東京書籍：Standard数学Ⅱ

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

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

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

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