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東京書籍:Advanced数学Ⅱ

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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

1節 三角関数


p.110 問1

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

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

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

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


p.111 問2

\({\small (1)}~140^\circ+360^\circ\times n\) (\(n\) は整数)
\({\small (2)}~280^\circ+360^\circ\times n\) (\(n\) は整数)
\({\small (3)}~70^\circ+360^\circ\times n\) (\(n\) は整数)
\({\small (4)}~250^\circ+360^\circ\times n\) (\(n\) は整数)
動径と一般角


p.111 問3

\({\large \frac{2}{3}}\pi~,~{\large \frac{5}{6}}\pi~,~-{\large \frac{2}{3}}\pi\)
\({\large \frac{9}{4}}\pi~,~{\Large \frac{1}{180}}\pi\)


p.112 問4

\(36^\circ~,~135^\circ\)
\(-450^\circ~,~-540^\circ\)


p.112 問5

\({\large \frac{9}{2}}\pi~,~{\large \frac{27}{2}}\pi\)
弧度法と扇形


p.113 問6

\({\small (1)}~\)
\(\sin{{\large \frac{5}{4}}\pi}=-{\large \frac{1}{\sqrt{2}}}~,~\cos{{\large \frac{5}{4}}\pi}=-{\large \frac{1}{\sqrt{2}}}\)
\(\tan{{\large \frac{5}{4}}\pi}=1\)
\({\small (2)}~\)
\(\sin{{\large \frac{11}{6}}\pi}=-{\large \frac{1}{2}}~,~\cos{{\large \frac{11}{6}}\pi}={\large \frac{\sqrt{3}}{2}}\)
\(\tan{{\large \frac{11}{6}}\pi}=-{\large \frac{1}{\sqrt{3}}}\)
\({\small (3)}~\)
\(\sin{(-3\pi)}=0~,~\cos{(-3\pi)}=-1\)
\(\tan{(-3\pi)}=0\)
三角関数の値(単位円)
【問題演習】三角関数の値(単位円)


p.114 問7

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


p.114 問8

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


p.116 問9

[証明]
\(\sin^2{\theta}+\cos^2{\theta}=1\) の両辺を \(\cos^2{\theta}\) で割ると、$$~~~{ \frac{\sin^2{\theta}}{\cos^2{\theta}}}+1={ \frac{1}{\cos^2{\theta}}}$$\({\large \frac{\sin{\theta}}{\cos{\theta}}}=\tan{\theta}\) より、$$~~~1+\tan^2{\theta}={ \frac{1}{\cos^2{\theta}}}$$[終]


p.117 問10

\(\cos{\theta}={\large \frac{2\sqrt{2}}{3}}~,~\tan{\theta}=-{\large \frac{\sqrt{2}}{4}}\)


p.117 問11

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


p.117 問12

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


p.117 問13

\({\small (1)}~\)[証明]
 (左辺)$$={ \frac{1}{\sin^2{\theta}}}\left({ \frac{1-\cos^2{\theta}}{\cos^2{\theta}}}\right)$$$$={ \frac{1}{\sin^2{\theta}}}\cdot{ \frac{\sin^2{\theta}}{\cos^2{\theta}}}$$$$={ \frac{1}{\cos^2{\theta}}}$$したがって、$$~{ \frac{1}{\sin^2{\theta}}}\left({ \frac{1}{\cos^2{\theta}}}-1\right)={ \frac{1}{\cos^2{\theta}}}$$[終]

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


p.118 問14

\(\sin{{\large \frac{27}{4}}\pi}={\large \frac{1}{\sqrt{2}}}~,~\cos{{\large \frac{27}{4}}\pi}=-{\large \frac{1}{\sqrt{2}}}\)
\(\tan{{\large \frac{27}{4}}\pi}=-1\)


p.118 問15

\(\sin{\left(-{\large \frac{5}{6}}\pi\right)}=-{\large \frac{1}{2}}\)
\(\cos{\left(-{\large \frac{5}{6}}\pi\right)}=-{\large \frac{\sqrt{3}}{2}}\)
\(\tan{\left(-{\large \frac{5}{6}}\pi\right)}={\large \frac{1}{\sqrt{3}}}\)


p.119 問16

\({\small (1)}~a\) \({\small (2)}~-a\) \({\small (3)}~-a\)
三角関数の性質①


p.119 問17

公式5
[証明]$$~~~~~\sin{\left({ \frac{\pi}{2}}-\theta\right)}$$$$=\sin{\left\{{ \frac{\pi}{2}}+(-\theta)\right\}}$$$$=\cos{(-\theta)}=\cos{\theta}$$
$$~~~~~\cos{\left({ \frac{\pi}{2}}-\theta\right)}$$$$=\cos{\left\{{ \frac{\pi}{2}}+(-\theta)\right\}}$$$$=-\sin{(-\theta)}=\sin{\theta}$$
$$~~~~~\tan{\left({ \frac{\pi}{2}}-\theta\right)}$$$$=\tan{\left\{{ \frac{\pi}{2}}+(-\theta)\right\}}$$$$=-{ \frac{1}{\tan{(-\theta)}}}={ \frac{1}{\tan{\theta}}}$$[終]

公式6
[証明]
 \(\sin{(\pi-\theta)}\)
\(=\sin{\{\pi+(-\theta)\}}\)
\(=-\sin{(-\theta)}=\sin{\theta}\)

 \(\cos{(\pi-\theta)}\)
\(=\cos{\{\pi+(-\theta)\}}\)
\(=-\cos{(-\theta)}=-\cos{\theta}\)

 \(\tan{(\pi-\theta)}\)
\(=\tan{\{\pi+(-\theta)\}}\)
\(=\tan{(-\theta)}=-\tan{\theta}\)
[終]
三角関数の性質②


p.123 問18

奇関数


p.123 問19

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

周期 \(2\pi\)
\({\small (2)}~\)

周期 \(2\pi\)
三角関数のグラフ②(縦幅の変化)


p.124 問20

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

周期 \(2\pi\)
\({\small (2)}~\)

周期 \(2\pi\)
三角関数のグラフ④(平行移動)


p.125 問21

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

周期 \(\pi\)
\({\small (2)}~\)

周期 \(3\pi\)
三角関数のグラフ③(周期の変化)


p.125 問22

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

周期 \(\pi\)
\({\small (2)}~\)

周期 \(4\pi\)
三角関数のグラフ⑤(式変形)


p.126 問23

\({\small (1)}~n\) を整数として、
 \(\theta={\large \frac{\pi}{3}}+2n\pi~,~{\large \frac{2}{3}}\pi+2n\pi\)
\({\small (2)}~n\) を整数として、
 \(\theta={\large \frac{\pi}{3}}+2n\pi~,~{\large \frac{5}{3}}\pi+2n\pi\)


p.126 問24

\({\small (1)}~n\) を整数として、
 \(\theta={\large \frac{\pi}{4}}+n\pi\)
\({\small (2)}~n\) を整数として、
 \(\theta={\large \frac{5}{6}}\pi+n\pi\)
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.127 問25

\({\small (1)}~\theta={\large \frac{3}{2}}\pi~,~{\large \frac{11}{6}}\pi\)
\({\small (2)}~\theta={\large \frac{7}{12}}\pi~,~{\large \frac{23}{12}}\pi\)
三角関数を含む方程式②(範囲変化)


p.128 問26

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


p.128 問27

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


p.129 問28

\({\small (1)}~{\large \frac{\pi}{3}}<\theta<{\large \frac{\pi}{2}}~,~{\large \frac{4}{3}}\pi<\theta<{\large \frac{3}{2}}\pi\)
\({\small (2)}~{\large \frac{\pi}{2}}<\theta≦{\large \frac{5}{6}}\pi~,~{\large \frac{2}{3}}\pi<\theta≦{\large \frac{11}{6}}\pi\)
三角関数を含む不等式①


p.129 問29

\(-{\large \frac{\pi}{3}}<\theta<{\large \frac{\pi}{4}}\)


p.130 問30

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


問題


p.131 1

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


p.131 2

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


p.131 3

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

\({\small (2)}~\)[証明]
\(\sin{\left({\large \frac{\pi}{2}}-\theta\right)}=\cos{\theta}\)
\(\cos{(\pi-\theta)}=-\cos{\theta}\)
また、$$~~~~~\sin{\left({ \frac{3}{2}\pi}+\theta\right)}$$$$=\sin{\left(\pi+{ \frac{\pi}{2}}+\theta\right)}$$$$=-\sin{\left({ \frac{\pi}{2}}+\theta\right)}$$$$=-\cos{\theta}$$これらより、
 (左辺)
\(=\cos{\theta}+\cos{\theta}-\cos{\theta}-\cos{\theta}=0\)
したがって、$$~\cos{\theta}+\sin{\left({ \frac{\pi}{2}}-\theta\right)}$$$$~~~~+\cos{(\pi-\theta)}+\sin{\left({ \frac{3}{2}\pi}+\theta\right)}=0$$[終]


p.131 4

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

周期 \(\pi\)
\({\small (2)}~\)

周期 \(4\pi\)
\({\small (3)}~\)

周期 \({\large \frac{2}{3}}\pi\)


p.131 5

\({\small (1)}~\theta={\large \frac{\pi}{3}}~,~{\large \frac{2}{3}}\pi~,~{\large \frac{4}{3}}\pi~,~{\large \frac{5}{3}}\pi\)
\({\small (2)}~\theta={\large \frac{\pi}{6}}~,~{\large \frac{\pi}{3}}~,~{\large \frac{7}{6}}\pi~,~{\large \frac{4}{3}}\pi\)
\({\small (3)}~\theta={\large \frac{3}{2}}\pi\)


p.131 6

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


p.131 7

\(\theta={\large \frac{\pi}{6}}~,~{\large \frac{5}{6}}\pi\) のとき、最大値 \({\large \frac{5}{4}}\)
\(\theta={\large \frac{3}{2}}\pi\) のとき、最小値 \(-1\)




2節 加法定理


p.134 問1

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


p.134 問2

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


p.134 問3

\({\small (1)}~-{\large \frac{77}{85}}\) \({\small (2)}~-{\large \frac{84}{85}}\) \({\small (3)}~{\large \frac{36}{85}}\)
加法定理と式の値


p.135 問4

[証明]
 \(\tan{(\alpha+\beta)}={\large \frac{\tan{\alpha}+\tan{\beta}}{1-\tan{\alpha}\tan{\beta}}}\)
これより、\(\beta~\to~-\beta\) とすると、
 \(\tan{(\alpha-\beta)}={\large \frac{\tan{\alpha}+\tan{(-\beta)}}{1-\tan{\alpha}\tan{(-\beta)}}}\)
ここで、\(\tan{(-\beta)}=-\tan{\beta}\) より、
 \(\tan{(\alpha-\beta)}={\large \frac{\tan{\alpha}-\tan{\beta}}{1+\tan{\alpha}\tan{\beta}}}\)
[終]


p.135 問5

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


p.135 問6

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


p.136 問7

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


p.136 問8

\(3\)
2直線のなす角


p.137 問9

\(\sin{2\alpha}=-{\large \frac{4\sqrt{2}}{9}}~,~\cos{2\alpha}=-{\large \frac{7}{9}}\)
\(\tan{2\alpha}={\large \frac{4\sqrt{2}}{7}}\)
2倍角の公式


p.137 問10

\({\small (1)}~\) [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.138 問11

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


p.138 問12

\(0≦\theta<{\large \frac{\pi}{6}}~,~{\large \frac{5}{6}}\pi<\theta<{\large \frac{3}{2}}\pi\)
\({\large \frac{3}{2}}\pi<\theta<2\pi\)


p.139 問13

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


p.139 問14

\({\large \frac{\sqrt{2+\sqrt{2}}}{2}}~,~\sqrt{2}-1\)


p.139 問15

\(-{\large \frac{\sqrt{6}}{6}}\)
半角の公式


p.140 問16

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


p.141 問17

\({\small (1)}~5\sin{(\theta+\alpha)}\)
ただし、\(\alpha\) は、
 \(\cos{\alpha}={\large \frac{3}{5}}~,~\sin{\alpha}={\large \frac{4}{5}}\)
\({\small (2)}~\sqrt{29}\sin{(\theta+\alpha)}\)
ただし、\(\alpha\) は、
 \(\cos{\alpha}={\large \frac{5}{\sqrt{29}}}~,~\sin{\alpha}=-{\large \frac{2}{\sqrt{29}}}\)
三角関数の合成


p.141 問18

\({\small (1)}~\)
最大値 \(2\sqrt{3}\)、最小値 \(-2\sqrt{3}\)
\({\small (2)}~\)
最大値 \(13\)、最小値 \(-13\)
三角関数の最大値・最小値


p.142 問19

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


問題


p.143 8

\({\small (1)}~{\large \frac{6-\sqrt{35}}{12}}\) \({\small (2)}~{\large \frac{-3\sqrt{5}+2\sqrt{7}}{12}}\)


p.143 9

\(0\)


p.143 10

\({\small (1)}~\)[証明]
加法定理より、
 (左辺)$$=\tan{\left({ \frac{\pi}{4}}+\theta\right)}$$$$={ \frac{\tan{{\large \frac{\pi}{4}}}+\tan{\theta}}{1-\tan{{\large \frac{\pi}{4}}}\tan{\theta}}}$$$$={ \frac{1+\tan{\theta}}{1-\tan{\theta}}}$$したがって、$$~~~\tan{\left({ \frac{\pi}{4}}+\theta\right)}={ \frac{1+\tan{\theta}}{1-\tan{\theta}}}$$[終]

\({\small (2)}~\)[証明]
2倍角の公式より、
 (左辺)$$={ \frac{\sin{2\theta}}{1+\cos{2\theta}}}$$$$={ \frac{2\sin{\theta}\cos{\theta}}{1+(2\cos^2{\theta}-1)}}$$$$={ \frac{2\sin{\theta}\cos{\theta}}{2\cos^2{\theta}}}$$$$={ \frac{\sin{\theta}}{\cos{\theta}}}=\tan{\theta}$$したがって、$$~~~{ \frac{\sin{2\theta}}{1+\cos{2\theta}}}=\tan{\theta}$$[終]


p.143 11

\(y={\large \frac{2}{3}}x~,~y=-{\large \frac{3}{2}}x\)


p.143 12

\({\small (1)}~-{\large \frac{17}{25}}\) \({\small (2)}~-{\large \frac{4\sqrt{21}}{25}}\)
\({\small (3)}~{\large \frac{\sqrt{70}}{10}}\)


p.143 13

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


p.143 14

\({\small (1)}~{\large \frac{\pi}{12}}<\theta<{\large \frac{7}{12}}\pi\)
\({\small (2)}~{\large \frac{5}{6}}\pi≦\theta≦{\large \frac{3}{2}}\pi\)


p.143 15

\(\theta={\large \frac{7}{6}}\pi~,~{\large \frac{11}{6}}\pi\) のとき、最大値 \({\large \frac{3}{2}}\)
\(\theta={\large \frac{\pi}{2}}\) のとき、最小値 \(-3\)


p.143 16

\(\theta={\large \frac{\pi}{4}}\) のとき、最大値 \(\sqrt{2}\)
\(\theta=\pi\) のとき、最小値 \(-1\)


p.144 発展1

[証明] 加法定理より、
 \(\sin{(\alpha+\beta)}\)
   \(=\sin{\alpha}\cos{\beta}+\cos{\alpha}\sin{\beta}\) …①
 \(\sin{(\alpha-\beta)}\)
   \(=\sin{\alpha}\cos{\beta}-\cos{\alpha}\sin{\beta}\) …②
①−②より、
 \(\sin{(\alpha+\beta)}-\sin{(\alpha-\beta)}\)
     \(=2\cos{\alpha}\sin{\beta}\)
したがって、
 \(\cos{\alpha}\sin{\beta}\)
    \(={\large \frac{1}{2}}\left\{\sin{(\alpha+\beta)}-\sin{(\alpha-\beta)}\right\}\)

また、加法定理より、
 \(\cos{(\alpha+\beta)}\)
   \(=\cos{\alpha}\cos{\beta}-\sin{\alpha}\sin{\beta}\) …③
 \(\cos{(\alpha-\beta)}\)
   \(=\cos{\alpha}\cos{\beta}+\sin{\alpha}\sin{\beta}\) …④
③+④より、
 \(\cos{(\alpha+\beta)}+\cos{(\alpha-\beta)}\)
     \(=2\cos{\alpha}\cos{\beta}\)
したがって、
 \(\cos{\alpha}\cos{\beta}\)
    \(={\large \frac{1}{2}}\left\{\cos{(\alpha+\beta)}+\cos{(\alpha-\beta)}\right\}\)

また、③−④より、
 \(\cos{(\alpha+\beta)}-\cos{(\alpha-\beta)}\)
     \(=-2\sin{\alpha}\sin{\beta}\)
したがって、
 \(\sin{\alpha}\sin{\beta}\)
  \(=-{\large \frac{1}{2}}\left\{\cos{(\alpha+\beta)}-\cos{(\alpha-\beta)}\right\}\)
[終]


p.144 発展2

\({\small (1)}~{\large \frac{\sqrt{2}}{4}}\) \({\small (2)}~{\large \frac{\sqrt{2}}{4}}\)


p.144 発展3

\({\small (1)}~{\large \frac{1}{2}}(\sin{4\theta}-\sin{3\theta})\)
\({\small (2)}~{\large \frac{1}{2}}(\cos{5\theta}+\cos{\theta})\)
\({\small (3)}~-{\large \frac{1}{2}}(\cos{5\theta}-\cos{3\theta})\)


p.145 発展4

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


p.145 発展5

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


p.145 発展6

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




練習問題 三角関数


練習問題A


p.146 1

\({\small (1)}~-k\) \({\small (2)}~-\sqrt{1-k^2}\)
\({\small (3)}~-{\large \frac{\sqrt{1-k^2}}{k}}\)


p.146 2

\({\small (1)}~-{\large \frac{5}{18}}\) \({\small (2)}~-{\large \frac{\sqrt{14}}{3}}\)


p.146 3

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


p.146 4

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


p.146 5

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

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


p.146 6

\({\small (1)}~\)[証明]
 \(2\theta=36^\circ~,~3\theta=54^\circ\)
これより、
 \(\sin{2\theta}\)
\(=\sin{36^\circ}=\sin{(90^\circ-54^\circ)}\)
\(=\cos{54^\circ}\)
\(=\cos{3\theta}\)
したがって、
 \(\sin{2\theta}=\cos{3\theta}\) [終]

\({\small (2)}~{\large \frac{-1+\sqrt{5}}{4}}\)


p.146 7

最大値 \(2\)、最小値 \(-2\)


p.146 8

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


練習問題B


p.147 9

\({\small (1)}~\theta={\large \frac{\pi}{8}}~,~{\large \frac{5}{8}}\pi~,~{\large \frac{9}{8}}\pi~,~{\large \frac{13}{8}}\pi\)
\({\small (2)}~0≦\theta<{\large \frac{\pi}{3}}~,~{\large \frac{2}{3}}\pi<\theta<2\pi\)


p.147 10

\({\small (1)}~\)最大値 \(\sqrt{7}\)、最小値 \(-\sqrt{7}\)
\({\small (2)}~\)最大値 \(1\)、最小値 \(-2\sqrt{2}-1\)


p.147 11

\(-{\large \frac{27}{32}}\)


p.147 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.147 13

\({\small (1)}~\)[証明]
\(\triangle {\rm ABQ}\) において、
 \(\cos{\theta}={\large \frac{{\rm AQ}}{{\rm AB}}}~\Leftrightarrow~{\rm AQ}=\cos{\theta}\)
\(\triangle {\rm DAP}\) において、
 \(\sin{\theta}={\large \frac{{\rm AP}}{{\rm AD}}}~\Leftrightarrow~{\rm AP}=\sin{\theta}\)
よって、1辺の長さ \({\rm PQ}\) は、
 \({\rm PQ=AQ-AP}=\cos{\theta}-\sin{\theta}\)
[終]

\({\small (2)}~{\large \frac{\pi}{12}}\)


 



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