オンライン家庭教師生徒募集中!詳しくはこちらから!

東京書籍:Standard数学Ⅱ

スポンサーリンク
スポンサーリンク

文字数が多く、重くなるのでページを分割しています。
各章は下のリンクまたはページ下の「次へ」をクリックしてください。

1章 方程式・式と証明
2章 図形と方程式
4章 指数関数・対数関数
5章 微分と積分

 

このページは非常に重くなるのでレイアウトを変更しております。



3章 三角関数


Readiness check


p.108 問1
\(\theta\) \(0^\circ\) \(30^\circ\) \(45^\circ\) \(60^\circ\) \(90^\circ\)
\(\sin{}\) \(0\) \({\large \frac{1}{2}}\) \({\large \frac{1}{\sqrt{2}}}\) \({\large \frac{\sqrt{3}}{2}}\) \(1\)
\(\cos{}\) \(1\) \({\large \frac{\sqrt{3}}{2}}\) \({\large \frac{1}{\sqrt{2}}}\) \({\large \frac{1}{2}}\) \(0\)
\(\tan{}\) \(0\) \({\large \frac{1}{\sqrt{3}}}\) \(1\) \(\sqrt{3}\)
\(\theta\) \(120^\circ\) \(135^\circ\) \(150^\circ\) \(180^\circ\)
\(\sin{}\) \({\large \frac{\sqrt{3}}{2}}\) \({\large \frac{1}{\sqrt{2}}}\) \({\large \frac{1}{2}}\) \(0\)
\(\cos{}\) \(-{\large \frac{1}{2}}\) \(-{\large \frac{1}{\sqrt{2}}}\) \(-{\large \frac{\sqrt{3}}{2}}\) \(-1\)
\(\tan{}\) \(-\sqrt{3}\) \(-1\) \(-{\large \frac{1}{\sqrt{3}}}\) \(0\)

p.108 問2

\({\small (1)}~\cos{22^\circ}\) \({\small (2)}~\sin{22^\circ}\)
\({\small (3)}~{\large \frac{1}{\tan{22^\circ}}}\) \({\small (4)}~\sin{31^\circ}\)
\({\small (5)}~-\cos{31^\circ}\) \({\small (6)}~-\tan{31^\circ}\)


p.109 問3

\({\small (1)}~45^\circ~,~135^\circ\) \({\small (2)}~150^\circ\) \({\small (3)}~135^\circ\)


p.109 問4

\({\small (1)}~\)
\(\sin{\theta}={\large \frac{3}{5}}~,~\tan{\theta}=-{\large \frac{3}{4}}\)
\({\small (2)}~\)
\(\theta\) が鋭角のとき
\(\cos{\theta}={\large \frac{\sqrt{6}}{3}}~,~\tan{\theta}={\large \frac{\sqrt{2}}{2}}\)
\(\theta\) が鈍角のとき
\(\cos{\theta}=-{\large \frac{\sqrt{6}}{3}}~,~\tan{\theta}=-{\large \frac{\sqrt{2}}{2}}\)
\({\small (3)}~\)
\(\cos{\theta}={\large \frac{\sqrt{5}}{3}}~,~\sin{\theta}={\large \frac{2}{3}}\)


1節 三角関数


p.110 問1

\({\small (1)}~\)

\({\small (2)}~\)

\({\small (3)}~\)

\({\small (4)}~\)


p.111 問2

\({\small (1)}~45^\circ+360^\circ\times n\) (\(n\) は整数)
\({\small (2)}~180^\circ+360^\circ\times n\) (\(n\) は整数)
\({\small (3)}~-30^\circ+360^\circ\times n\) (\(n\) は整数)


p.111 問3

\({\small (1)}~60^\circ\) \({\small (2)}~10^\circ\)
\({\small (3)}~150^\circ\) \({\small (4)}~45^\circ\)
動径と一般角


p.112 問4

\({\large \frac{3}{2}}\pi~,~{\large \frac{7}{4}}\pi~,~-{\large \frac{\pi}{3}}~,~-{\large \frac{5}{4}}\pi\)


p.113 問5

\(36^\circ~,~210^\circ~,~-330^\circ~,~-270^\circ\)


p.113 問6

\(6\pi~,~24\pi\)
弧度法と扇形


p.115 問7

\({\small (1)}~\)
\(\sin{{\large \frac{7}{6}}\pi}=-{\large \frac{1}{2}}~,~\cos{{\large \frac{7}{6}}\pi}=-{\large \frac{\sqrt{3}}{2}}\)
\(\tan{{\large \frac{7}{6}}\pi}={\large \frac{1}{\sqrt{3}}}\)
\({\small (2)}~\)
\(\sin{{\large \frac{9}{4}}\pi}={\large \frac{1}{\sqrt{2}}}~,~\cos{{\large \frac{9}{4}}\pi}={\large \frac{1}{\sqrt{2}}}\)
\(\tan{{\large \frac{9}{4}}\pi}=1\)
\({\small (3)}~\)
\(\sin{\left(-{\large \frac{\pi}{3}}\right)}=-{\large \frac{\sqrt{3}}{2}}~,~\cos{\left(-{\large \frac{\pi}{3}}\right)}={\large \frac{1}{2}}\)
\(\tan{\left(-{\large \frac{\pi}{3}}\right)}=-\sqrt{3}\)
\({\small (4)}~\)
\(\sin{(-3\pi)}=0~,~\cos{(-3\pi)}=-1\)
\(\tan{(-3\pi)}=0\)
三角関数の値(単位円)
【問題演習】三角関数の値(単位円)


p.115 問8

\({\small (1)}~\)第4象限
\({\small (2)}~\)第2象限


p.117 問9

\({\small (1)}~\)
\(\sin{\theta}=-{\large \frac{2\sqrt{2}}{3}}~,~\tan{\theta}=-2\sqrt{2}\)
\({\small (2)}~\)
\(\cos{\theta}=-{\large \frac{\sqrt{6}}{3}}~,~\tan{\theta}={\large \frac{\sqrt{2}}{2}}\)


p.117 問10

\(\cos{\theta}=-{\large \frac{2\sqrt{5}}{5}}~,~\sin{\theta}=-{\large \frac{\sqrt{5}}{5}}\)
三角関数の相互関係の公式


p.118 問11

\({\large \frac{12}{25}}~,~{\large \frac{37}{125}}\)
三角関数の式の値


p.118 問12

[証明] (左辺)
\(={\large \frac{\cos{\theta}}{1+\sin{\theta}}}+\tan{\theta}\)
\(={\large \frac{\cos{\theta}}{1+\sin{\theta}}}\cdot{\large \frac{1-\sin{\theta}}{1-\sin{\theta}}}+\tan{\theta}\)
\(={\large \frac{\cos{\theta}(1-\sin{\theta})}{\cos^2{\theta}}}+{\large \frac{\sin{\theta}}{\cos{\theta}}}\)
\(={\large \frac{1-\sin{\theta}+\sin{\theta}}{\cos{\theta}}}\)
\(={\large \frac{1}{\cos{\theta}}}=\) (右辺)
したがって、
\({\large \frac{\cos{\theta}}{1+\sin{\theta}}}+\tan{\theta}={\large \frac{1}{\cos{\theta}}}\) [終]
三角関数の等式の証明


p.119 問13

\(\sin{\left(-{\large \frac{\pi}{6}}\right)}=-{\large \frac{1}{2}}~,~\cos{\left(-{\large \frac{\pi}{6}}\right)}={\large \frac{\sqrt{3}}{2}}\)
\(\tan{\left(-{\large \frac{\pi}{6}}\right)}=-{\large \frac{1}{\sqrt{3}}}\)


p.120 問14

\(\sin{{\large \frac{5}{4}}\pi}=-{\large \frac{1}{\sqrt{2}}}~,~\cos{{\large \frac{5}{4}}\pi}=-{\large \frac{1}{\sqrt{2}}}\)
\(\tan{{\large \frac{5}{4}}\pi}=1\)
三角関数の性質①


p.121 問15

\({\small (1)}~{\large \frac{3}{8}}\pi\) \({\small (2)}~{\large \frac{\pi}{3}}\) \({\small (3)}~{\large \frac{\pi}{4}}\)
\({\small (4)}~{\large \frac{\pi}{6}}\) \({\small (5)}~{\large \frac{\pi}{8}}\) \({\small (6)}~{\large \frac{\pi}{4}}\)
三角関数の性質②


p.125 問16

周期 \(2\pi\)

三角関数のグラフ②(縦幅の変化)


p.125 問17

周期 \(2\pi\)

三角関数のグラフ②(縦幅の変化)


p.126 問18

周期 \(2\pi\)

三角関数のグラフ④(平行移動)


p.127 問19

周期 \(\pi\)

三角関数のグラフ③(周期の変化)


p.127 問20

周期 \({\large \frac{\pi}{2}}\)

三角関数のグラフ⑤(式変形)


p.128 問21

\({\small (1)}~\theta={\large \frac{4}{3}}\pi~,~{\large \frac{5}{3}}\pi\)
\({\small (2)}~\theta={\large \frac{\pi}{4}}~,~{\large \frac{7}{4}}\pi\)
\({\small (3)}~\theta={\large \frac{3}{4}}\pi~,~{\large \frac{7}{4}}\pi\)
三角関数を含む方程式①
【問題演習】三角関数を含む方程式


p.129 問22

\({\small (1)}~\theta={\large \frac{\pi}{2}}~,~\pi\)
\({\small (2)}~\theta={\large \frac{\pi}{2}}~,~{\large \frac{5}{6}}\pi\)
三角関数を含む方程式②(範囲変化)


p.130 問23

\({\small (1)}~{\large \frac{2}{3}}\pi≦\theta≦{\large \frac{4}{3}}\pi\)
\({\small (2)}~0≦\theta<{\large \frac{\pi}{3}}~,~{\large \frac{5}{3}}\pi<\theta<2\pi\)
\({\small (3)}~0≦\theta≦{\large \frac{5}{4}}\pi~,~{\large \frac{7}{4}}\pi≦\theta<2\pi\)


p.131 問24

\({\small (1)}~-{\large \frac{\pi}{4}}<\theta<{\large \frac{\pi}{6}}\)
\({\small (2)}~-{\large \frac{\pi}{2}}<\theta≦{\large \frac{\pi}{3}}\)
三角関数を含む不等式①


p.132 問1

Challenge
\({\small (1)}~\)
\(\theta={\large \frac{\pi}{3}}~,~{\large \frac{5}{3}}\pi\) で最大値 \({\large \frac{9}{4}}\)
\(\theta=\pi\) で最小値 \(0\)
\({\small (2)}~\)
\(\theta={\large \frac{\pi}{6}}~,~{\large \frac{5}{6}}\pi\) で最大値 \({\large \frac{5}{4}}\)
\(\theta={\large \frac{3}{2}}\pi\) で最小値 \(-1\)
三角関数を含む2次関数


Training


p.133 1

\({\large \frac{8}{3}}\pi~,~{\large \frac{16}{3}}\pi\)


p.133 2

\({\small (1)}~\)
\(\sin{{\large \frac{11}{6}}\pi}=-{\large \frac{1}{2}}~,~\cos{{\large \frac{11}{6}}\pi}={\large \frac{\sqrt{3}}{2}}\)
\(\tan{{\large \frac{11}{6}}\pi}=-{\large \frac{1}{\sqrt{3}}}\)
\({\small (2)}~\)
\(\sin{{\large \frac{11}{4}}\pi}={\large \frac{1}{\sqrt{2}}}~,~\cos{{\large \frac{11}{4}}\pi}=-{\large \frac{1}{\sqrt{2}}}\)
\(\tan{{\large \frac{11}{4}}\pi}=-1\)
\({\small (3)}~\)
\(\sin{\left(-{\large \frac{2}{3}}\pi\right)}=-{\large \frac{\sqrt{3}}{2}}\)
\(\cos{\left(-{\large \frac{2}{3}}\pi\right)}=-{\large \frac{1}{2}}\)
\(\tan{\left(-{\large \frac{2}{3}}\pi\right)}=\sqrt{3}\)
\({\small (4)}~\)
\(\sin{5\pi}=0~,~\cos{5\pi}=-1\)
\(\tan{5\pi}=0\)


p.133 3

\({\small (1)}~\)
\(\sin{\theta}=-{\large \frac{\sqrt{6}}{3}}~,~\tan{\theta}=-\sqrt{2}\)
\({\small (2)}~\)
\(\cos{\theta}=-{\large \frac{3\sqrt{10}}{10}}~,~\sin{\theta}=-{\large \frac{\sqrt{10}}{10}}\)


p.133 4

\({\small (1)}~{\large \frac{7}{18}}\)
\({\small (2)}~{\large \frac{4}{3}}\)
\({\small (3)}~{\large \frac{25\sqrt{2}}{54}}\)


p.133 5

[証明]
 (左辺)$$={\frac{\cos{\theta}}{1+\sin{\theta}}}+{\frac{\cos{\theta}}{1-\sin{\theta}}}$$$$~={\frac{\cos{\theta}(1-\sin{\theta})+\cos{\theta}(1+\sin{\theta})}{(1+\sin{\theta})(1-\sin{\theta})}}$$$$~={\frac{2\cos{\theta}}{\cos^2{\theta}}}$$$$~={\frac{2}{\cos{\theta}}}$$ \(=\)(右辺)
したがって、
\({\large \frac{\cos{\theta}}{1+\sin{\theta}}}+{\large \frac{\cos{\theta}}{1-\sin{\theta}}}={\large \frac{2}{\cos{\theta}}}\)
[終]


p.133 6

\({\small (1)}~b\) \({\small (2)}~-b\) \({\small (3)}~-{\large \frac{b}{a}}\)


p.133 7

\({\small (1)}~\)周期 \(\pi\)

\({\small (2)}~\)周期 \(2\pi\)

\({\small (3)}~\)周期 \({\large \frac{2}{3}}\pi\)


p.133 8

\({\small (1)}~\theta={\large \frac{7}{6}}\pi~,~{\large \frac{11}{6}}\pi\)
\({\small (2)}~\theta={\large \frac{5}{6}}\pi~,~{\large \frac{7}{6}}\pi\)
\({\small (3)}~\theta={\large \frac{\pi}{6}}~,~{\large \frac{7}{6}}\pi\)


p.133 9

\({\small (1)}~\theta={\large \frac{\pi}{12}}~,~{\large \frac{5}{12}}\pi\)
\({\small (2)}~\theta={\large \frac{\pi}{2}}~,~{\large \frac{3}{2}}\pi\)


p.133 10

\({\small (1)}~\)
\({\large \frac{\pi}{3}}≦\theta≦{\large \frac{2}{3}}\pi\)
\({\small (2)}~\)
\(0≦\theta<{\large \frac{\pi}{4}}~,~{\large \frac{7}{4}}\pi<\theta<2\pi\)
\({\small (3)}~\)
\(0≦\theta<{\large \frac{\pi}{2}}~,~{\large \frac{5}{6}}\pi≦\theta<{\large \frac{3}{2}}\pi\)
\({\large \frac{11}{6}}\pi≦\theta<2\pi\)




2節 加法定理


p.134 問1

[証明]
図より、
 \({\rm OH}=\cos{(\alpha+\beta)}\) …①
また、
 \({\rm OH=OC+HC}\) …②
ここで、
 \({\rm OC}={\rm OB}\cos{\beta}\)
また、\({\rm OB}=\cos{\alpha}\) より、
 \({\rm OC}=\cos{\alpha}\cos{\beta}\) …③
次に、
 \({\rm HC=AI}={\rm AB}\sin{\beta}\)
また、\({\rm AB}=\sin{\alpha}\) より、
 \({\rm HC}=\sin{\alpha}\sin{\beta}\) …④
したがって、②に①、③、④を代入すると、
 \(\cos{(\alpha+\beta)}\)
   \(=\cos{\alpha}\cos{\beta}-\sin{\alpha}\sin{\beta}\)
[終]


p.134 問2

[証明]
\(\cos{(\alpha+\beta)}=\cos{\alpha}\cos{\beta}-\sin{\alpha}\sin{\beta}\)
この式において、\(\beta\) を \(-\beta\) に置き換えると、
(左辺) \(=\cos{(\alpha-\beta)}\)
 (右辺)
\(=\cos{\alpha}\cos{(-\beta)}-\sin{\alpha}\sin{(-\beta)}\)
ここで、\(\cos{(-\beta)}=\cos{\beta}\) と \(\sin{(-\beta)}=-\sin{\beta}\) より、
\(=\cos{\alpha}\cos{\beta}+\sin{\alpha}\sin{\beta}\)
したがって、
\(\cos{(\alpha-\beta)}=\cos{\alpha}\cos{\beta}+\sin{\alpha}\sin{\beta}\)
[終]


p.135 問3

\({\large \frac{\sqrt{6}-\sqrt{2}}{4}}~,~{\large \frac{\sqrt{6}+\sqrt{2}}{4}}\)


p.135 問4

\({\large \frac{\sqrt{2}+\sqrt{6}}{4}}~,~{\large \frac{\sqrt{2}-\sqrt{6}}{4}}\)
加法定理


p.136 問5

\({\small (1)}~-{\large \frac{56}{65}}\) \({\small (2)}~-{\large \frac{63}{65}}\) \({\small (3)}~{\large \frac{33}{65}}\)
加法定理と式の値


p.137 問6

\(2+\sqrt{3}~,~2-\sqrt{3}\)


p.137 問7

\({\large \frac{\pi}{4}}\)
2直線のなす角


p.138 問8

[証明] (左辺)
\(=(\sin{\alpha}+\cos{\alpha})^2\)
\(=\sin^2{\alpha}+2\sin{\alpha}\cos{\alpha}+\cos^2{\alpha}\)
\(=1+2\sin{\alpha}\cos{\alpha}\)
2倍角の公式より、
\(=1+2\sin{2\alpha}=\)(右辺)
したがって、
\((\sin{\alpha}+\cos{\alpha})^2=1+\sin{2\alpha}\)
[終]


p.138 問9

\({\large \frac{24}{25}}~,~{\large \frac{7}{25}}\)
2倍角の公式


p.139 問10

\(\theta=0~,~{\large \frac{2}{3}}\pi~,~{\large \frac{4}{3}}\pi\)


p.139 問11

\({\large \frac{\sqrt{2+\sqrt{2}}}{2}}~,~{\large \frac{\sqrt{2+\sqrt{2}}}{2}}\)
半角の公式


p.141 問12

\({\small (1)}~2\sqrt{3}\sin{\left(\theta+{\large \frac{\pi}{3}}\right)}\)
\({\small (2)}~\sqrt{2}\sin{\left(\theta-{\large \frac{3}{4}}\pi\right)}\)
三角関数の合成


p.141 問13

\({\small (1)}~\)
最大値 \(2\)、最小値 \(-2\)
\({\small (2)}~\)
最大値 \(13\)、最小値 \(-13\)
三角関数の最大値・最小値


p.142 問1

Challenge
\({\small (1)}~\theta={\large \frac{\pi}{12}}~,~{\large \frac{17}{12}}\pi\)
\({\small (2)}~\theta=\pi~,~{\large \frac{5}{3}}\pi\)
合成を用いる方程式と不等式


Training


p.143 11

\({\small (1)}~-{\large \frac{63}{65}}\) \({\small (2)}~-{\large \frac{16}{65}}\)


p.143 12

\({\small (1)}~\sqrt{3}\cos{\theta}\) \({\small (2)}~\sqrt{3}\cos{\theta}\)
\({\small (3)}~-\sin{\theta}\) \({\small (4)}~1\)


p.143 13

\({\small (1)}~3\) \({\small (2)}~{\large \frac{1}{2}}\)


p.143 14

\({\small (1)}~-{\large \frac{4\sqrt{2}}{9}}\) \({\small (2)}~{\large \frac{7}{9}}\)
\({\small (3)}~-{\large \frac{4\sqrt{2}}{7}}\)


p.143 15

\({\small (1)}~\theta={\large \frac{\pi}{6}}~,~{\large \frac{\pi}{2}}~,~{\large \frac{5}{6}}\pi~,~{\large \frac{3}{2}}\pi\)
\({\small (2)}~\theta=0~,~{\large \frac{\pi}{3}}~,~\pi~,~{\large \frac{5}{3}}\pi\)


p.143 16

\({\small (1)}~\sqrt{2}\sin{\left(\theta+{\large \frac{3}{4}}\pi\right)}\)
\({\small (2)}~2\sqrt{3}\sin{\left(\theta-{\large \frac{\pi}{6}}\right)}\)


p.143 17

最大値 \(5\)、最小値 \(-5\)
三角関数の最大値・最小値




Level Up 三角関数


p.144 1

\({\small (1)}~-{\large \frac{\sqrt{6}}{2}}\) \({\small (2)}~-{\large \frac{3\sqrt{6}}{8}}\)


p.144 2

\({\small (1)}~2\sin{\theta}\) \({\small (2)}~0\)


p.144 3

\({\small (1)}~\)周期 \(\pi\)

\({\small (2)}~\)周期 \(\pi\)

\({\small (3)}~\)周期 \(4\pi\)


p.144 4

\(r=2~,~a=3~,~b={\large \frac{\pi}{2}}\)


p.144 5

\({\small (1)}~\theta={\large \frac{\pi}{3}}~,~{\large \frac{\pi}{2}}~,~{\large \frac{4}{3}}\pi~,~{\large \frac{3}{2}}\pi\)
\({\small (2)}~\)
\(0≦\theta≦{\large \frac{\pi}{2}}~,~{\large \frac{7}{6}}\pi≦\theta<2\pi\)


p.144 6

\({\small (1)}~\)
\({\large \frac{\theta}{2}}<\theta<{\large \frac{3}{4}}\pi~,~{\large \frac{5}{4}}\pi<\theta<{\large \frac{3}{2}}\pi\)
\({\small (2)}~\)
\(0≦\theta≦{\large \frac{\pi}{6}}~,~{\large \frac{5}{6}}\pi≦\theta≦{\large \frac{7}{6}}\pi\)
\({\large \frac{11}{6}}\pi≦\theta<2\pi\)


p.144 7

\({\large \frac{4}{3}}\)


p.144 8

\({\small (1)}~\) [証明]
\(~~~~~\sin{3\alpha}\)
\(=\sin{(2\alpha+\alpha)}\)
\(=\sin{2\alpha}\cos{\alpha}+\cos{2\alpha}\sin{\alpha}\)
\(=2\sin{\alpha}\cos^2{\alpha}\)
   \(+(1-2\sin^2{\alpha})\sin{\alpha}\)
\(=2\sin{\alpha}(1-\sin^2{\alpha})\)
   \(+\sin{\alpha}-2\sin^3{\alpha}\)
\(=3\sin{\alpha}-4\sin^3{\alpha}\)
したがって、
\(\sin{3\alpha}=3\sin{\alpha}-4\sin^3{\alpha}\) [終]
\({\small (2)}~\) [証明]
\(~~~~~\cos{3\alpha}\)
\(=\cos{(2\alpha+\alpha)}\)
\(=\cos{2\alpha}\cos{\alpha}-\sin{2\alpha}\sin{\alpha}\)
\(=(2\cos^2{\alpha}-1)\cos{\alpha}\)
   \(-2\sin^2{\alpha}\cos{\alpha}\)
\(=2\cos^3{\alpha}-\cos{\alpha}\)
   \(-2(1-\cos^2{\alpha})\cos{\alpha}\)
\(=4\cos^3{\alpha}-3\cos{\alpha}\)
したがって、
\(\cos{3\alpha}=4\cos^3{\alpha}-3\cos{\alpha}\) [終]


p.144 9

\({\small (1)}~\theta={\large \frac{\pi}{2}}~,~{\large \frac{2}{3}}\pi~,~{\large \frac{4}{3}}\pi~,~{\large \frac{3}{2}}\pi\)
\({\small (2)}~\theta={\large \frac{\pi}{6}}~,~{\large \frac{5}{6}}\pi\)


p.144 10

\({\small (1)}~{\large \frac{1}{2}}\sin{2\theta}\)

\({\small (2)}~-{\large \frac{1}{2}}\cos{2\theta}+{\large \frac{1}{2}}\)


p.144 11

\(\theta={\large \frac{\pi}{3}}\) で最大値 \(2\sqrt{3}\)
\(\theta=\pi\) で最小値 \(-\sqrt{3}\)


p.144 12

\({\small (1)}~y={\large \frac{1}{2}}t^2+t-{\large \frac{1}{2}}\)
\({\small (2)}~-1≦y≦{\large \frac{1}{2}}+\sqrt{2}\)


p.144 13

\({\small (1)}~\sqrt{2}\sin{\left(2\theta-{\large \frac{\pi}{4}}\right)}\)
\({\small (2)}~{\large \frac{\pi}{4}}~,~{\large \frac{\pi}{2}}\)


p.144 14

\({\small (1)}~y=-\sin{2\theta}+\cos{2\theta}+3\)
\({\small (2)}~\)
\(\theta={\large \frac{7}{8}}\pi~,~{\large \frac{15}{8}}\pi\) で、
 最大値 \(\sqrt{2}+3\)
\(\theta={\large \frac{3}{8}}\pi~,~{\large \frac{11}{8}}\pi\) で、
 最小値 \(-\sqrt{2}+3\)


 



次のページ「4章 指数関数・対数関数」