac{\mathrm{d}}{\mathrm{d}x}\left(2\csc^2{\left(x\right)}\cot{\left(x\right)}\right) \\&=&2\frac{\mathrm{d}}{\mathrm{d}x}\left(\csc^2{\left(x\right)}\cot{\left(x\right)}\right) \\&=&2\left\{ \left(\frac{\mathrm{d}}{\mathrm{d}x}\csc^2{\left(x\right)}\right)\cot{\left(x\right)} +\csc^2{\left(x\right)}\left(\frac{\mathrm{d}}{\mathrm{d}x}\cot{\left(x\right)}\right) \right\} \;\cdots\;(uv)^\prime=u^\prime v+u v^\prime \\&=&2\left\{ \left(-2\csc^2{\left(x\right)}\cot{\left(x\right)}\right)\cot{\left(x\right)} +\csc^2{\left(x\right)}\left(-\csc^2{\left(x\right)}\right) \right\} \;\cdots\;\frac{\mathrm{d}}{\mathrm{d}x}\csc^2{\left(x\right)}=-2\csc^2{\left(x\right)}\cot{\left(x\right)} ,\frac{\mathrm{d}}{\mathrm{d}x}\cot{\left(x\right)}=-\csc^2{\left(x\right)} \\&=&-2\left( 2\csc^2{\left(x\right)}\cot^2{\left(x\right)} +\csc^4{\left(x\right)} \right) \\\; \\\frac{\mathrm{d}^4}{\mathrm{d}x^4}\cot{\left(x\right)} &=&\frac{\mathrm{d}}{\mathrm{d}x} \left(-2\left( 2\csc^2{\left(x\right)}\cot^2{\left(x\right)} +\csc^4{\left(x\right)} \right)\right) \\&=&-2\left[ \frac{\mathrm{d}}{\mathrm{d}x}\left( 2\csc^2{\left(x\right)}\cot^2{\left(x\right)} +\csc^4{\left(x\right)} \right)\right] \\&=&-2\left[ \frac{\mathrm{d}}{\mathrm{d}x}\left( 2\csc^2{\left(x\right)}\cot^2{\left(x\right)} \right) +\frac{\mathrm{d}}{\mathrm{d}x}\left( \csc^4{\left(x\right)} \right) \right] \;\cdots\;(u+v)^\prime=u^\prime+v^\prime \\&=&-2\left[2\left\{ \left(\frac{\mathrm{d}}{\mathrm{d}x}\csc^2{\left(x\right)}\right) \cot^2{\left(x\right)} +\csc^2{\left(x\right)}\left(\frac{\mathrm{d}}{\mathrm{d}x}\cot^2{\left(x\right)}\right) \right\} + 4\csc^3{\left(x\right)} \left( -\csc{\left(x\right)}\cot{\left(x\right)} \right) \right] \;\cdots\;(uv)^\prime=u^\prime v+u v^\prime ,u=\csc{\left(x\right)},f=u^4,\frac{\mathrm{d}f}{\mathrm{d}x}=\frac{\mathrm{d}f}{\mathrm{d}u}\frac{\mathrm{d}u}{\mathrm{d}x}=4u^3\frac{\mathrm{d}u}{\mathrm{d}x} ,\frac{\mathrm{d}}{\mathrm{d}x}\csc{\left(x\right)}=-\csc{\left(x\right)}\cot{\left(x\right)} \\&=&-2\left[2\left\{ (-2\csc^2{\left(x\right)}\cot{\left(x\right)}) \cot^2{\left(x\right)} +\csc^2{\left(x\right)}(2\cot{\left(x\right)}\left(-\csc^2{\left(x\right)}\right)) \right\} -4\csc^4{\left(x\right)}\cot{\left(x\right)} \right] \;\cdots\;\frac{\mathrm{d}}{\mathrm{d}x}\csc^2{\left(x\right)}=-2\csc^2{\left(x\right)}\cot{\left(x\right)} ,u=\cot{\left(x\right)},f=u^2,\frac{\mathrm{d}f}{\mathrm{d}x}=\frac{\mathrm{d}f}{\mathrm{d}u}\frac{\mathrm{d}u}{\mathrm{d}x}=2u\frac{\mathrm{d}u}{\mathrm{d}x} ,\frac{\mathrm{d}}{\mathrm{d}x}\cot{\left(x\right)}=-\csc^2{\left(x\right)} \\&=&-2\left[2\left\{ -2\csc^2{\left(x\right)}\cot^3{\left(x\right)} -2\csc^4{\left(x\right)}\cot{\left(x\right)} \right\} -4\csc^4{\left(x\right)}\cot{\left(x\right)} \right] \\&=&-2\left[ -4\csc^2{\left(x\right)}\cot^3{\left(x\right)} -4\csc^4{\left(x\right)}\cot{\left(x\right)} -4\csc^4{\left(x\right)}\cot{\left(x\right)} \right] \\&=&-2\left[ -4\csc^2{\left(x\right)}\cot^3{\left(x\right)} -8\csc^4{\left(x\right)}\cot{\left(x\right)} \right] \\&=&8\csc^2{\left(x\right)}\cot{\left(x\right)}\left(\cot^2{\left(x\right)}+2\csc^2{\left(x\right)}\right) \\\; \\\frac{\mathrm{d}^5}{\mathrm{d}x^5}\cot{\left(x\right)} &=&\frac{\mathrm{d}}{\mathrm{d}x}\left\{ 8\csc^2{\left(x\right)}\cot{\left(x\right)}\left(\cot^2{\left(x\right)}+2\csc^2{\left(x\right)}\right) \right\} \\&=&4\frac{\mathrm{d}}{\mathrm{d}x} \left\{ 2\csc^2{\left(x\right)}\cot{\left(x\right)}\left(\cot^2{\left(x\right)}+2\csc^2{\left(x\right)}\right) \right\} \\&=&4\left\{ \left(\frac{\mathrm{d}}{\mathrm{d}x}2\csc^2{\left(x\right)}\cot{\left(x\right)}\right) \left(\cot^2{\left(x\right)}+2\csc^2{\left(x\right)}\right) + 2\csc^2{\left(x\right)}\cot{\left(x\right)} \left(\frac{\mathrm{d}}{\mathrm{d}x}\left(\cot^2{\left(x\right)}+2\csc^2{\left(x\right)}\right)\right) \right\} \\&=&4\left[ \left\{-2\left( 2\csc^2{\left(x\right)}\cot^2{\left(x\right)} +\csc^4{\left(x\right)} \right)\right\} \left(\cot^2{\left(x\right)}+2\csc^2{\left(x\right)}\right) + 2\csc^2{\left(x\right)}\cot{\left(x\right)} \left(\frac{\mathrm{d}}{\mathrm{d}x}\cot^2{\left(x\right)}+2\frac{\mathrm{d}}{\mathrm{d}x}\csc^2{\left(x\right)}\right) \right] \\&=&4\left[ \left( -4\csc^2{\left(x\right)}\cot^2{\left(x\right)} -2\csc^4{\left(x\right)} \right) \left(\cot^2{\left(x\right)}+2\csc^2{\left(x\right)}\right) + 2\csc^2{\left(x\right)}\cot{\left(x\right)} \left\{ \left(-2\csc^2{\left(x\right)}\cot{\left(x\right)}\right) +2\left(-2\csc^2{\left(x\right)}\cot{\left(x\right)}\right) \right\} \right] \\&=&4\left\{ \left( -4\csc^2{\left(x\right)}\cot^2{\left(x\right)}\left(\cot^2{\left(x\right)}+2\csc^2{\left(x\right)}\right) -2\csc^4{\left(x\right)}\left(\cot^2{\left(x\right)}+2\csc^2{\left(x\right)}\right) \right) + 2\csc^2{\left(x\right)}\cot{\left(x\right)}\cdot-2\csc^2{\left(x\right)}\cot{\left(x\right)} +2\csc^2{\left(x\right)}\cot{\left(x\right)}\cdot2\left(-2\csc^2{\left(x\right)}\cot{\left(x\right)}\right) \right\} \\&=&4\left( -4\csc^2{\left(x\right)}\cot^4{\left(x\right)}-8\csc^4{\left(x\right)}\cot^2{\left(x\right)} -2\csc^4{\left(x\right)}\cot^2{\left(x\right)}-4\csc^6{\left(x\right)} -4\csc^4{\left(x\right)}\cot^2{\left(x\right)} -8\csc^4{\left(x\right)}\cot^2{\left(x\right)} \right) \\&=&4\left( -4\csc^2{\left(x\right)}\cot^4{\left(x\right)} -4\csc^6{\left(x\right)} -22\csc^4{\left(x\right)}\cot^2{\left(x\right)} \right) \\&=&-8\left( 2\csc^6{\left(x\right)} +2\csc^2{\left(x\right)}\cot^4{\left(x\right)} +11\csc^4{\left(x\right)}\cot^2{\left(x\right)} \right) \end{eqnarray}$$\(x=\frac{\pi}{2}\)でのテーラー展開を求めておく
$$\begin{eqnarray} \cot{\left(x\right)}&=& \frac{1}{0!}\left[\left.\frac{\mathrm{d}^0}{\mathrm{d}x^0}\cot{\left(x\right)}\right|_{x=\frac{\pi}{2}}\right]\left(x-\frac{\pi}{2}\right)^0 \\&&+\frac{1}{1!}\left[\left.\frac{\mathrm{d}^1}{\mathrm{d}x^1}\cot{\left(x\right)}\right|_{x=\frac{\pi}{2}}\right]\left(x-\frac{\pi}{2}\right)^1 \\&&+\frac{1}{2!}\left[\left.\frac{\mathrm{d}^2}{\mathrm{d}x^2}\cot{\left(x\right)}\right|_{x=\frac{\pi}{2}}\right]\left(x-\frac{\pi}{2}\right)^2 \\&&+\frac{1}{3!}\left[\left.\frac{\mathrm{d}^3}{\mathrm{d}x^3}\cot{\left(x\right)}\right|_{x=\frac{\pi}{2}}\right]\left(x-\frac{\pi}{2}\right)^3 \\&&+\frac{1}{4!}\left[\left.\frac{\mathrm{d}^4}{\mathrm{d}x^4}\cot{\left(x\right)}\right|_{x=\frac{\pi}{2}}\right]\left(x-\frac{\pi}{2}\right)^4 \\&&+\frac{1}{5!}\left[\left.\frac{\mathrm{d}^5}{\mathrm{d}x^5}\cot{\left(x\right)}\right|_{x=\frac{\pi}{2}}\right]\left(x-\frac{\pi}{2}\right)^5 \\&&+\cdots \\&=& \frac{1}{1} \left[\cot{\left(\frac{\pi}{2}\right)}\right]\left(x-\frac{\pi}{2}\right)^0 \\&&+\frac{1}{1}\left[-\csc^2{\left(\frac{\pi}{2}\right)}\right]\left(x-\frac{\pi}{2}\right)^1 \\&&+\frac{1}{2}\left[2\cot{\left(\frac{\pi}{2}\right)}\csc^2{\left(\frac{\pi}{2}\right)}\right]\left(x-\frac{\pi}{2}\right)^2 \\&&+\frac{1}{6}\left[-2\left(2\csc^2{\left(\frac{\pi}{2}\right)}\cot^2{\left(\frac{\pi}{2}\right)}+\csc^4{\left(\frac{\pi}{2}\right)}\right)\right]\left(x-\frac{\pi}{2}\right)^3 \\&&+\frac{1}{24}\left[8\csc^2{\left(\frac{\pi}{2}\right)}\cot{\left(\frac{\pi}{2}\right)}\left(\cot^2{\left(\frac{\pi}{2}\right)}+2\csc^2{\left(\frac{\pi}{2}\right)}\right)\right]\left(x-\frac{\pi}{2}\right)^4 \\&&+\frac{1}{120}\left[-8\left( 2\csc^6{\left(\frac{\pi}{2}\right)} +2\csc^2{\left(\frac{\pi}{2}\right)}\cot^4{\left(\frac{\pi}{2}\right)} +11\csc^4{\left(\frac{\pi}{2}\right)}\cot^2{\left(\frac{\pi}{2}\right)} \right)\right]\left(x-\frac{\pi}{2}\right)^5 \\&&+\cdots \\&=& \left[0\right]\cdot 1 \\&&+\left[-1\cdot1^2\right]\left(x-\frac{\pi}{2}\right) \\&&+\frac{1}{2}\left[2\cdot 0 \cdot 1^2\right]\left(x-\frac{\pi}{2}\right)^2 \\&&+\frac{1}{6}\left[-2\left(2 \cdot 1^2 \cdot 0^2+ 1^4\right)\right]\left(x-\frac{\pi}{2}\right)^3 \\&&+\frac{1}{24}\left[8\cdot1^2 \cdot0\left(0^2+2\cdot1^2\right)\right]\left(x-\frac{\pi}{2}\right)^4 \\&&+\frac{1}{120}\left[-8\left(2\cdot 1^6+2\cdot 1^2\cdot0^4 +11\cdot1^4\cdot0^2\right)\right]\left(x-\frac{\pi}{2}\right)^5 \\&&+\cdots \\&=& -\left(x-\frac{\pi}{2}\right)-\frac{2}{6}\left(x-\frac{\pi}{2}\right)^3-\frac{16}{120}\left(x-\frac{\pi}{2}\right)^5+\cdots \\&=& -\left(x-\frac{\pi}{2}\right)-\frac{1}{3}\left(x-\frac{\pi}{2}\right)^3-\frac{2}{15}\left(x-\frac{\pi}{2}\right)^5+\cdots \end{eqnarray}$$\(\lim_{x\rightarrow \frac{\pi}{2}} \cot{\left(x\right)}\)を求める
$$\begin{eqnarray} \\\lim_{x\rightarrow \frac{\pi}{2}} \cot{\left(x\right)}&=& \lim_{x\rightarrow \frac{\pi}{2}} \left[ -\left(x-\frac{\pi}{2}\right)-\frac{1}{3}\left(x-\frac{\pi}{2}\right)^3-\frac{2}{15}\left(x-\frac{\pi}{2}\right)^5+\cdots \right] \\&=&0 \end{eqnarray}$$tan(θ)=?
original: https://www.youtube.com/watch?v=rtUBRi2q4mo
\(\theta=\)?
問い(\(\tan{\left(\theta\right)=}\))
\(\overline{AC}\)に垂直な補助線\(\overline{BD}\)を引く
$$\begin{eqnarray}
\angle C &=& \pi - \left(\angle A + \angle B\right)
\\ &=& \pi-\left\{\theta+\left(\frac{\pi}{2}+\theta\right)\right\}
\\ &=& \frac{\pi}{2}-2\theta
\end{eqnarray}$$
$$\begin{eqnarray}
\overline{BD}&=&2\sin{\left(\frac{\pi}{2}-2\theta\right)}
\\\overline{CD}&=&2\cos{\left(\frac{\pi}{2}-2\theta\right)}
\\\overline{AD}&=&\overline{AC}-\overline{CD}
\\&=&5-2\cos{\left(\frac{\pi}{2}-2\theta\right)}
\end{eqnarray}$$
$$\begin{eqnarray}
\tan{\left(\theta\right)}&=&\frac{\overline{BD}}{\overline{AD}}
\\&=&\frac{2\sin{\left(\frac{\pi}{2}-2\theta\right)}}{5-2\cos{\left(\frac{\pi}{2}-2\theta\right)}}
\\&=&\frac{2\cos{\left(2\theta\right)}}{5-2\sin{\left(2\theta\right)}}
\;\cdots\;\cos{\left(\frac{\pi}{2}-x\right)}=\sin{\left(x\right)},\;\sin{\left(\frac{\pi}{2}-x\right)}=\cos{\left(x\right)}
\\&=&\frac{2}{5}\;\cdots\;\tan{\left(\theta\right)}=\frac{B\cos{\left(2\theta\right)}}{A-B\sin{\left(2\theta\right)}}ならば\tan{\left(\theta\right)}=\frac{B}{A}(下記)
\end{eqnarray}$$
$$\begin{eqnarray}
\tan{\left(\theta\right)}&=&\frac{B\cos{\left(2\theta\right)}}{A-B\sin{\left(2\theta\right)}}
\\\frac{1}{\tan{\left(\theta\right)}}&=&\frac{A-B\sin{\left(2\theta\right)}}{B\cos{\left(2\theta\right)}}
\\&=&\frac{1}{\cos{\left(2\theta\right)}}\left\{\frac{A}{B}-\frac{\cancel{B}}{\cancel{B}}\sin{\left(2\theta\right)}\right\}
\\\frac{\cos{\left(2\theta\right)}}{\tan{\left(\theta\right)}}&=&\frac{A}{B}-\sin{\left(2\theta\right)}
\\\frac{\cos{\left(2\theta\right)}}{\tan{\left(\theta\right)}}+\sin{\left(2\theta\right)}&=&\frac{A}{B}
\\\frac{\cos{\left(2\theta\right)}+\tan{\left(\theta\right)}\sin{\left(2\theta\right)}}{\tan{\left(\theta\right)}}&=&\frac{A}{B}
\\\frac{\cos^2{\left(\theta\right)}-\sin^2{\left(\theta\right)}+2\tan{\left(\theta\right)}\sin{\left(\theta\right)}\cos{\left(\theta\right)}}{\tan{\left(\theta\right)}}&=&\frac{A}{B}
\;\cdots\;\cos{\left(2\theta\right)}=\cos^2{\left(\theta\right)}-\sin^2{\left(\theta\right)},\;\sin{\left(2\theta\right)}=2\sin{\left(\theta\right)}\cos{\left(\theta\right)}
\\\frac{\cos^2{\left(\theta\right)}-\sin^2{\left(\theta\right)}+2\frac{\sin{\left(\theta\right)}}{\cancel{\cos{\left(\theta\right)}}}\sin{\left(\theta\right)}\cancel{\cos{\left(\theta\right)}}}{\tan{\left(\theta\right)}}&=&\frac{A}{B}
\;\cdots\;\tan{\left(\theta\right)}=\frac{\sin{\left(\theta\right)}}{\cos{\left(\theta\right)}}
\\\frac{\cos^2{\left(\theta\right)}-\sin^2{\left(\theta\right)}+2\sin^2{\left(\theta\right)}}{\tan{\left(\theta\right)}}&=&\frac{A}{B}
\\\frac{\cos^2{\left(\theta\right)}+\sin^2{\left(\theta\right)}}{\tan{\left(\theta\right)}}&=&\frac{A}{B}
\\\frac{1}{\tan{\left(\theta\right)}}&=&\frac{A}{B}\;\cdots\;\cos^2{\left(\theta\right)}+\sin^2{\left(\theta\right)}=1
\\\tan{\left(\theta\right)}&=&\frac{B}{A}
\end{eqnarray}$$
sin(mx)sin(nx), cos(mx)cos(nx), sin(mx)cos(nx) の0から2πまでの積分
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
$$\begin{eqnarray} m,n\in\mathbb{N} \\\int_{0}^{2\pi}\sin{\left(m\;x\right)}\sin{\left(n\;x\right)}\mathrm{d}x&=& \begin{cases} \pi\;\cdots\;m=n \\0\;\cdots\;m\ne n \end{cases} \\\int_{0}^{2\pi}\cos{\left(m\;x\right)}\cos{\left(n\;x\right)}\mathrm{d}x&=& \begin{cases} \pi\;\cdots\;m=n \\0\;\cdots\;m\ne n \end{cases} \\\int_{0}^{2\pi}\sin{\left(m\;x\right)}\cos{\left(n\;x\right)}\mathrm{d}x&=&0 \end{eqnarray}$$(sin(x)+3)/(cos(x)+2)の最大値最小値 (一階微分を用いて極値として解く)
問い
(問いを知った動画) $$ \begin{eqnarray} \frac{\sin{\left(\theta\right)}+3}{\cos{\left(\theta\right)}+2}の\frac{-\pi}{2}\leq\theta\leq\frac{\pi}{2}における最大値最小値を求めよ \end{eqnarray} $$一階微分の解で求める
(別解:直線の傾きとして解く)極値を得るため一階微分を求める
$$ \begin{eqnarray} f(\theta)&=&\frac{\sin{\left(\theta\right)}+3}{\cos{\left(\theta\right)}+2} \\f^{\prime}(\theta)&=&\left\{\left(\sin{\left(\theta\right)}+3\right)\left(\cos{\left(\theta\right)}+2\right)^{-1}\right\}^{\prime} \\&=&\left(\sin{\left(\theta\right)}+3\right)\left\{\left(\cos{\left(\theta\right)}+2\right)^{-1}\right\}^{\prime} +\left(\sin{\left(\theta\right)}+3\right)^{\prime}\left(\cos{\left(\theta\right)}+2\right)^{-1} \;\cdots\;\href{https://shikitenkai.blogspot.com/2020/02/blog-post.html}{(fg)^{\prime}=fg^{\prime}+f^{\prime}g} \\&=&\left(\sin{\left(\theta\right)}+3\right)\left\{(-1)\left(\cos{\left(\theta\right)}+2\right)^{-2}\left(\cos{\left(\theta\right)}+2\right)^{\prime}\right\} +\cos{\left(\theta\right)}\left(\cos{\left(\theta\right)}+2\right)^{-1} \\&=&\left(\sin{\left(\theta\right)}+3\right)\left\{(-1)\left(\cos{\left(\theta\right)}+2\right)^{-2}\left(-\sin{\left(\theta\right)}\right)\right\} +\cos{\left(\theta\right)}\left(\cos{\left(\theta\right)}+2\right)^{-1} \\&=&\left(\sin{\left(\theta\right)}+3\right)\frac{\sin{\left(\theta\right)}}{\left(\cos{\left(\theta\right)}+2\right)^{2}} +\frac{\cos{\left(\theta\right)}}{\cos{\left(\theta\right)}+2} \\&=&\frac{\left(\sin{\left(\theta\right)}+3\right)\sin{\left(\theta\right)}}{\left(\cos{\left(\theta\right)}+2\right)^{2}} +\frac{\cos{\left(\theta\right)}}{\cos{\left(\theta\right)}+2}\frac{\cos{\left(\theta\right)+2}}{\cos{\left(\theta\right)}+2} \\&=&\frac{\left(\sin{\left(\theta\right)}+3\right)\sin{\left(\theta\right)}}{\left(\cos{\left(\theta\right)}+2\right)^{2}} +\frac{\cos{\left(\theta\right)}\left(\cos{\left(\theta\right)}+2\right)}{\left(\cos{\left(\theta\right)}+2\right)^2} \\&=&\frac{\left(\sin{\left(\theta\right)}+3\right)\sin{\left(\theta\right)}+\cos{\left(\theta\right)}\left(\cos{\left(\theta\right)}+2\right)}{\left(\cos{\left(\theta\right)}+2\right)^{2}} \\&=&\frac{\sin{\left(\theta\right)}^2+3\sin{\left(\theta\right)}+\cos{\left(\theta\right)}^2+2\cos{\left(\theta\right)}}{\left(\cos{\left(\theta\right)}+2\right)^{2}} \\&=&\frac{1+3\sin{\left(\theta\right)}+2\cos{\left(\theta\right)}}{\left(\cos{\left(\theta\right)}+2\right)^{2}} \;\cdots\;\sin{\left(\theta\right)}^2+\cos{\left(\theta\right)}^2=1 \\&&\;\cdots\;\cos{\left(\theta\right)}は\pm1の範囲なので分母が0や\inftyの心配はない \end{eqnarray} $$極値を得るため一階微分が0となる解を求める
$$ \begin{eqnarray} \\f^{\prime}(\theta)&=&0 \;\cdots\;fの極大極小値となる\thetaを求める \\1+3\sin{\left(\theta\right)}+2\cos{\left(\theta\right)}&=&0 \;\cdots\;f^{\prime}の分子が0になればf^{\prime}自体も0であるので分子のみ \\1+\sqrt{3^2+2^2}\sin{\left(\theta + \tan^{-1}{\left(\frac{2}{3}\right)}\right)}&=&0 \;\cdots\;三角凾数の合成\;a\sin{\left(\theta\right)}+b\cos{\left(\theta\right)}=\sqrt{a^2+b^2}\sin{\left(\theta + \tan^{-1}{\left(\frac{b}{a}\right)}\right)} \\1+\sqrt{13}\sin{\left(\theta + \tan^{-1}{\left(\frac{2}{3}\right)}\right)}&=&0 \\\sqrt{13}\sin{\left(\theta + \tan^{-1}{\left(\frac{2}{3}\right)}\right)}&=&-1 \\\sin{\left(\theta + \tan^{-1}{\left(\frac{2}{3}\right)}\right)}&=&\frac{-1}{\sqrt{13}} \end{eqnarray} $$\(\sin\)の逆凾数
$$ \begin{eqnarray} \\\theta + \tan^{-1}{\left(\frac{2}{3}\right)}&=&\sin^{-1}\left(\frac{-1}{\sqrt{13}}\right),-\pi-\sin^{-1}\left(\frac{-1}{\sqrt{13}}\right) \\&&\;\cdots\;三角凾数は単射ではないので，逆凾数を得るために制限が加えられる \\&&\;\cdots\;y=\sin^{-1}\left(x\right)とすると定義域-1\leq x \leq 1，値域\frac{\pi}{2}\leq y\leq-\frac{-\pi}{2} \\&&\;\cdots\;制限がなければ一つのy軸の値(\sin{\left(x\right)})に対してy軸からx軸方向の正負に等距離となる単位円上の2つの点に対する角度(\theta_1,\theta_2)であり2つ値を持つ \\&&\;\cdots\;\frac{-1}{\sqrt{13}}は負なので制限のない状態では単位円第三,四象限上の点に対する角度(\theta_1,\theta_2)である \\&&\;\cdots\;制限を考慮すると\sin^{-1}\left(\frac{-1}{\sqrt{13}}\right)は単位円第四象限上の点への角度(\theta_1)の値のみを表す \\&&\;\cdots\;このため第三象限の点への角度(\theta_2)の値は-\piから\sin^{-1}\left(\frac{-1}{\sqrt{13}}\right)の値を引いたものとして得るようにする \end{eqnarray} $$\(\sin^{-1}\left(\frac{-1}{\sqrt{13}}\right)\)のケース
$$ \begin{eqnarray} \theta_1 + \tan^{-1}{\left(\frac{2}{3}\right)}&=&\sin^{-1}\left(\frac{-1}{\sqrt{13}}\right) \\\theta_1 &=&\sin^{-1}\left(\frac{-1}{\sqrt{13}}\right)-\tan^{-1}{\left(\frac{2}{3}\right)} \\ &=&\tan^{-1}\left( \frac{\frac{-1}{\sqrt{13}}}{\sqrt{1-\left(\frac{-1}{\sqrt{13}}\right)^2}} \right)-\tan^{-1}{\left(\frac{2}{3}\right)} \\ &=&\tan^{-1}\left( \frac{\frac{-1}{\sqrt{13}}}{\sqrt{1-\frac{1}{13}}} \right)-\tan^{-1}{\left(\frac{2}{3}\right)} \\ &=&\tan^{-1}\left( \frac{\frac{-1}{\sqrt{13}}}{\sqrt{\frac{13}{13}-\frac{1}{13}}} \right)-\tan^{-1}{\left(\frac{2}{3}\right)} \\ &=&\tan^{-1}\left( \frac{\frac{-1}{\sqrt{13}}}{\sqrt{\frac{12}{13}}} \right)-\tan^{-1}{\left(\frac{2}{3}\right)} \\ &=&\tan^{-1}\left( \frac{\frac{-1}{\sqrt{13}}}{\frac{\sqrt{12}}{\sqrt{13}}}\frac{\sqrt{13}}{\sqrt{13}} \right)-\tan^{-1}{\left(\frac{2}{3}\right)} \\ &=&\tan^{-1}\left( \frac{-1}{\sqrt{12}} \right)-\tan^{-1}{\left(\frac{2}{3}\right)} \\ &=&\tan^{-1}\left( \frac{-1}{2\sqrt{3}} \right)-\tan^{-1}{\left(\frac{2}{3}\right)} \\ &=&\tan^{-1}\left(\frac{\frac{-1}{2\sqrt{3}}-\frac{2}{3}}{1+\frac{-1}{2\sqrt{3}}\frac{2}{3}}\right) \;\cdots\;\tan^{-1}{\left(u\right)}\pm\tan^{-1}{\left(v\right)}=\tan^{-1}{\left(\frac{u\pm v}{1\mp uv}\right)} \\ &=&\tan^{-1}\left(\frac{\frac{-3-4\sqrt{3}}{6\sqrt{3}}}{1+\frac{-2}{6\sqrt{3}}}\right) \;\cdots\;\frac{-1}{2\sqrt{3}}\frac{2}{3}=\frac{-2}{6\sqrt{3}}=-0.192… \\ &=&\tan^{-1}\left(\frac{\frac{-3-4\sqrt{3}}{6\sqrt{3}}}{\frac{-2+6\sqrt{3}}{6\sqrt{3}}}\frac{6\sqrt{3}}{6\sqrt{3}}\right) \\ &=&\tan^{-1}\left(\frac{-3-4\sqrt{3}}{-2+6\sqrt{3}}\frac{-2-6\sqrt{3}}{-2+6\sqrt{3}}\right) \\ &=&\tan^{-1}\left(\frac{6+18\sqrt{3}+8\sqrt{3}+72}{4-108}\right) \\ &=&\tan^{-1}\left(\frac{78+26\sqrt{3}}{-104}\right) \\ &=&\tan^{-1}\left(\frac{-3-\sqrt{3}}{4}\right) \\\tan\left(\theta_1\right)&=&\frac{-3-\sqrt{3}}{4}\lt0 \\&&\;\cdots\;\tan\left(\theta_1\right)が負の値ということは\theta_1は単位円第二,四象限上の点の角度となる \\&&\;\cdots\;\tan^{-1}の制限から\theta_1は単位円第四象限上の点の角度の値である \\&&\;\cdots\;これは問いの\thetaの条件の範囲になっている． \end{eqnarray} $$\(-\pi-\sin^{-1}\left(\frac{-1}{\sqrt{13}}\right)\)のケース
$$ \begin{eqnarray} \theta_2 + \tan^{-1}{\left(\frac{2}{3}\right)}&=&-\pi-\sin^{-1}\left(\frac{-1}{\sqrt{13}}\right) \\\theta_2 &=&-\pi-\sin^{-1}\left(\frac{-1}{\sqrt{13}}\right)-\tan^{-1}{\left(\frac{2}{3}\right)} \\ &=&-\pi-\tan^{-1}\left( \frac{\frac{-1}{\sqrt{13}}}{\sqrt{1-\left(\frac{-1}{\sqrt{13}}\right)^2}} \right)-\tan^{-1}{\left(\frac{2}{3}\right)} \\ &=&-\pi-\tan^{-1}\left( \frac{\frac{-1}{\sqrt{13}}}{\sqrt{1-\frac{1}{13}}} \right)-\tan^{-1}{\left(\frac{2}{3}\right)} \\ &=&-\pi-\tan^{-1}\left( \frac{\frac{-1}{\sqrt{13}}}{\sqrt{\frac{13}{13}-\frac{1}{13}}} \right)-\tan^{-1}{\left(\frac{2}{3}\right)} \\ &=&-\pi-\tan^{-1}\left( \frac{\frac{-1}{\sqrt{13}}}{\sqrt{\frac{12}{13}}} \right)-\tan^{-1}{\left(\frac{2}{3}\right)} \\ &=&-\pi-\tan^{-1}\left( \frac{\frac{-1}{\sqrt{13}}}{\frac{\sqrt{12}}{\sqrt{13}}}\frac{\sqrt{13}}{\sqrt{13}} \right)-\tan^{-1}{\left(\frac{2}{3}\right)} \\ &=&-\pi-\tan^{-1}\left( \frac{-1}{\sqrt{12}} \right)-\tan^{-1}{\left(\frac{2}{3}\right)} \\ &=&-\pi-\tan^{-1}\left( \frac{-1}{2\sqrt{3}} \right)-\tan^{-1}{\left(\frac{2}{3}\right)} \\ &=&-\pi-\left(\tan^{-1}\left( \frac{-1}{2\sqrt{3}} \right)+\tan^{-1}{\left(\frac{2}{3}\right)}\right) \\ &=&-\pi-\tan^{-1}\left(\frac{\frac{-1}{2\sqrt{3}}+\frac{2}{3}}{1-\frac{-1}{2\sqrt{3}}\frac{2}{3}}\right) \;\cdots\;\tan^{-1}{\left(u\right)}\pm\tan^{-1}{\left(v\right)}=\tan^{-1}{\left(\frac{u\pm v}{1\mp uv}\right)} \\ &=&-\pi-\tan^{-1}\left(\frac{\frac{-3+4\sqrt{3}}{6\sqrt{3}}}{1+\frac{2}{6\sqrt{3}}}\right) \;\cdots\;\frac{1}{2\sqrt{3}}\frac{2}{3}=\frac{2}{6\sqrt{3}}=0.192… \\&=&-\pi-\tan^{-1}\left(\frac{\frac{-3+4\sqrt{3}}{6\sqrt{3}}}{\frac{2+6\sqrt{3}}{6\sqrt{3}}}\right) \\&=&-\pi-\tan^{-1}\left(\frac{\frac{-3+4\sqrt{3}}{6\sqrt{3}}}{\frac{2+6\sqrt{3}}{6\sqrt{3}}}\frac{6\sqrt{3}}{6\sqrt{3}}\right) \\&=&-\pi-\tan^{-1}\left(\frac{-3+4\sqrt{3}}{6\sqrt{3}+2}\frac{2-6\sqrt{3}}{2-6\sqrt{3}}\right) \\&=&-\pi-\tan^{-1}\left(\frac{-6+18\sqrt{3}+8\sqrt{3}-72}{4-108}\right) \\&=&-\pi-\tan^{-1}\left(\frac{-78+26\sqrt{3}}{-104}\right) \\&=&-\pi-\tan^{-1}\left(\frac{3-\sqrt{3}}{4}\right) \\&=&-\pi+\tan^{-1}\left(\frac{-3+\sqrt{3}}{4}\right) \;\cdots\;-\tan^{-1}{\left(x\right)}=\tan^{-1}{\left(-x\right)} \\\theta_2+\pi&=&\tan^{-1}\left(\frac{-3+\sqrt{3}}{4}\right) \\\tan\left(\theta_2+\pi\right)&=&\frac{-3+\sqrt{3}}{4}\lt0 \\&&\;\cdots\;\tan\left(\theta_2+\pi\right)が負の値ということは\theta_2+\piは単位円第二,四象限上の点の角度となる \\&&\;\cdots\;\tan^{-1}の制限から\theta_2+\piは単位円第四象限上の点の角度の値である \\&&\;\cdots\;\theta_2は-\piに第四象限の点の角度の値(負の値)を加えたものであり単位円第二象限上の点の角度の値となる \\&&\;\cdots\;これは問いの\thetaの条件の範囲になっていない \end{eqnarray} $$\(\tan\)の値から\(\sin,\cos\)の値を求める
$$ \begin{eqnarray} \tan\left(\theta\right)&=&\frac{\sin\left(\theta\right)}{\cos\left(\theta\right)} \\&=&\frac{\sin\left(\theta\right)}{\sqrt{1-\sin^2\left(\theta\right)}} \;\cdots\;\cos^2\left(\theta\right)+\sin^2\left(\theta\right)=1,\cos\left(\theta\right)=\sqrt{1-\sin^2\left(\theta\right)} \\\tan\left(\theta\right)\sqrt{1-\sin^2\left(\theta\right)}&=&\sin\left(\theta\right) \\\tan^2\left(\theta\right)\left(1-\sin^2\left(\theta\right)\right)&=&\sin^2\left(\theta\right) \\\tan^2\left(\theta\right)-\tan^2\left(\theta\right)\sin^2\left(\theta\right)&=&\sin^2\left(\theta\right) \\\tan^2\left(\theta\right)&=&\sin^2\left(\theta\right)+\tan^2\left(\theta\right)\sin^2\left(\theta\right) \\\tan^2\left(\theta\right)&=&\sin^2\left(\theta\right)\left(1+\tan^2\left(\theta\right)\right) \\\frac{\tan^2\left(\theta\right)}{\left(1+\tan^2\left(\theta\right)\right)}&=&\sin^2\left(\theta\right) \\\sin\left(\theta\right)&=&\frac{\tan\left(\theta\right)}{\sqrt{1+\tan^2\left(\theta\right)}} \\\cos\left(\theta\right)&=&\frac{\sin\left(\theta\right)}{\tan\left(\theta\right)} \\&=&\frac{\frac{\tan\left(\theta\right)}{\sqrt{1+\tan^2\left(\theta\right)}}}{\tan\left(\theta\right)} \\&=&\frac{\tan\left(\theta\right)}{\tan\left(\theta\right)}\frac{1}{\sqrt{1+\tan^2\left(\theta\right)}} \\&=&\frac{1}{\sqrt{1+\tan^2\left(\theta\right)}} \end{eqnarray} $$\(\sin,\cos\)の値 / \(\theta_1\)のケース
$$ \begin{eqnarray} \\\sin\left(\theta_1\right)&=&\frac{\tan\left(\theta_1\right)}{\sqrt{1+\tan^2\left(\theta_1\right)}} \\&=&\frac{\frac{-3-\sqrt{3}}{4}}{\sqrt{1+\left(\frac{-3-\sqrt{3}}{4}\right)^2}} \\&=&\frac{\frac{-3-\sqrt{3}}{4}}{\sqrt{1+\frac{9+6\sqrt{3}+3}{16}}} \\&=&\frac{\frac{-3-\sqrt{3}}{4}}{\sqrt{\frac{16+12+6\sqrt{3}}{16}}} \\&=&\frac{\frac{-3-\sqrt{3}}{4}}{\sqrt{\frac{28+6\sqrt{3}}{16}}} \\&=&\frac{\frac{-3-\sqrt{3}}{4}}{\frac{1}{4}\sqrt{28+6\sqrt{3}}} \\&=&\frac{-3-\sqrt{3}}{\sqrt{28+6\sqrt{3}}} \\&=&\frac{-3-\sqrt{3}}{\sqrt{27+6\sqrt{3}+1}} \\&=&\frac{-3-\sqrt{3}}{\sqrt{\left(3\sqrt{3}\right)^2+6\sqrt{3}+1}} \\&=&\frac{-3-\sqrt{3}}{\sqrt{\left(1+3\sqrt{3}\right)^2}} \\&=&\frac{-3-\sqrt{3}}{1+3\sqrt{3}} \\&=&\frac{-3-\sqrt{3}}{1+3\sqrt{3}}\frac{1-3\sqrt{3}}{1-3\sqrt{3}} \\&=&\frac{-3+9\sqrt{3}-\sqrt{3}+9}{1-27} \\&=&\frac{6+8\sqrt{3}}{26} \\&=&\frac{3+2\sqrt{3}}{13} \\\cos\left(\theta_1\right)&=&\frac{1}{\sqrt{1+\tan^2\left(\theta_1\right)}} \\&=&\frac{1}{\sqrt{1+\left(\frac{-3-\sqrt{3}}{4}\right)^2}} \\&=&\frac{1}{\sqrt{1+\frac{3+6\sqrt{3}+9}{16}}} \\&=&\frac{1}{\sqrt{\frac{16+3+6\sqrt{3}+9}{16}}} \\&=&\frac{1}{\sqrt{\frac{28+6\sqrt{3}}{16}}} \\&=&\frac{1}{\frac{1}{4}\sqrt{28+6\sqrt{3}}} \\&=&\frac{4}{\sqrt{28+6\sqrt{3}}} \\&=&\frac{4}{\sqrt{27+6\sqrt{3}+1}} \\&=&\frac{4}{\sqrt{(3\sqrt{3})^2+6\sqrt{3}+1}} \\&=&\frac{4}{\sqrt{(1+3\sqrt{3})^2}} \\&=&\frac{4}{1+3\sqrt{3}} \\&=&\frac{4}{1+3\sqrt{3}}\frac{1-3\sqrt{3}}{1-3\sqrt{3}} \\&=&\frac{4\left(1-3\sqrt{3}\right)}{1-27} \\&=&\frac{4\left(1-3\sqrt{3}\right)}{-26} \\&=&\frac{2\left(-1+3\sqrt{3}\right)}{13} \\&=&\frac{-2+6\sqrt{3}}{13} \end{eqnarray} $$\(\frac{\sin{\left(\theta_1\right)}+3}{\cos{\left(\theta_1\right)}+2}\)の値
$$ \begin{eqnarray} \frac{\sin{\left(\theta_1\right)}+3}{\cos{\left(\theta_1\right)}+2} &=&\frac{\frac{-3-4\sqrt{3}}{13}+3}{-2+\frac{6\sqrt{3}}{13}+2} \\&=&\frac{\frac{-3-4\sqrt{3}+39}{13}}{\frac{-2+6\sqrt{3}+26}{13}} \\&=&\frac{\frac{36-4\sqrt{3}+39}{13}}{\frac{24+6\sqrt{3}+26}{13}}\frac{13}{13} \\&=&\frac{36-4\sqrt{3}}{24+6\sqrt{3}} \\&=&\frac{4(9-\sqrt{3})}{6(4+\sqrt{3})} \\&=&\frac{4}{6}\frac{9-\sqrt{3}}{4+\sqrt{3}} \\&=&\frac{2}{3}\frac{9-\sqrt{3}}{4+\sqrt{3}}\frac{4-\sqrt{3}}{4-\sqrt{3}} \\&=&\frac{2}{3}\frac{36-9\sqrt{3}-4\sqrt{3}+3}{16-3} \\&=&\frac{2}{3}\frac{39-13\sqrt{3}}{-13} \\&=&\frac{2}{3}\left(3-\sqrt{3}\right) \end{eqnarray} $$\(\sin,\cos\)の値 / \(\theta_2\)のケース
$$ \begin{eqnarray} \\\sin\left(\theta_2+\pi\right)&=&\frac{\tan\left(\theta_2+\pi\right)}{\sqrt{1+\tan^2\left(\theta_2+\pi\right)}} \\&=&\frac{\frac{-3+\sqrt{3}}{4}}{\sqrt{1+\left(\frac{-3+\sqrt{3}}{4}\right)^2}} \\&=&\frac{\frac{-3+\sqrt{3}}{4}}{\sqrt{1+\frac{9-6\sqrt{3}+3}{16}}} \\&=&\frac{\frac{-3+\sqrt{3}}{4}}{\sqrt{\frac{16+9-6\sqrt{3}+3}{16}}} \\&=&\frac{\frac{-3+\sqrt{3}}{4}}{\sqrt{\frac{28-6\sqrt{3}}{16}}} \\&=&\frac{\frac{-3+\sqrt{3}}{4}}{\frac{1}{4}\sqrt{28-6\sqrt{3}}}\frac{4}{4} \\&=&\frac{-3+\sqrt{3}}{\sqrt{28-6\sqrt{3}}} \\&=&\frac{-3+\sqrt{3}}{\sqrt{1-6\sqrt{3}+27}} \\&=&\frac{-3+\sqrt{3}}{\sqrt{1-6\sqrt{3}+\left(3\sqrt{3}\right)^2}} \\&=&\frac{-3+\sqrt{3}}{\sqrt{\left(1-3\sqrt{3}\right)^2}} \\&=&\frac{-3+\sqrt{3}}{\left|1-3\sqrt{3}\right|} \\&=&\frac{-3+\sqrt{3}}{-\left(1-3\sqrt{3}\right)} \;\cdots\;3\sqrt{3}\gt1より1-3\sqrt{3}\lt0,A\lt0の時\left|A\right|=-A \\&=&\frac{-3+\sqrt{3}}{-1+3\sqrt{3}} \\&=&\frac{-3+\sqrt{3}}{-1+3\sqrt{3}}\frac{-1-3\sqrt{3}}{-1-3\sqrt{3}} \\&=&\frac{3+9\sqrt{3}-\sqrt{3}-9}{1-27} \\&=&\frac{-6+8\sqrt{3}}{-26} \\&=&\frac{3-4\sqrt{3}}{13} \\\sin\left(\theta_2\right)&=&-\frac{3-4\sqrt{3}}{13} \;\cdots\;\sin\left(x+\pi\right)=-\sin\left(x\right)=\sin\left(-x\right) \\&=&\frac{-3+4\sqrt{3}}{13} \\\cos\left(\theta_2+\pi\right)&=&\frac{1}{\sqrt{1+\tan^2\left(\theta_2\right)}} \\&=&\frac{1}{\sqrt{1+\left(\frac{-3+\sqrt{3}}{4}\right)^2}} \\&=&\frac{1}{\sqrt{1+\frac{9-6\sqrt{3}+3}{16}}} \\&=&\frac{1}{\sqrt{\frac{16+9-6\sqrt{3}+3}{16}}} \\&=&\frac{1}{\sqrt{\frac{28-6\sqrt{3}}{16}}} \\&=&\frac{1}{\frac{1}{4}\sqrt{28-6\sqrt{3}}} \\&=&\frac{1}{\frac{1}{4}\sqrt{28-6\sqrt{3}}}\frac{4}{4} \\&=&\frac{4}{\sqrt{28-6\sqrt{3}}} \\&=&\frac{4}{\sqrt{1-6\sqrt{3}+27}} \\&=&\frac{4}{\sqrt{1-6\sqrt{3}+\left(3\sqrt{3}\right)^2}} \\&=&\frac{4}{\sqrt{\left(1-3\sqrt{3}\right)^2}} \\&=&\frac{4}{\left|1-3\sqrt{3}\right|} \\&=&\frac{4}{-\left(1-3\sqrt{3}\right)} \;\cdots\;3\sqrt{3}\gt1より1-3\sqrt{3}\lt0,A\lt0の時\left|A\right|=-A \\&=&\frac{4}{-1+3\sqrt{3}} \\&=&\frac{4}{-1+3\sqrt{3}}\frac{-1-3\sqrt{3}}{-1-3\sqrt{3}} \\&=&\frac{4\left(-1-3\sqrt{3}\right)}{1-27} \\&=&\frac{4\left(-1-3\sqrt{3}\right)}{-26} \\&=&\frac{-2\left(-1-3\sqrt{3}\right)}{13} \\&=&\frac{2+6\sqrt{3}}{13} \\\cos\left(\theta_2\right)&=&-\frac{2+6\sqrt{3}}{13} \\&=&\frac{-2-6\sqrt{3}}{13} \end{eqnarray} $$\(\frac{\sin{\left(\theta_2\right)}+3}{\cos{\left(\theta_2\right)}+2}\)の値
$$ \begin{eqnarray} \frac{\sin{\left(\theta_2\right)}+3}{\cos{\left(\theta_2\right)}+2} &=&\frac{\frac{-3+4\sqrt{3}}{13}+3}{\frac{-2-6\sqrt{3}}{13}+2} \\&=&\frac{\frac{-3+4\sqrt{3}+39}{13}}{\frac{-2-6\sqrt{3}+26}{13}} \\&=&\frac{\frac{36+4\sqrt{3}}{13}}{\frac{24-6\sqrt{3}}{13}} \\&=&\frac{36+4\sqrt{3}}{24-6\sqrt{3}} \\&=&\frac{4}{6}\frac{9+\sqrt{3}}{4-\sqrt{3}} \\&=&\frac{2}{3}\frac{9+\sqrt{3}}{4-\sqrt{3}} \\&=&\frac{2}{3}\frac{9+\sqrt{3}}{4-\sqrt{3}}\frac{4+\sqrt{3}}{4+\sqrt{3}} \\&=&\frac{2}{3}\frac{36+9\sqrt{3}+4\sqrt{3}+3}{16-3} \\&=&\frac{2}{3}\frac{39+13\sqrt{3}}{13} \\&=&\frac{2}{3}\left(3+\sqrt{3}\right) \end{eqnarray} $$\(\frac{\sin{\left(\theta_1,2\right)}+3}{\cos{\left(\theta_1,2\right)}+2}\)の比較
$$ \begin{eqnarray} \frac{\sin{\left(\theta_1\right)}+3}{\cos{\left(\theta_1\right)}+2}&=&\frac{2}{3}\left(3-\sqrt{3}\right) \\\frac{\sin{\left(\theta_2\right)}+3}{\cos{\left(\theta_2\right)}+2}&=&\frac{2}{3}\left(3+\sqrt{3}\right) \\より \\\frac{\sin{\left(\theta_1\right)}+3}{\cos{\left(\theta_1\right)}+2}\lt\frac{\sin{\left(\theta_2\right)}+3}{\cos{\left(\theta_2\right)}+2} \end{eqnarray} $$ よって\(\theta_1\)が最小値，\(\theta_2\)が最大値の角度となるが，\(\theta_2\)は範囲外のため，範囲内での最大を求める必要がある． $$ \begin{eqnarray} \theta=\frac{\pi}{2}の場合&:&\frac{\sin{\left(\frac{\pi}{2}\right)}+3}{\cos{\left(\frac{\pi}{2}\right)}+2} &=&\frac{1+3}{0+2}=\frac{4}{2}=2 \\\theta=\frac{-\pi}{2}の場合&:&\frac{\sin{\left(\frac{-\pi}{2}\right)}+3}{\cos{\left(\frac{-\pi}{2}\right)}+2} &=&\frac{-1+3}{0+2}=\frac{2}{2}=1 \end{eqnarray} $$ \(\theta=\frac{\pi}{2}\)の値の方が大きいのでこれが範囲内での最大値となる．答え
以上より与式の最大値最小値はそれぞれ，最大値\(2\)，最小値\(\frac{2}{3}\left(3-\sqrt{3}\right)\)となる．(sin(x)+3)/(cos(x)+2)の最大値最小値 (直線の傾きとして解く)
問い
(問いを知った動画) $$ \begin{eqnarray} \frac{\sin{\left(\theta\right)}+3}{\cos{\left(\theta\right)}+2}の\frac{-\pi}{2}\leq\theta\leq\frac{\pi}{2}における最大値最小値 \end{eqnarray} $$直線の傾きとして求める
(別解: 一階微分を用いて極値として解く)問いを知った動画の通り解いてみる．
与式を直線の傾きとみて幾何の問題として解く
$$ \begin{eqnarray} k&=&\frac{\sin{\left(\theta\right)}+3}{\cos{\left(\theta\right)}+2} \\&=&\frac{\sin{\left(\theta\right)}-(-3)}{\cos{\left(\theta\right)}-(-2)} \\&&\;\cdots\;点(\cos{\left(\theta\right)},\sin{\left(\theta\right)})と点(-2,-3)を通る直線lの傾き \end{eqnarray} $$直線の方程式
$$ \begin{eqnarray} \\y-(-3)&=&k(x-(-2)) \\0&=&kx-y+2k-3 \;\cdots\;ax+by+c=0の形での直線lの式 \end{eqnarray} $$ \(点(\cos{\left(\theta\right)},\sin{\left(\theta\right)})\)は原点を中心とした半径1の円(単位円)上の点を意味する．よってこの直線\(l\)は，点(-2,-3)から半径1の円(単位円)への接線ということになる．
この接線は円の中心点(原点)からの距離が1となる直線である．
点と直線の距離
$$ \begin{eqnarray} \\d&=&\frac{\left|ap+bq+c\right|}{\sqrt{a^2+b^2}}\;\cdots\;\href{https://shikitenkai.blogspot.com/2020/06/blog-post_11.html}{直線ax+by+c=0と点(p,q)との距離dの式} \\1&=&\frac{\left|k\cdot0-1\cdot0+2k-3\right|}{\sqrt{k^2+(-1)^2}}\;\cdots\;ここでは(p,q)は原点なので(0,0) \\&=&\frac{\left|2k-3\right|}{\sqrt{k^2+1}} \\\sqrt{k^2+1}&=&\left|2k-3\right| \\k^2+1&=&(2k-3)^2 \\&=&4k^2-12k+9 \\0&=&3k^2-12k+8 \;\cdots\;単位円に円の外の点から接線を引くと2つの直線l_{1,2}が引けるため傾きも2つ得られる \end{eqnarray} $$二次方程式の解(の公式)
$$ \begin{eqnarray} \\k_{l_{1,2}}&=&\frac{-(-12)\pm\sqrt{(-12)^2-4\cdot3\cdot8}}{2\cdot3} \;\cdots\;\href{https://shikitenkai.blogspot.com/2020/11/blog-post.html}{ax^2+bx+c=0においてx=\frac{-b\pm\sqrt{b^2-4ac}}{2a}} \\&=&\frac{12\pm\sqrt{144-96}}{6} \\&=&\frac{12\pm\sqrt{48}}{6} \\&=&\frac{12\pm4\sqrt{3}}{6} \\&=&\frac{6\pm2\sqrt{3}}{3} \\&=&\frac{2}{3}(3\pm\sqrt{3}) \end{eqnarray} $$解と図との対応
$$ \begin{eqnarray} \\k_{l_1}&=&\frac{2}{3}(3+\sqrt{3}) \\k_{l_2}&=&\frac{2}{3}(3-\sqrt{3}) \\k_{l_1}&\gt&k_{l_2}\;\cdots\;よって傾きの大小関係から図中l_1の傾きk_{l_1}であり,l_2の傾きがk_{l_2}である \end{eqnarray} $$ 図より\(l_1\)の接点位置の角度\(\theta_1\)は\(\frac{\pi}{2}\)を超えているので範囲外となり，点\((0,1)\)を通る直線\(j\)の傾きが最大値となる． $$ \begin{eqnarray} k_{j}&=&\frac{\sin{\left(\theta\right)}+3}{\cos{\left(\theta\right)}+2} &=&\frac{\sin{\left(\frac{\pi}{2}\right)}+3}{\cos{\left(\frac{\pi}{2}\right)}+2} &=&\frac{1+3}{0+2} &=&\frac{4}{2} &=&2 \end{eqnarray} $$答え
以上より与式の最大値最小値はそれぞれ，最大値\(2\)，最小値\(\frac{2}{3}\left(3-\sqrt{3}\right)\)となる．確認:角度\(\theta_{1,2}\)を求める
直交する直線の傾き
直線\(l\)と直交する直線\(l^{\prime}\)の傾き\(s\)が\(\tan{\left(\theta\right)}\)となる． また直交する直線同士の傾きは掛け合わせると-1となる． $$ \begin{eqnarray} k_{1}\cdot s_{1}&=&-1 \\s_1&=&\frac{-1}{k_{1}} \\k_{2}\cdot s_{2}&=&-1 \\s_2&=&\frac{-1}{k_{2}} \end{eqnarray} $$\(s_1\)から\(\theta_{1}\)を求める
直線\(l_1\)と直交する直線\(l_1^{\prime}\)の傾き\(s_1\)を求める． $$ \begin{eqnarray} \\s_1&=&\frac{-1}{k_1} \\&=&\frac{-1}{\frac{2}{3}(3+\sqrt{3})} \;\cdots\;k_1=\frac{2}{3}(3+\sqrt{3}) \\&=&\frac{3}{2}\frac{-1}{3+\sqrt{3}} \\&=&\frac{3}{2}\frac{-1}{3+\sqrt{3}}\frac{3-\sqrt{3}}{3-\sqrt{3}} \\&=&\frac{3}{2}\frac{-3+\sqrt{3}}{9-3} \\&=&\frac{3}{2}\frac{-3+\sqrt{3}}{6} \\&=&\frac{3}{2}\frac{1}{6}\left(-3+\sqrt{3}\right) \\&=&\frac{1}{4}\left(-3+\sqrt{3}\right) \\&=&\frac{-3+\sqrt{3}}{4}\lt0 \\\theta_1&=&\tan^{-1}\left(\frac{-3+\sqrt{3}}{4}\right),\tan^{-1}\left(\frac{-3+\sqrt{3}}{4}\right)+\pi \;\cdots\;\tan^{-1}の値域を\frac{-\pi}{2}から\frac{\pi}{2}とする \end{eqnarray} $$ \(s_1\)が負の値ということは\(s_1=\tan\left(\theta_1\right)\)とした際の\(\theta_1\)は単位円第二,四象限上の点の角度となる．が，図よりこの角度は単位円第二象限上の点の角度なので\(\theta_1=\tan^{-1}\left(s_1\right)+\pi=\tan^{-1}\left(\frac{-3+\sqrt{3}}{4}\right)+\pi\)となる．
\(s_2\)から\(\theta_{2}\)を求める
直線\(l_2\)と直交する直線\(l_2^{\prime}\)の傾き\(s_2\)を求める． $$ \begin{eqnarray} \\s_2&=&\frac{-1}{k_2} \\&=&\frac{-1}{\frac{2}{3}(3-\sqrt{3})} \;\cdots\;k_2=\frac{2}{3}(3-\sqrt{3}) \\&=&\frac{3}{2}\frac{-1}{3-\sqrt{3}} \\&=&\frac{3}{2}\frac{-1}{3-\sqrt{3}}\frac{3+\sqrt{3}}{3+\sqrt{3}} \\&=&\frac{3}{2}\frac{-3-\sqrt{3}}{9-3} \\&=&\frac{3}{2}\frac{-3-\sqrt{3}}{6} \\&=&\frac{3}{2}\frac{1}{6}\left(-3-\sqrt{3}\right) \\&=&\frac{1}{4}\left(-3-\sqrt{3}\right) \\&=&\frac{-3-\sqrt{3}}{4}\lt0 \\\theta_2&=&\tan^{-1}\left(\frac{-3-\sqrt{3}}{4}\right),\tan^{-1}\left(\frac{-3-\sqrt{3}}{4}\right)+\pi \;\cdots\;\tan^{-1}の値域を\frac{-\pi}{2}から\frac{\pi}{2}とする \end{eqnarray} $$ \(s_2\)が負の値ということは\(s_2=\tan\left(\theta_2\right)\)とした際の\(\theta_2\)は単位円第二,四象限上の点の角度となる．が，図よりこの角度は単位円第四象限上の点の角度なので\(\theta_2=\tan^{-1}\left(s_2\right)=\tan^{-1}\left(\frac{-3-\sqrt{3}}{4}\right)\)となる．
回転行列と三角凾数の加法定理
角度\(\alpha\)と角度\(\beta\)の回転行列を掛ける
$$\begin{array}{rcl} \begin{pmatrix} cos(\alpha) & -sin(\alpha) \\ sin(\alpha) & cos(\alpha) \\ \end{pmatrix} \begin{pmatrix} cos(\beta) & -sin(\beta) \\ sin(\beta) & cos(\beta) \\ \end{pmatrix} \begin{pmatrix} x\\y \end{pmatrix} &=& \begin{pmatrix} cos(\alpha)cos(\beta)-sin(\alpha)sin(\beta) & -cos(\alpha)sin(\beta)-sin(\alpha)cos(\beta) \\ sin(\alpha)cos(\beta)+cos(\alpha)sin(\beta) & -sin(\alpha)sin(\beta)+cos(\alpha)cos(\beta) \\ \end{pmatrix} \begin{pmatrix} x\\y \end{pmatrix}\\ &=& \begin{pmatrix} cos(\alpha)cos(\beta)-sin(\alpha)sin(\beta) & -(sin(\alpha)cos(\beta)+cos(\alpha)sin(\beta)) \\ sin(\alpha)cos(\beta)+cos(\alpha)sin(\beta) & -sin(\alpha)sin(\beta)+cos(\alpha)cos(\beta) \\ \end{pmatrix} \begin{pmatrix} x\\y \end{pmatrix}\\ &=& \begin{pmatrix} cos(\alpha+\beta) & -sin(\alpha+\beta) \\ sin(\alpha+\beta) & cos(\alpha+\beta) \\ \end{pmatrix} \begin{pmatrix} x\\y \end{pmatrix}\,\dotso\,これが角度(\alpha+\beta)の回転行列と等しい．\\ \end{array}$$ 対応する要素同士が加法定理となる． $$\begin{array}{rcl} cos(\alpha+\beta)&=&cos(\alpha)cos(\beta)-sin(\alpha)sin(\beta)\\ sin(\alpha+\beta)&=&sin(\alpha)cos(\beta)+cos(\alpha)sin(\beta)\\ \end{array}$$角度\(\alpha\)と角度\(-\beta\)の回転行列を掛ける
$$\begin{array}{rcl} \begin{pmatrix} cos(\alpha) & -sin(\alpha) \\ sin(\alpha) & cos(\alpha) \\ \end{pmatrix} \begin{pmatrix} cos(-\beta) & -sin(-\beta) \\ sin(-\beta) & cos(-\beta) \\ \end{pmatrix} \begin{pmatrix} x\\y \end{pmatrix} &=& \begin{pmatrix} cos(\alpha)cos(-\beta)-sin(\alpha)sin(-\beta) & -cos(\alpha)sin(-\beta)-sin(\alpha)cos(-\beta) \\ sin(\alpha)cos(-\beta)+cos(\alpha)sin(-\beta) & -sin(\alpha)sin(-\beta)+cos(\alpha)cos(-\beta) \\ \end{pmatrix} \begin{pmatrix} x\\y \end{pmatrix}\\ &=& \begin{pmatrix} cos(\alpha)cos(\beta)+sin(\alpha)sin(\beta) & cos(\alpha)sin(\beta)-sin(\alpha)cos(\beta) \\ sin(\alpha)cos(\beta)-cos(\alpha)sin(\beta) & sin(\alpha)sin(\beta)+cos(\alpha)cos(\beta) \\ \end{pmatrix} \begin{pmatrix} x\\y \end{pmatrix}\,\dotso\,cos(-\beta)=cos(\beta), sin(-\beta)=-sin(\beta)を適用．\\ &=& \begin{pmatrix} cos(\alpha)cos(\beta)+sin(\alpha)sin(\beta) & -(sin(\alpha)cos(\beta)-cos(\alpha)sin(\beta)) \\ sin(\alpha)cos(\beta)-cos(\alpha)sin(\beta) & cos(\alpha)cos(\beta)+sin(\alpha)sin(\beta) \\ \end{pmatrix} \begin{pmatrix} x\\y \end{pmatrix}\\ &=& \begin{pmatrix} cos(\alpha-\beta) & -sin(\alpha-\beta) \\ sin(\alpha-\beta) & cos(\alpha-\beta) \\ \end{pmatrix} \begin{pmatrix} x\\y \end{pmatrix}\,\dotso\,これが角度(\alpha-\beta)の回転行列と等しい．\\ \end{array}$$ $$\begin{array}{rcl} cos(\alpha -\beta)&=&cos(\alpha)cos(\beta)+sin(\alpha)sin(\beta)\\ sin(\alpha-\beta)&=&sin(\alpha)cos(\beta)-cos(\alpha)sin(\beta)\\ \end{array}$$角度\(\alpha\)と角度\(\beta\)が一致(\(\alpha=\beta=\theta\))
$$\begin{array}{rcl} \begin{pmatrix} cos(\theta) & -sin(\theta) \\ sin(\theta) & cos(\theta) \\ \end{pmatrix} \begin{pmatrix} cos(\theta) & -sin(\theta) \\ sin(\theta) & cos(\theta) \\ \end{pmatrix} \begin{pmatrix} x\\y \end{pmatrix} &=& \begin{pmatrix} cos(\theta)cos(\theta)-sin(\theta)sin(\theta) & -cos(\theta)sin(\theta)-sin(\theta)cos(\theta) \\ sin(\theta)cos(\theta)+cos(\theta)sin(\theta) & -sin(\theta)sin(\theta)+cos(\theta)cos(\theta) \\ \end{pmatrix} \begin{pmatrix} x\\y \end{pmatrix}\\ &=& \begin{pmatrix} cos(\theta)^2-sin(\theta)^2 & -2sin(\theta)cos(\theta) \\ 2sin(\theta)cos(\theta) & cos(\theta)^2-sin(\theta)^2 \\ \end{pmatrix} \begin{pmatrix} x\\y \end{pmatrix}\\ &=& \begin{pmatrix} cos(2\theta) & -sin(2\theta) \\ sin(2\theta) & cos(2\theta) \\ \end{pmatrix} \begin{pmatrix} x\\y \end{pmatrix}\,\dotso\,これが角度(2\theta)の回転行列と等しい．\\ \end{array}$$ 対応する要素同士が倍角公式となる． $$\begin{array}{rcl} cos(2\theta)&=&cos(\theta)^2-sin(\theta)^2\\ sin(2\theta)&=&2sin(\theta)cos(\theta)\\ \end{array}$$ド・モアブルの定理(de Moivre's theorem)と三角凾数(trigonometric function)の3倍角の公式
ド・モアブルの定理
$$\displaystyle (\cos\theta+i\sin\theta)^n=\cos(n\theta)+i\sin(n\theta)$$3倍角の公式(Triple-angle formulae)
$$\begin{array}{rcl} \displaystyle (\cos\theta+i\sin\theta)^3 &=& \displaystyle \cos^3\theta+3\cos^2\theta(i\sin\theta)+3\cos\theta(i\sin\theta)^2+(i\sin\theta)^3\\ &=& \displaystyle \cos^3\theta+i3\cos^2\theta\sin\theta-3\cos\theta\sin^2\theta-i\sin^3\theta\\ &=& \displaystyle \left(\cos^3\theta-3\cos\theta\sin^2\theta\right)+i\left(3\cos^2\theta\sin\theta-\sin^3\theta\right)\\ &=& \displaystyle \cos(3\theta)+i\sin(3\theta)\,\dotso\,これがド・モアブルの定理の結果と等しい． \end{array}$$ $$\begin{array}{rcl} \displaystyle \cos(3\theta)&=& \cos^3\theta-3\cos\theta\sin^2\theta\\ &=& \cos^3\theta-3\cos\theta(1-\cos^2\theta)\\ &=& \cos^3\theta-3\cos\theta+3\cos^3\theta\\ &=& 4\cos^3\theta-3\cos\theta\\ \displaystyle \sin(3\theta)&=& 3\cos^2\theta\sin\theta-\sin^3\theta\\ &=& 3\sin\theta(1-\sin^2\theta)-\sin^3\theta\\ &=& 3\sin\theta-3\sin^3\theta-\sin^3\theta\\ &=& 3\sin\theta-4\sin^3\theta\\ \end{array}$$4倍角の場合
$$\begin{array}{rcl} \displaystyle (\cos\theta+i\sin\theta)^4 &=& \displaystyle \cos^4\theta+4\cos^3\theta(i\sin\theta)+6\cos^2\theta(i\sin\theta)^2+4\cos\theta(i\sin\theta)^3+(i\sin\theta)^4\\ &=& \displaystyle \cos^4\theta+i4\cos^3\theta\sin\theta-6\cos^2\theta\sin^2\theta-i4\cos\theta\sin^3\theta+\sin^4\theta\\ &=& \displaystyle \left(\cos^4\theta-6\cos^2\theta\sin^2\theta+\sin^4\theta\right)+i\left(4\cos^3\theta\sin\theta-4\cos\theta\sin^3\theta\right)\\ &=& \displaystyle \cos(4\theta)+i\sin(4\theta)\,\dotso\,これがド・モアブルの定理の結果と等しい． \end{array}$$ $$\begin{array}{rcl} \displaystyle \cos(4\theta)&=&\cos^4\theta-6\cos^2\theta\sin^2\theta+\sin^4\theta\\ &=&\cos^4\theta-6\cos^2\theta(1-\cos^2\theta)+(1-\cos^2\theta)^2\\ &=&\cos^4\theta-6\cos^2\theta+6\cos^4\theta+1-2\cos^2\theta+\cos^4\theta\\ &=&\cos^4\theta+6\cos^4\theta+\cos^4\theta-6\cos^2\theta-2\cos^2\theta+1\\ &=&8\cos^4\theta-8\cos^2\theta+1\\ \displaystyle \sin(4\theta)&=&4\cos^3\theta\sin\theta-4\cos\theta\sin^3\theta\\ &=&\cos\theta(4\cos^2\theta\sin\theta-4\sin^3\theta)\\ &=&\cos\theta(4\sin\theta(1-\sin^2\theta)-4\sin^3\theta)\\ &=&\cos\theta(4\sin\theta-4\sin^3\theta-4\sin^3\theta)\\ &=&\cos\theta(4\sin\theta-8\sin^2\theta)\\ \end{array}$$
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