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．
0 件のコメント:
コメントを投稿