数研出版：改訂版高等学校数学Ⅱ

p.129 1
\(\sin{\theta}={\large \frac{2}{\sqrt{5}}}~,~\cos{\theta}=-{\large \frac{1}{\sqrt{5}}}\)
または
\(\sin{\theta}=-{\large \frac{2}{\sqrt{5}}}~,~\cos{\theta}={\large \frac{1}{\sqrt{5}}}\)

p.129 2
\({\small (1)}~{\large \frac{1-a^2}{2}}\)
\({\small (2)}~{\large \frac{a(3-a^2)}{2}}\)

p.129 3
\({\small (1)}~\)

周期 \(\pi\)
\({\small (2)}~\)

周期 \(4\pi\)
\({\small (3)}~\)

周期 \(2\pi\)
\({\small (4)}~\)

周期 \({\large \frac{2}{3}}\pi\)

p.129 4
\({\small (1)}~\theta={\large \frac{\pi}{6}}~,~{\large \frac{11}{6}}\pi\)
\({\small (2)}~\theta={\large \frac{5}{6}}\pi~,~{\large \frac{11}{6}}\pi\)
\({\small (3)}~{\large \frac{4}{3}}\pi<\theta<{\large \frac{5}{3}}\pi\)
\({\small (4)}~{\large \frac{\pi}{2}}<\theta≦{\large \frac{2}{3}}\pi~,~{\large \frac{3}{2}}\pi<\theta≦{\large \frac{5}{3}}\pi\)
\({\small (5)}~\theta={\large \frac{\pi}{2}}~,~{\large \frac{5}{6}}\pi\)
\({\small (6)}~0≦\theta<{\large \frac{\pi}{2}}~,~{\large \frac{5}{6}}\pi<\theta<2\pi\)
→ 三角関数を含む不等式②（範囲変化）

p.129 5
\(\theta={\large \frac{2}{3}}\pi~,~{\large \frac{4}{3}}\pi\) で最大値 \({\large \frac{5}{4}}\)
\(\theta=0\) で最小値 \(-1\)

p.129 6
ア:\(2\)　イ:\(5\)　ウ:\(2\)　エ:\(2\)　オ:\(7\)
カ:\(6\)　キ:\(1\)　ク:\(1\)　ケ:\(6\)
→ 三角関数を含む２次方程式

p.132 練習24
\({\large \frac{\sqrt{6}-\sqrt{2}}{4}}\)

p.132 練習25
\({\large \frac{\sqrt{6}-\sqrt{2}}{4}}~,~{\large \frac{\sqrt{6}+\sqrt{2}}{4}}\)
→ 加法定理

p.132 練習26
\({\large \frac{6+4\sqrt{5}}{15}}~,~-{\large \frac{3\sqrt{5}+8}{15}}\)
→ 加法定理と式の値

p.133 練習27
\(-2-\sqrt{3}\)

p.133 練習28
\(2-\sqrt{3}\)

p.134 練習29
\({\large \frac{\pi}{4}}\)
→ ２直線のなす角

p.135 研究1
\(\left({\large \frac{4\sqrt{3}-3}{2}},{\large \frac{3\sqrt{3}+4}{2}}\right)\)

p.136 練習30
\({\small (1)}~-{\large \frac{4\sqrt{5}}{9}}\)
\({\small (2)}~{\large \frac{1}{9}}\)
\({\small (3)}~-4\sqrt{5}\)
→ ２倍角の公式

p.136 練習31
\({\small (1)}~\) [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.137 練習32
\({\small (1)}~{\large \frac{\sqrt{2-\sqrt{2}}}{2}}\)
\({\small (2)}~{\large \frac{\sqrt{2+\sqrt{2}}}{2}}\)
\({\small (3)}~{\large \frac{\sqrt{2-\sqrt{2}}}{2}}\)

p.137 練習33
\({\small (1)}~{\large \frac{3}{\sqrt{10}}}\)
\({\small (2)}~{\large \frac{1}{\sqrt{10}}}\)
\({\small (3)}~3\)
→ 半角の公式

p.138 練習34
\({\small (1)}~\theta=0~,~{\large \frac{\pi}{6}}~,~{\large \frac{5}{6}}\pi~,~\pi\)
\({\small (2)}~0<\theta<{\large \frac{\pi}{6}}~,~{\large \frac{5}{6}}\pi<\theta<\pi\)
→ ２倍角を含む方程式・不等式

p.139 練習35
\({\small (1)}~2\sin{\left(\theta+{\large \frac{\pi}{6}}\right)}\)
\({\small (2)}~\sqrt{2}\sin{\left(\theta-{\large \frac{\pi}{4}}\right)}\)
→ 三角関数の合成

p.140 練習36
最大値 \(2\)、最小値 \(-2\)
→ 三角関数の最大値・最小値

p.141 練習37
\(x={\large \frac{\pi}{2}}~,~{\large \frac{11}{6}}\pi\)
→ 合成を用いる方程式と不等式

p.143 7
\({\small (1)}~{\large \frac{33}{65}}\)　\({\small (2)}~{\large \frac{16}{65}}\)

p.143 8
\(y=(\sqrt{3}+2)x~,~y=(\sqrt{3}-2)x\)

p.144 9
\({\small (1)}~\)[証明]
　(左辺)$$=(\sin{\alpha}+\cos{\alpha})^2$$$$=\sin^2{\alpha}+2\sin{\alpha}\cos{\alpha}+\cos^2{\alpha}$$$$=1+2\sin{\alpha}\cos{\alpha}$$　(右辺)$$=1+\sin{2\alpha}$$$$=1+2\sin{\alpha}\cos{\alpha}$$したがって、
　\((\sin{\alpha}+\cos{\alpha})^2=1+\sin{2\alpha}\)
[終]
\({\small (2)}~\)[証明]
　(左辺)$$=\frac{\sin{2\alpha}}{1+\cos{2\alpha}}$$$$=\frac{2\sin{\alpha}\cos{\alpha}}{1+2\cos^2{\alpha}-1}$$$$=\frac{2\sin{\alpha}\cos{\alpha}}{2\cos^2{\alpha}}$$$$=\frac{\sin{\alpha}}{\cos{\alpha}}$$$$=\tan{\alpha}$$したがって、$$\frac{\sin{2\alpha}}{1+\cos{2\alpha}}=\tan{\alpha}$$[終]

p.143 10
\({\small (1)}~\theta=0~,~{\large \frac{\pi}{3}}~,~\pi~,~{\large \frac{5}{3}}\pi\)
\({\small (2)}~0<\theta<\pi~,~{\large \frac{7}{6}}\pi<\theta<{\large \frac{11}{6}}\pi\)
→ 三角関数を含む２次不等式

p.143 11
\({\small (1)}~x={\large \frac{\pi}{2}}~,~{\large \frac{5}{6}}\pi\)
\({\small (2)}~0≦x≦{\large \frac{\pi}{2}}~,~{\large \frac{5}{6}}\pi≦x<2\pi\)

p.143 12
ア:\(5\)　イ:\(4\)　ウ:\(5\)　エ:\(3\)　オ:\(5\)
カ:\(5\)　キ:\(5\)

p.144 1
\({\small (1)}~-{\large \frac{\sqrt{3}}{2}}\)　\({\small (2)}~0\)　\({\small (3)}~{\large \frac{1}{\sqrt{3}}}\)

p.144 2
\({\small (1)}~-{\large \frac{\sqrt{6}}{2}}\)　\({\small (2)}~-{\large \frac{3\sqrt{6}}{8}}\)

p.144 3
\({\small (1)}~\)[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.144 4
\({\small (1)}~x={\large \frac{\pi}{6}}~,~{\large \frac{5}{6}}\pi~,~{\large \frac{3}{2}}\pi\)
\({\small (2)}~{\large \frac{\pi}{6}}<x<{\large \frac{5}{6}}\pi\)

p.144 5
\({\small (1)}~\)$$~~~~~\sin{(\pi-\theta)}$$$$~=\sin{\pi}\cos{\theta}-\cos{\pi}\sin{\theta}$$$$~=0\cdot\cos{\theta}-(-1)\cdot\sin{\theta}$$$$=\sin{\theta}$$
$$~~~~~~\cos{(\pi-\theta)}$$$$~=\cos{\pi}\cos{\theta}+\sin{\pi}\sin{\theta}$$$$~=(-1)\cdot\cos{\theta}+0\cdot\sin{\theta}=-\cos{\theta}$$
$$~~~~~~\tan{(\pi-\theta)}$$$$~=\frac{\tan{\pi}-\tan{\theta}}{1+\tan{\pi}\tan{\theta}}$$$$~=\frac{0-\tan{\theta}}{1+0\cdot\tan{\theta}}$$$$=-\tan{\theta}$$
\({\small (2)}~\)$$~~~~~\sin{\left(\frac{\pi}{2}-\theta\right)}$$$$~=\sin{\frac{\pi}{2}}\cos{\theta}-\cos{\frac{\pi}{2}}\sin{\theta}$$$$~=1\cdot\cos{\theta}-0\cdot\sin{\theta}$$$$=\cos{\theta}$$
$$~~~~~~\cos{\left(\frac{\pi}{2}-\theta\right)}$$$$~=\cos{\frac{\pi}{2}}\cos{\theta}+\sin{\frac{\pi}{2}}\sin{\theta}$$$$~=0\cdot\cos{\theta}+1\cdot\sin{\theta}=\sin{\theta}$$
$$~~~~~~\tan{\left(\frac{\pi}{2}-\theta\right)}$$$$~=\frac{\sin{\left(\frac{\pi}{2}-\theta\right)}}{\cos{\left(\frac{\pi}{2}-\theta\right)}}$$$$~=\frac{\cos{\theta}}{\sin{\theta}}$$$$~=\frac{1}{\tan{\theta}}$$

p.144 6
\(\theta={\large \frac{3}{2}}\pi\) で最小値 \(-2\)

p.144 7
\({\small (1)}~\)
\(x={\large \frac{5}{6}}\pi\) 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.145 8
\({\small (1)}~-{\large \frac{59}{72}}\)　\({\small (2)}~1\)

p.145 9
[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.145 10
\((\sqrt{3}-3,3\sqrt{3}+1)\)

p.145 11
\({\small (1)}~\)

周期 \(\pi\)
\({\small (2)}~\)

周期 \(\pi\)

p.145 12
\(x={\large \frac{\pi}{8}}\) で最大値 \({\large \frac{\sqrt{2}}{2}}\)
\(x={\large \frac{5}{8}}\pi\) で最小値 \(-{\large \frac{\sqrt{2}}{2}}\)

p.145 13
\({\small (1)}~y=t^2+t-1\)
\({\small (2)}~-\sqrt{2}≦t≦\sqrt{2}\)
\({\small (3)}~\)
最大値 \(1+\sqrt{2}\)
最小値 \(-{\large \frac{5}{4}}\)