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。
第１章 式と証明
第２章 複素数と方程式
第３章 図形と方程式
第５章 指数関数と対数関数
第６章 微分法と積分法
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。
第４章 三角関数
第１節 三角関数
p.105 練習1
\({\small (1)}~\)
\({\small (2)}~\)
\({\small (3)}~\)
\({\small (4)}~\)
\({\small (5)}~\)
\({\small (1)}~\)
\({\small (2)}~\)
\({\small (3)}~\)
\({\small (4)}~\)
\({\small (5)}~\)
p.106 練習3
\({\small (1)}~\)
[証明]
弧の長さ \(1\) に対する中心角の大きさが \(1\) ラジアンである
中心角 \(180^\circ\) の弧は半円となるので、弧の長さは、
\(2\pi \times {\large \frac{1}{2}}=\pi\)
したがって、弧の長さ \(\pi\) に対する中心角の大きさは \(\pi\) ラジアンである
[終]
\({\small (2)}~\)
[証明]
\(1\) ラジアンに対する中心角の大きさを \(a^\circ\) とする
(1) の結果より、弧の長さ \(\pi\) に対する中心角の大きさは \(\pi\) ラジアンであるので、
\(\pi:180^\circ=1:a^\circ\)
よって、
\(a={\large \frac{180}{\pi}}\)
したがって、
\(1\) ラジアンは \(\left({\large \frac{180}{\pi}}\right)^\circ\)
[終]
\({\small (1)}~\)
[証明]
弧の長さ \(1\) に対する中心角の大きさが \(1\) ラジアンである
中心角 \(180^\circ\) の弧は半円となるので、弧の長さは、
\(2\pi \times {\large \frac{1}{2}}=\pi\)
したがって、弧の長さ \(\pi\) に対する中心角の大きさは \(\pi\) ラジアンである
[終]
\({\small (2)}~\)
[証明]
\(1\) ラジアンに対する中心角の大きさを \(a^\circ\) とする
(1) の結果より、弧の長さ \(\pi\) に対する中心角の大きさは \(\pi\) ラジアンであるので、
\(\pi:180^\circ=1:a^\circ\)
よって、
\(a={\large \frac{180}{\pi}}\)
したがって、
\(1\) ラジアンは \(\left({\large \frac{180}{\pi}}\right)^\circ\)
[終]
p.106 練習4
\({\small (1)}~{\large \frac{7}{6}}\pi\) \({\small (2)}~{\large \frac{4}{3}}\pi\) \({\small (3)}~{\large \frac{11}{6}}\pi\)
\({\small (4)}~225^\circ\) \({\small (5)}~270^\circ\)
\({\small (1)}~{\large \frac{7}{6}}\pi\) \({\small (2)}~{\large \frac{4}{3}}\pi\) \({\small (3)}~{\large \frac{11}{6}}\pi\)
\({\small (4)}~225^\circ\) \({\small (5)}~270^\circ\)
p.107 練習5
\({\small (1)}~l={\large \frac{4}{3}}\pi~,~S={\large \frac{8}{3}}\pi\)
\({\small (2)}~l=7\pi~,~S=21\pi\)
→ 弧度法と扇形
\({\small (1)}~l={\large \frac{4}{3}}\pi~,~S={\large \frac{8}{3}}\pi\)
\({\small (2)}~l=7\pi~,~S=21\pi\)
→ 弧度法と扇形
p.109 練習6
\({\small (1)}~\)
\(\sin{{\large \frac{5}{4}}\pi}=-{\large \frac{1}{\sqrt{2}}}~,~\cos{{\large \frac{5}{4}}\pi}=-{\large \frac{1}{\sqrt{2}}}\)
\(\tan{{\large \frac{5}{4}}\pi}=1\)
\({\small (2)}~\)
\(\sin{{\large \frac{11}{6}}\pi}=-{\large \frac{1}{2}}~,~\cos{{\large \frac{11}{6}}\pi}={\large \frac{\sqrt{3}}{2}}\)
\(\tan{{\large \frac{11}{6}}\pi}=-{\large \frac{1}{\sqrt{3}}}\)
\({\small (3)}~\)
\(\sin{\left(-{\large \frac{\pi}{3}}\right)}=-{\large \frac{\sqrt{3}}{2}}~,~\cos{\left(-{\large \frac{\pi}{3}}\right)}={\large \frac{1}{2}}\)
\(\tan{\left(-{\large \frac{\pi}{3}}\right)}=-\sqrt{3}\)
→ 三角関数の値（単位円）
→ 【問題演習】三角関数の値（単位円）
\({\small (1)}~\)
\(\sin{{\large \frac{5}{4}}\pi}=-{\large \frac{1}{\sqrt{2}}}~,~\cos{{\large \frac{5}{4}}\pi}=-{\large \frac{1}{\sqrt{2}}}\)
\(\tan{{\large \frac{5}{4}}\pi}=1\)
\({\small (2)}~\)
\(\sin{{\large \frac{11}{6}}\pi}=-{\large \frac{1}{2}}~,~\cos{{\large \frac{11}{6}}\pi}={\large \frac{\sqrt{3}}{2}}\)
\(\tan{{\large \frac{11}{6}}\pi}=-{\large \frac{1}{\sqrt{3}}}\)
\({\small (3)}~\)
\(\sin{\left(-{\large \frac{\pi}{3}}\right)}=-{\large \frac{\sqrt{3}}{2}}~,~\cos{\left(-{\large \frac{\pi}{3}}\right)}={\large \frac{1}{2}}\)
\(\tan{\left(-{\large \frac{\pi}{3}}\right)}=-\sqrt{3}\)
→ 三角関数の値（単位円）
→ 【問題演習】三角関数の値（単位円）
p.109 練習7
\({\small (1)}~\)第４象限 \({\small (2)}~\)第３象限
\({\small (1)}~\)第４象限 \({\small (2)}~\)第３象限
p.110 練習8
\(\cos{\theta}={\large \frac{2\sqrt{2}}{3}}~,~\tan{\theta}=-{\large \frac{1}{2\sqrt{2}}}\)
\(\cos{\theta}={\large \frac{2\sqrt{2}}{3}}~,~\tan{\theta}=-{\large \frac{1}{2\sqrt{2}}}\)
p.110 練習9
\(\cos{\theta}=-{\large \frac{1}{\sqrt{10}}}~,~\sin{\theta}=-{\large \frac{3}{\sqrt{10}}}\)
→ 三角関数の相互関係の公式
\(\cos{\theta}=-{\large \frac{1}{\sqrt{10}}}~,~\sin{\theta}=-{\large \frac{3}{\sqrt{10}}}\)
→ 三角関数の相互関係の公式
p.112 練習12
\({\small (1)}~\)
[証明]
(左辺)
\(=(\sin{\theta}+\cos{\theta})^2+(\sin{\theta}-\cos{\theta})^2\)
\(=(\sin^2{\theta}+2\sin{\theta}\cos{\theta}+\cos^2{\theta})\)
\(+(\sin^2{\theta}-2\sin{\theta}\cos{\theta}+\cos^2{\theta})\)
\(=2(\sin^2{\theta}+\cos^2{\theta})\)
\(=2\)
したがって、
\((\sin{\theta}+\cos{\theta})^2+(\sin{\theta}-\cos{\theta})^2=2\)
[終]
\({\small (2)}~\)
[証明]
(左辺)$$=\tan^2{\theta}-\sin^2{\theta}$$$$=\frac{\sin^2{\theta}}{\cos^2{\theta}}-\sin^2{\theta}$$$$=\frac{\sin^2{\theta}-\sin^2{\theta}\cos^2{\theta}}{\cos^2{\theta}}$$$$=\frac{\sin^2{\theta}(1-\cos^2{\theta})}{\cos^2{\theta}}$$$$=\frac{\sin^2{\theta}\cdot\sin^2{\theta}}{\cos^2{\theta}}$$$$=\frac{\sin^4{\theta}}{\cos^2{\theta}}$$ (右辺)$$=\tan^2{\theta}\sin^2{\theta}$$$$=\frac{\sin^2{\theta}}{\cos^2{\theta}}\cdot\sin^2{\theta}$$$$=\frac{\sin^4{\theta}}{\cos^2{\theta}}$$したがって、
\(\tan^2{\theta}-\sin^2{\theta}=\tan^2{\theta}\sin^2{\theta}\)
[終]
→ 三角関数の等式の証明
\({\small (1)}~\)
[証明]
(左辺)
\(=(\sin{\theta}+\cos{\theta})^2+(\sin{\theta}-\cos{\theta})^2\)
\(=(\sin^2{\theta}+2\sin{\theta}\cos{\theta}+\cos^2{\theta})\)
\(+(\sin^2{\theta}-2\sin{\theta}\cos{\theta}+\cos^2{\theta})\)
\(=2(\sin^2{\theta}+\cos^2{\theta})\)
\(=2\)
したがって、
\((\sin{\theta}+\cos{\theta})^2+(\sin{\theta}-\cos{\theta})^2=2\)
[終]
\({\small (2)}~\)
[証明]
(左辺)$$=\tan^2{\theta}-\sin^2{\theta}$$$$=\frac{\sin^2{\theta}}{\cos^2{\theta}}-\sin^2{\theta}$$$$=\frac{\sin^2{\theta}-\sin^2{\theta}\cos^2{\theta}}{\cos^2{\theta}}$$$$=\frac{\sin^2{\theta}(1-\cos^2{\theta})}{\cos^2{\theta}}$$$$=\frac{\sin^2{\theta}\cdot\sin^2{\theta}}{\cos^2{\theta}}$$$$=\frac{\sin^4{\theta}}{\cos^2{\theta}}$$ (右辺)$$=\tan^2{\theta}\sin^2{\theta}$$$$=\frac{\sin^2{\theta}}{\cos^2{\theta}}\cdot\sin^2{\theta}$$$$=\frac{\sin^4{\theta}}{\cos^2{\theta}}$$したがって、
\(\tan^2{\theta}-\sin^2{\theta}=\tan^2{\theta}\sin^2{\theta}\)
[終]
→ 三角関数の等式の証明
p.117 練習13
\({\small (1)}~\)
周期 \(2\pi\)
\({\small (2)}~\)
周期 \(2\pi\)
\({\small (3)}~\)
周期 \(\pi\)
→ 三角関数のグラフ④（平行移動）
\({\small (1)}~\)
周期 \(2\pi\)
\({\small (2)}~\)
周期 \(2\pi\)
\({\small (3)}~\)
周期 \(\pi\)
→ 三角関数のグラフ④（平行移動）
p.118 練習15
\({\small (1)}~\)
周期 \(\pi\)
\({\small (2)}~\)
周期 \(4\pi\)
\({\small (3)}~\)
周期 \({\large \frac{\pi}{2}}\)
→ 三角関数のグラフ③（周期の変化）
\({\small (1)}~\)
周期 \(\pi\)
\({\small (2)}~\)
周期 \(4\pi\)
\({\small (3)}~\)
周期 \({\large \frac{\pi}{2}}\)
→ 三角関数のグラフ③（周期の変化）
p.120 練習16
\({\small (1)}~-{\large \frac{1}{2}}\) \({\small (2)}~{\large \frac{\sqrt{3}}{2}}\) \({\small (3)}~-1\)
→ 三角関数の性質①
→ 三角関数の性質②
\({\small (1)}~-{\large \frac{1}{2}}\) \({\small (2)}~{\large \frac{\sqrt{3}}{2}}\) \({\small (3)}~-1\)
→ 三角関数の性質①
→ 三角関数の性質②
p.121 練習17
\({\small (1)}~\theta={\large \frac{\pi}{3}}~,~{\large \frac{2}{3}}\pi\)
\({\small (2)}~\theta={\large \frac{2}{3}}\pi~,~{\large \frac{4}{3}}\pi\)
\({\small (3)}~\theta={\large \frac{3}{2}}\pi\)
\({\small (1)}~\theta={\large \frac{\pi}{3}}~,~{\large \frac{2}{3}}\pi\)
\({\small (2)}~\theta={\large \frac{2}{3}}\pi~,~{\large \frac{4}{3}}\pi\)
\({\small (3)}~\theta={\large \frac{3}{2}}\pi\)
p.121 練習18
\({\small (1)}~\theta={\large \frac{4}{3}}\pi+2n\pi~,~{\large \frac{5}{3}}\pi+2n\pi\)
ただし、\(n\) は整数
\({\small (2)}~\theta={\large \frac{3}{4}}\pi+2n\pi~,~{\large \frac{5}{4}}\pi+2n\pi\)
ただし、\(n\) は整数
\({\small (1)}~\theta={\large \frac{4}{3}}\pi+2n\pi~,~{\large \frac{5}{3}}\pi+2n\pi\)
ただし、\(n\) は整数
\({\small (2)}~\theta={\large \frac{3}{4}}\pi+2n\pi~,~{\large \frac{5}{4}}\pi+2n\pi\)
ただし、\(n\) は整数
p.122 練習19
\({\small (1)}~\)
\(\theta={\large \frac{\pi}{6}}~,~{\large \frac{7}{6}}\pi\)
\(\theta={\large \frac{\pi}{6}}+n\pi\)
ただし、\(n\) は整数
\({\small (2)}~\)
\(\theta={\large \frac{2}{3}}\pi~,~{\large \frac{5}{3}}\pi\)
\(\theta={\large \frac{2}{3}}\pi+n\pi\)
ただし、\(n\) は整数
\({\small (1)}~\)
\(\theta={\large \frac{\pi}{6}}~,~{\large \frac{7}{6}}\pi\)
\(\theta={\large \frac{\pi}{6}}+n\pi\)
ただし、\(n\) は整数
\({\small (2)}~\)
\(\theta={\large \frac{2}{3}}\pi~,~{\large \frac{5}{3}}\pi\)
\(\theta={\large \frac{2}{3}}\pi+n\pi\)
ただし、\(n\) は整数
p.123 練習20
\({\small (1)}~\theta={\large \frac{\pi}{6}}~,~{\large \frac{5}{6}}\pi\)
\({\small (2)}~\theta=0~,~{\large \frac{2}{3}}\pi~,~{\large \frac{4}{3}}\pi\)
→ 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
\({\small (1)}~\theta={\large \frac{\pi}{6}}~,~{\large \frac{5}{6}}\pi\)
\({\small (2)}~\theta=0~,~{\large \frac{2}{3}}\pi~,~{\large \frac{4}{3}}\pi\)
→ 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.123 練習21
\({\small (1)}~{\large \frac{\pi}{3}}≦\theta≦{\large \frac{5}{3}}\pi\)
\({\small (2)}~{\large \frac{\pi}{4}}<\theta<{\large \frac{3}{4}}\pi\)
→ 三角関数を含む不等式①
\({\small (1)}~{\large \frac{\pi}{3}}≦\theta≦{\large \frac{5}{3}}\pi\)
\({\small (2)}~{\large \frac{\pi}{4}}<\theta<{\large \frac{3}{4}}\pi\)
→ 三角関数を含む不等式①
補充問題
p.124 1
\({\small (1)}~\theta={\large \frac{\pi}{3}}~,~\pi\)
\({\small (2)}~\theta={\large \frac{\pi}{2}}~,~{\large \frac{5}{6}}\pi\)
\({\small (3)}~\theta={\large \frac{\pi}{2}}~,~{\large \frac{11}{6}}\pi\)
→ 三角関数を含む方程式②（範囲変化）
\({\small (1)}~\theta={\large \frac{\pi}{3}}~,~\pi\)
\({\small (2)}~\theta={\large \frac{\pi}{2}}~,~{\large \frac{5}{6}}\pi\)
\({\small (3)}~\theta={\large \frac{\pi}{2}}~,~{\large \frac{11}{6}}\pi\)
→ 三角関数を含む方程式②（範囲変化）
p.124 2
\({\small (1)}~0≦\theta<{\large \frac{5}{6}}\pi~,~{\large \frac{7}{6}}\pi<\theta<2\pi\)
\({\small (2)}~{\large \frac{7}{6}}\pi≦\theta≦{\large \frac{11}{6}}\pi\)
\({\small (3)}~{\large \frac{\pi}{4}}≦\theta<{\large \frac{\pi}{2}}~,~{\large \frac{5}{4}}\pi≦\theta<{\large \frac{3}{2}}\pi\)
\({\small (4)}~0≦\theta<{\large \frac{\pi}{3}}~,~{\large \frac{\pi}{2}}<\theta<{\large \frac{4}{3}}\pi\)
\({\large \frac{3}{2}}\pi<\theta<2\pi\)
\({\small (1)}~0≦\theta<{\large \frac{5}{6}}\pi~,~{\large \frac{7}{6}}\pi<\theta<2\pi\)
\({\small (2)}~{\large \frac{7}{6}}\pi≦\theta≦{\large \frac{11}{6}}\pi\)
\({\small (3)}~{\large \frac{\pi}{4}}≦\theta<{\large \frac{\pi}{2}}~,~{\large \frac{5}{4}}\pi≦\theta<{\large \frac{3}{2}}\pi\)
\({\small (4)}~0≦\theta<{\large \frac{\pi}{3}}~,~{\large \frac{\pi}{2}}<\theta<{\large \frac{4}{3}}\pi\)
\({\large \frac{3}{2}}\pi<\theta<2\pi\)
p.124 3
\({\small (1)}~y=(x+1)^2-1\)
\({\small (2)}~\)
\(\theta={\large \frac{\pi}{2}}\) で最大値 \(3\)
\(\theta={\large \frac{3}{2}}\pi\) で最小値 \(-1\)
→ 三角関数を含む２次関数
\({\small (1)}~y=(x+1)^2-1\)
\({\small (2)}~\)
\(\theta={\large \frac{\pi}{2}}\) で最大値 \(3\)
\(\theta={\large \frac{3}{2}}\pi\) で最小値 \(-1\)
→ 三角関数を含む２次関数
第２節 加法定理
p.125 練習22
①において、\(\beta\) を \(-\beta\) とすると、
(左辺)
\(=\sin{(\alpha-\beta)}\)
(右辺)
\(=\sin{\alpha}\cos{(-\beta)}+\cos{\alpha}\sin{(-\beta)}\)
\(=\sin{\alpha}\cos{\beta}-\cos{\alpha}\sin{\beta}\)
したがって、
\(\sin{(\alpha-\beta)}\)
\(=\sin{\alpha}\cos{\beta}-\cos{\alpha}\sin{\beta}\)
①において、\(\beta\) を \(-\beta\) とすると、
(左辺)
\(=\sin{(\alpha-\beta)}\)
(右辺)
\(=\sin{\alpha}\cos{(-\beta)}+\cos{\alpha}\sin{(-\beta)}\)
\(=\sin{\alpha}\cos{\beta}-\cos{\alpha}\sin{\beta}\)
したがって、
\(\sin{(\alpha-\beta)}\)
\(=\sin{\alpha}\cos{\beta}-\cos{\alpha}\sin{\beta}\)
②において、\(\beta\) を \(-\beta\) とすると、
(左辺)
\(=\cos{(\alpha-\beta)}\)
(右辺)
\(=\cos{\alpha}\cos{(-\beta)}-\sin{\alpha}\sin{(-\beta)}\)
\(=\cos{\alpha}\cos{\beta}+\sin{\alpha}\sin{\beta}\)
したがって、
\(\cos{(\alpha-\beta)}\)
\(=\cos{\alpha}\cos{\beta}+\sin{\alpha}\sin{\beta}\)
p.126 練習23
\({\large \frac{\sqrt{6}-\sqrt{2}}{4}}\)
\({\large \frac{\sqrt{6}-\sqrt{2}}{4}}\)
p.126 練習24
\({\large \frac{\sqrt{6}+\sqrt{2}}{4}}\)
\({\large \frac{\sqrt{6}+\sqrt{2}}{4}}\)
p.127 練習25
\(-2-\sqrt{3}\)
\(-2-\sqrt{3}\)
p.129 練習28
\({\small (1)}~{\large \frac{2}{3}}\) \({\small (2)}~-{\large \frac{4\sqrt{5}}{9}}\) \({\small (3)}~{\large \frac{1}{9}}\)
→ ２倍角の公式
\({\small (1)}~{\large \frac{2}{3}}\) \({\small (2)}~-{\large \frac{4\sqrt{5}}{9}}\) \({\small (3)}~{\large \frac{1}{9}}\)
→ ２倍角の公式
p.129 練習29
\({\small (1)}~\) [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]
\({\small (1)}~\) [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.130 練習30
\({\small (1)}~{\large \frac{\sqrt{2-\sqrt{2}}}{2}}\)
\({\small (2)}~{\large \frac{\sqrt{2+\sqrt{2}}}{2}}\)
\({\small (3)}~{\large \frac{\sqrt{2-\sqrt{2}}}{2}}\)
→ 半角の公式
\({\small (1)}~{\large \frac{\sqrt{2-\sqrt{2}}}{2}}\)
\({\small (2)}~{\large \frac{\sqrt{2+\sqrt{2}}}{2}}\)
\({\small (3)}~{\large \frac{\sqrt{2-\sqrt{2}}}{2}}\)
→ 半角の公式
p.130 練習31
[証明]$$~~~~~~\tan{2\alpha}$$$$~=\tan{(\alpha+\alpha)}$$$$~=\frac{\tan{\alpha}+\tan{\alpha}}{1-\tan{\alpha}\tan{\alpha}}$$$$~=\frac{2\tan{\alpha}}{1-\tan^2{\alpha}}$$[終]
[証明]$$~~~~~~\tan^2{\frac{\alpha}{2}}$$$$~=\frac{\sin^2{\frac{\alpha}{2}}}{\cos^2{\frac{\alpha}{2}}}$$$$~=\frac{\frac{1-\cos{\alpha}}{2}}{\frac{1+\cos{\alpha}}{2}}$$$$~=\frac{1-\cos{\alpha}}{1+\cos{\alpha}}$$[終]
\({\small (1)}~-{\large \frac{3}{4}}\)
\({\small (2)}~{\large \frac{1}{\sqrt{5}}}\)
→ ２倍角の公式
→ 半角の公式
[証明]$$~~~~~~\tan{2\alpha}$$$$~=\tan{(\alpha+\alpha)}$$$$~=\frac{\tan{\alpha}+\tan{\alpha}}{1-\tan{\alpha}\tan{\alpha}}$$$$~=\frac{2\tan{\alpha}}{1-\tan^2{\alpha}}$$[終]
[証明]$$~~~~~~\tan^2{\frac{\alpha}{2}}$$$$~=\frac{\sin^2{\frac{\alpha}{2}}}{\cos^2{\frac{\alpha}{2}}}$$$$~=\frac{\frac{1-\cos{\alpha}}{2}}{\frac{1+\cos{\alpha}}{2}}$$$$~=\frac{1-\cos{\alpha}}{1+\cos{\alpha}}$$[終]
\({\small (1)}~-{\large \frac{3}{4}}\)
\({\small (2)}~{\large \frac{1}{\sqrt{5}}}\)
→ ２倍角の公式
→ 半角の公式
p.131 練習32
\({\small (1)}~\theta=0~,~{\large \frac{\pi}{6}}~,~{\large \frac{5}{6}}\pi~,~\pi\)
\({\small (2)}~\theta={\large \frac{\pi}{2}}~,~{\large \frac{7}{6}}\pi~,~{\large \frac{3}{2}}\pi~,~{\large \frac{11}{6}}\pi\)
→ ２倍角を含む方程式・不等式
\({\small (1)}~\theta=0~,~{\large \frac{\pi}{6}}~,~{\large \frac{5}{6}}\pi~,~\pi\)
\({\small (2)}~\theta={\large \frac{\pi}{2}}~,~{\large \frac{7}{6}}\pi~,~{\large \frac{3}{2}}\pi~,~{\large \frac{11}{6}}\pi\)
→ ２倍角を含む方程式・不等式
p.133 練習33
\({\small (1)}~2\sin{\left(\theta+{\large \frac{\pi}{6}}\right)}\)
\({\small (2)}~\sqrt{2}\sin{\left(\theta-{\large \frac{\pi}{4}}\right)}\)
→ 三角関数の合成
\({\small (1)}~2\sin{\left(\theta+{\large \frac{\pi}{6}}\right)}\)
\({\small (2)}~\sqrt{2}\sin{\left(\theta-{\large \frac{\pi}{4}}\right)}\)
→ 三角関数の合成
補充問題
p.136 4
\({\small (1)}~2~,~4\)
\({\small (2)}~(-\sqrt{2},3\sqrt{2})\)
\({\small (1)}~2~,~4\)
\({\small (2)}~(-\sqrt{2},3\sqrt{2})\)
p.136 5
\({\large \frac{\pi}{4}}\)
\({\large \frac{\pi}{4}}\)
p.136 6
\({\small (1)}~0≦\theta≦{\large \frac{\pi}{3}}~,~\pi≦\theta≦{\large \frac{5}{3}}\pi\)
\({\small (2)}~0<\theta<\pi~,~{\large \frac{7}{6}}\pi<\theta<{\large \frac{11}{6}}\pi\)
\({\small (1)}~0≦\theta≦{\large \frac{\pi}{3}}~,~\pi≦\theta≦{\large \frac{5}{3}}\pi\)
\({\small (2)}~0<\theta<\pi~,~{\large \frac{7}{6}}\pi<\theta<{\large \frac{11}{6}}\pi\)
p.136 7
最大値 \(\sqrt{5}\)、最小値 \(-\sqrt{5}\)
最大値 \(\sqrt{5}\)、最小値 \(-\sqrt{5}\)
章末問題 三角関数
章末問題Ａ
p.137 1
\({\small (1)}~-{\large \frac{\sqrt{3}}{2}}\) \({\small (2)}~0\) \({\small (3)}~{\large \frac{1}{\sqrt{3}}}\)
\({\small (1)}~-{\large \frac{\sqrt{3}}{2}}\) \({\small (2)}~0\) \({\small (3)}~{\large \frac{1}{\sqrt{3}}}\)
p.137 2
\({\small (1)}~\)
周期 \(4\pi\)
\({\small (2)}~\)
周期 \(2\pi\)
\({\small (1)}~\)
周期 \(4\pi\)
\({\small (2)}~\)
周期 \(2\pi\)
p.137 3
\({\small (1)}~-{\large \frac{\sqrt{6}}{2}}\) \({\small (2)}~-{\large \frac{3\sqrt{6}}{8}}\)
\({\small (1)}~-{\large \frac{\sqrt{6}}{2}}\) \({\small (2)}~-{\large \frac{3\sqrt{6}}{8}}\)
p.137 4
\({\small (1)}~0\) \({\small (2)}~-{\large \frac{7}{25}}\)
\({\small (1)}~0\) \({\small (2)}~-{\large \frac{7}{25}}\)
p.137 5
\({\small (1)}~\)[証明]
(左辺)$$=\frac{1}{1+\cos{\theta}}+\frac{1}{1-\cos{\theta}}$$$$=\frac{1-\cos{\theta}+1+\cos{\theta}}{(1+\cos{\theta})(1-\cos{\theta})}$$$$=\frac{2}{1-\cos^2{\theta}}$$$$=\frac{2}{\sin^2{\theta}}$$したがって、$$\frac{1}{1+\cos{\theta}}+\frac{1}{1-\cos{\theta}}=\frac{2}{\sin^2{\theta}}$$[終]
\({\small (2)}~\)[証明]
(左辺)$$=\frac{1}{\tan{\theta}}-\tan{\theta}$$$$=\frac{\cos{\theta}}{\sin{\theta}}-\frac{\sin{\theta}}{\cos{\theta}}$$$$=\frac{\cos^2{\theta}-\sin^2{\theta}}{\sin{\theta}\cos{\theta}}$$ (右辺)$$=\frac{2\cos{2\theta}}{\sin{2\theta}}$$$$=\frac{2(\cos^2{\theta}-\sin^2{\theta})}{2\sin{\theta}\cos{\theta}}$$$$=\frac{\cos^2{\theta}-\sin^2{\theta}}{\sin{\theta}\cos{\theta}}$$したがって、$$\frac{1}{\tan{\theta}}-\tan{\theta}=\frac{2\cos{2\theta}}{\sin{2\theta}}$$[終]
\({\small (1)}~\)[証明]
(左辺)$$=\frac{1}{1+\cos{\theta}}+\frac{1}{1-\cos{\theta}}$$$$=\frac{1-\cos{\theta}+1+\cos{\theta}}{(1+\cos{\theta})(1-\cos{\theta})}$$$$=\frac{2}{1-\cos^2{\theta}}$$$$=\frac{2}{\sin^2{\theta}}$$したがって、$$\frac{1}{1+\cos{\theta}}+\frac{1}{1-\cos{\theta}}=\frac{2}{\sin^2{\theta}}$$[終]
\({\small (2)}~\)[証明]
(左辺)$$=\frac{1}{\tan{\theta}}-\tan{\theta}$$$$=\frac{\cos{\theta}}{\sin{\theta}}-\frac{\sin{\theta}}{\cos{\theta}}$$$$=\frac{\cos^2{\theta}-\sin^2{\theta}}{\sin{\theta}\cos{\theta}}$$ (右辺)$$=\frac{2\cos{2\theta}}{\sin{2\theta}}$$$$=\frac{2(\cos^2{\theta}-\sin^2{\theta})}{2\sin{\theta}\cos{\theta}}$$$$=\frac{\cos^2{\theta}-\sin^2{\theta}}{\sin{\theta}\cos{\theta}}$$したがって、$$\frac{1}{\tan{\theta}}-\tan{\theta}=\frac{2\cos{2\theta}}{\sin{2\theta}}$$[終]
p.137 6
\({\small (1)}~x={\large \frac{\pi}{6}}~,~{\large \frac{5}{6}}\pi~,~{\large \frac{3}{2}}\pi\)
\({\small (2)}~x=0~,~{\large \frac{2}{3}}\pi~,~{\large \frac{4}{3}}\pi\)
\({\small (1)}~x={\large \frac{\pi}{6}}~,~{\large \frac{5}{6}}\pi~,~{\large \frac{3}{2}}\pi\)
\({\small (2)}~x=0~,~{\large \frac{2}{3}}\pi~,~{\large \frac{4}{3}}\pi\)
p.137 7
\({\small (1)}~\)$$~~~~~\sin{(\pi-\theta)}$$$$~=\sin{\pi}\cos{\theta}-\cos{\pi}\sin{\theta}$$$$~=0\cdot\cos{\theta}-(-1)\cdot\sin{\theta}$$$$=\sin{\theta}$$
$$~~~~~~\cos{(\pi-\theta)}$$$$~=\cos{\pi}\cos{\theta}+\sin{\pi}\sin{\theta}$$$$~=(-1)\cdot\cos{\theta}+0\cdot\sin{\theta}$$$$~=-\cos{\theta}$$
$$~~~~~~\tan{(\pi-\theta)}$$$$~=\frac{\tan{\pi}-\tan{\theta}}{1+\tan{\pi}\tan{\theta}}$$$$~=\frac{0-\tan{\theta}}{1+0\cdot\tan{\theta}}$$$$=-\tan{\theta}$$
\({\small (2)}~\)$$~~~~~\sin{\left(\frac{\pi}{2}-\theta\right)}$$$$~=\sin{\frac{\pi}{2}}\cos{\theta}-\cos{\frac{\pi}{2}}\sin{\theta}$$$$~=1\cdot\cos{\theta}-0\cdot\sin{\theta}$$$$=\cos{\theta}$$
$$~~~~~~\cos{\left(\frac{\pi}{2}-\theta\right)}$$$$~=\cos{\frac{\pi}{2}}\cos{\theta}+\sin{\frac{\pi}{2}}\sin{\theta}$$$$~=0\cdot\cos{\theta}+1\cdot\sin{\theta}$$$$~=\sin{\theta}$$
$$~~~~~~\tan{\left(\frac{\pi}{2}-\theta\right)}$$$$~=\frac{\sin{\left(\frac{\pi}{2}-\theta\right)}}{\cos{\left(\frac{\pi}{2}-\theta\right)}}$$$$~=\frac{\cos{\theta}}{\sin{\theta}}$$$$~=\frac{1}{\tan{\theta}}$$
\({\small (1)}~\)$$~~~~~\sin{(\pi-\theta)}$$$$~=\sin{\pi}\cos{\theta}-\cos{\pi}\sin{\theta}$$$$~=0\cdot\cos{\theta}-(-1)\cdot\sin{\theta}$$$$=\sin{\theta}$$
$$~~~~~~\cos{(\pi-\theta)}$$$$~=\cos{\pi}\cos{\theta}+\sin{\pi}\sin{\theta}$$$$~=(-1)\cdot\cos{\theta}+0\cdot\sin{\theta}$$$$~=-\cos{\theta}$$
$$~~~~~~\tan{(\pi-\theta)}$$$$~=\frac{\tan{\pi}-\tan{\theta}}{1+\tan{\pi}\tan{\theta}}$$$$~=\frac{0-\tan{\theta}}{1+0\cdot\tan{\theta}}$$$$=-\tan{\theta}$$
\({\small (2)}~\)$$~~~~~\sin{\left(\frac{\pi}{2}-\theta\right)}$$$$~=\sin{\frac{\pi}{2}}\cos{\theta}-\cos{\frac{\pi}{2}}\sin{\theta}$$$$~=1\cdot\cos{\theta}-0\cdot\sin{\theta}$$$$=\cos{\theta}$$
$$~~~~~~\cos{\left(\frac{\pi}{2}-\theta\right)}$$$$~=\cos{\frac{\pi}{2}}\cos{\theta}+\sin{\frac{\pi}{2}}\sin{\theta}$$$$~=0\cdot\cos{\theta}+1\cdot\sin{\theta}$$$$~=\sin{\theta}$$
$$~~~~~~\tan{\left(\frac{\pi}{2}-\theta\right)}$$$$~=\frac{\sin{\left(\frac{\pi}{2}-\theta\right)}}{\cos{\left(\frac{\pi}{2}-\theta\right)}}$$$$~=\frac{\cos{\theta}}{\sin{\theta}}$$$$~=\frac{1}{\tan{\theta}}$$
章末問題Ｂ
p.138 8
\(\theta={\large \frac{7}{6}}\pi~,~{\large \frac{11}{6}}\pi\) で最小値 \(-{\large \frac{3}{2}}\)
\(\theta={\large \frac{7}{6}}\pi~,~{\large \frac{11}{6}}\pi\) で最小値 \(-{\large \frac{3}{2}}\)
p.138 9
\({\small (1)}~-{\large \frac{59}{72}}\) \({\small (2)}~1\)
\({\small (1)}~-{\large \frac{59}{72}}\) \({\small (2)}~1\)
p.138 10
[証明]
(左辺)$$~=\frac{\sin{(\alpha-\beta)}}{\sin{(\alpha+\beta)}}$$$$~=\frac{\sin{\alpha}\cos{\beta}-\cos{\alpha}\sin{\beta}}{\sin{\alpha}\cos{\beta}+\cos{\alpha}\sin{\beta}}$$両辺を \(\cos{\alpha}\cos{\beta}\) で割ると、$$~=\frac{\frac{\sin{\alpha}}{\cos{\alpha}}-\frac{\sin{\beta}}{\cos{\beta}}}{\frac{\sin{\alpha}}{\cos{\alpha}}+\frac{\sin{\beta}}{\cos{\beta}}}$$$$~=\frac{\tan{\alpha}-\tan{\beta}}{\tan{\alpha}+\tan{\beta}}$$したがって、$$\frac{\sin{(\alpha-\beta)}}{\sin{(\alpha+\beta)}}=\frac{\tan{\alpha}-\tan{\beta}}{\tan{\alpha}+\tan{\beta}}$$[終]
[証明]
(左辺)$$~=\frac{\sin{(\alpha-\beta)}}{\sin{(\alpha+\beta)}}$$$$~=\frac{\sin{\alpha}\cos{\beta}-\cos{\alpha}\sin{\beta}}{\sin{\alpha}\cos{\beta}+\cos{\alpha}\sin{\beta}}$$両辺を \(\cos{\alpha}\cos{\beta}\) で割ると、$$~=\frac{\frac{\sin{\alpha}}{\cos{\alpha}}-\frac{\sin{\beta}}{\cos{\beta}}}{\frac{\sin{\alpha}}{\cos{\alpha}}+\frac{\sin{\beta}}{\cos{\beta}}}$$$$~=\frac{\tan{\alpha}-\tan{\beta}}{\tan{\alpha}+\tan{\beta}}$$したがって、$$\frac{\sin{(\alpha-\beta)}}{\sin{(\alpha+\beta)}}=\frac{\tan{\alpha}-\tan{\beta}}{\tan{\alpha}+\tan{\beta}}$$[終]
p.138 11
\({\small (1)}~\)
周期 \(\pi\)
\({\small (2)}~\)
周期 \(\pi\)
\({\small (1)}~\)
周期 \(\pi\)
\({\small (2)}~\)
周期 \(\pi\)
p.138 12
\({\small (1)}~\)
\(x={\large \frac{5}{6}}\pi\) で最大値 \(2\)
\(x={\large \frac{11}{6}}\pi\) で最小値 \(-2\)
\({\small (2)}~x={\large \frac{\pi}{3}}~,~{\large \frac{4}{3}}\pi\)
\({\small (3)}~0≦x≦{\large \frac{\pi}{3}}~,~{\large \frac{4}{3}}\pi≦x<2\pi\)
\({\small (1)}~\)
\(x={\large \frac{5}{6}}\pi\) で最大値 \(2\)
\(x={\large \frac{11}{6}}\pi\) で最小値 \(-2\)
\({\small (2)}~x={\large \frac{\pi}{3}}~,~{\large \frac{4}{3}}\pi\)
\({\small (3)}~0≦x≦{\large \frac{\pi}{3}}~,~{\large \frac{4}{3}}\pi≦x<2\pi\)
p.138 13
最大値 \({\large \frac{\sqrt{5}}{2}}+1\)
最小値 \({\large \frac{3}{2}}\)
最大値 \({\large \frac{\sqrt{5}}{2}}+1\)
最小値 \({\large \frac{3}{2}}\)
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