【新課程】東京書籍：Standard数学Ⅱ[702]

p.121 問1\({\small (1)}~\)

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

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

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

p.121 問2以下 \(n\) は整数$${\small (1)}~45^\circ+360^\circ\times n$$$${\small (2)}~180^\circ+360^\circ\times n$$$${\small (3)}~-30^\circ+360^\circ\times n$$

p.121 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.121 問4$$~~~{ \frac{\,3\,}{2}}\pi~,~{ \frac{\,7\,}{4}}\pi~,~-{ \frac{\,\pi\,}{3}}~,~-{ \frac{\,5\,}{4}}\pi$$

p.123 問5$$~~~36^\circ~,~210^\circ~,~-330^\circ~,~-270^\circ$$

p.123 問6$$~~~l=6\pi~,~S=24\pi$$→ 弧度法と扇形

p.125 問7$${\small (1)}~\sin{{ \frac{\,7\,}{6}}\pi}=-{ \frac{1}{\,2\,}}$$$$~~~~~\cos{{ \frac{\,7\,}{6}}\pi}=-{ \frac{\,\sqrt{3}\,}{2}}$$$$~~~~~\tan{{ \frac{\,7\,}{6}}\pi}={ \frac{1}{\,\sqrt{3}\,}}$$$${\small (2)}~\sin{{ \frac{\,9\,}{4}}\pi}={ \frac{1}{\,\sqrt{2}\,}}$$$$~~~~~\cos{{ \frac{\,9\,}{4}}\pi}={ \frac{1}{\,\sqrt{2}\,}}$$$$~~~~~\tan{{ \frac{\,9\,}{4}}\pi}=1$$$${\small (3)}~\sin{\left(-{ \frac{\,\pi\,}{3}}\right)}=-{ \frac{\,\sqrt{3}\,}{2}}$$$$~~~~~\cos{\left(-{ \frac{\,\pi\,}{3}}\right)}={ \frac{1}{\,2\,}}$$$$~~~~~\tan{\left(-{ \frac{\,\pi\,}{3}}\right)}=-\sqrt{3}$$$${\small (4)}~\sin{(-3\pi)}=0$$$$~~~~~\cos{(-3\pi)}=-1$$$$~~~~~\tan{(-3\pi)}=0$$→ 三角関数の値（単位円）
→ 【問題演習】三角関数の値（単位円）

p.125 問8\({\small (1)}~\)第４象限
\({\small (2)}~\)第２象限

p.127 問9$${\small (1)}~\sin{\theta}=-{ \frac{\,2\sqrt{2}\,}{3}}~,~\tan{\theta}=-2\sqrt{2}$$$${\small (2)}~\cos{\theta}=-{ \frac{\,\sqrt{6}\,}{3}}~,~\tan{\theta}={ \frac{\,\sqrt{2}\,}{2}}$$

p.127 問10$$~~~\cos{\theta}=-{ \frac{\,2\sqrt{5}\,}{5}}~,~\sin{\theta}=-{ \frac{\,\sqrt{5}\,}{5}}$$→ 三角関数の相互関係の公式

p.128 問11$$~~~{ \frac{12}{\,25\,}}~,~{ \frac{37}{\,125\,}}$$→ 三角関数の式の値

p.128 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.130 問13$$~~~\sin{\left(-{ \frac{\,\pi\,}{6}}\right)}=-{ \frac{1}{\,2\,}}$$$$~~~\cos{\left(-{ \frac{\,\pi\,}{6}}\right)}={ \frac{\,\sqrt{3}\,}{2}}$$$$~~~\tan{\left(-{ \frac{\,\pi\,}{6}}\right)}=-{ \frac{1}{\,\sqrt{3}\,}}$$

p.130 問14$$~~~\sin{{ \frac{\,5\,}{4}}\pi}=-{ \frac{1}{\,\sqrt{2}\,}}$$$$~~~\cos{{ \frac{\,5\,}{4}}\pi}=-{ \frac{1}{\,\sqrt{2}\,}}$$$$~~~\tan{{ \frac{\,5\,}{4}}\pi}=1$$→ 三角関数の性質①

p.131 問15$${\small (1)}~{ \frac{3}{\,8\,}}\pi$$$${\small (2)}~{ \frac{\,\pi\,}{3}}$$$${\small (3)}~{ \frac{\,\pi\,}{4}}$$$${\small (4)}~{ \frac{\,\pi\,}{6}}$$$${\small (5)}~{ \frac{\,\pi\,}{8}}$$$${\small (6)}~{ \frac{\,\pi\,}{4}}$$→ 三角関数の性質②

p.136 問16周期 \(2\pi\)

→ 三角関数のグラフ②（縦幅の変化）

p.136 問17周期 \(2\pi\)

→ 三角関数のグラフ②（縦幅の変化）

p.137 問18\(y=\sin{\theta}\) のグラフを \(\theta\) 軸方向に \(-{\large \frac{\,\pi\,}{\,4\,}}\) だけ平行移動したグラフ

p.137 問19周期 \(2\pi\)

→ 三角関数のグラフ④（平行移動）

p.138 問20周期 \(\pi\)

→ 三角関数のグラフ③（周期の変化）

p.138 問21周期 \({\large \frac{\pi}{2}}\)

\(y=\tan{\theta}\) のグラフは周期が \(\pi\) で、漸近線の位置が$$~~~\theta=-\frac{\,\pi\,}{\,2\,}~,~\frac{\,\pi\,}{\,2\,}~,~\frac{\,3\,}{\,2\,}\pi~,~\cdots$$であり、
\(y=\tan{2\theta}\) のグラフでは周期が \({\large \frac{\,\pi\,}{\,2\,}}\) で、漸近線の位置が$$~~~\theta=-\frac{\,\pi\,}{\,4\,}~,~\frac{\,\pi\,}{\,4\,}~,~\frac{\,3\,}{\,4\,}\pi~,~\cdots$$となることより、グラフをかく
→ 三角関数のグラフ⑤（式変形）

p.139 問22$${\small (1)}~\theta={ \frac{\,4\,}{3}}\pi~,~{ \frac{\,5\,}{3}}\pi$$$${\small (2)}~\theta={ \frac{\,\pi\,}{4}}~,~{ \frac{\,7\,}{4}}\pi$$$${\small (3)}~\theta={ \frac{3}{\,4\,}}\pi~,~{ \frac{\,7\,}{4}}\pi$$→ 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.140 問23$${\small (1)}~\theta={ \frac{\,\pi\,}{2}}~,~\pi$$$${\small (2)}~\theta={ \frac{\,\pi\,}{6}}~,~{ \frac{\,3\,}{2}}\pi$$→ 三角関数を含む方程式②（範囲変化）

p.141 問24$${\small (1)}~{ \frac{2}{\,3\,}}\pi≦\theta≦{ \frac{\,4\,}{3}}\pi$$$${\small (2)}~0≦\theta<{ \frac{\,\pi\,}{3}}~,~{ \frac{\,5\,}{3}}\pi<\theta<2\pi$$$${\small (3)}~0≦\theta≦{ \frac{\,5\,}{4}}\pi~,~{ \frac{\,7\,}{4}}\pi≦\theta<2\pi$$

p.142 問25$${\small (1)}~-{ \frac{\,\pi\,}{4}}<\theta<{ \frac{\,\pi\,}{6}}$$$${\small (2)}~-{ \frac{\,\pi\,}{2}}<\theta≦{ \frac{\,\pi\,}{3}}$$→ 三角関数を含む不等式①

p.143 Challenge 問1\({\small (1)}~\)\(\theta=\pi\) で最小値 \(4\)
　　\(\theta={\large \frac{\,\pi\,}{3}}~,~{\large \frac{\,5\,}{3}}\pi\) で最大値 \({\large \frac{7}{\,4\,}}\)
\({\small (2)}~\)\(\theta={\large \frac{\,\pi\,}{6}}~,~{\large \frac{5}{\,6\,}}\pi\) 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

p.144 Training 1$${\small (1)}~\sin{{ \frac{\,11\,}{6}}\pi}=-{ \frac{1}{\,2\,}}$$$$~~~~~\cos{{ \frac{\,11\,}{6}}\pi}={ \frac{\,\sqrt{3}\,}{2}}$$$$~~~~~\tan{{ \frac{\,11\,}{6}}\pi}=-{ \frac{1}{\,\sqrt{3}\,}}$$$${\small (2)}~\sin{{ \frac{\,11\,}{4}}\pi}={ \frac{1}{\,\sqrt{2}\,}}$$$$~~~~~\cos{{ \frac{\,11\,}{4}}\pi}=-{ \frac{1}{\,\sqrt{2}\,}}$$$$~~~~~\tan{{ \frac{\,11\,}{4}}\pi}=-1$$$${\small (3)}~\sin{\left(-{ \frac{2}{\,3\,}}\pi\right)}=-{ \frac{\,\sqrt{3}\,}{2}}$$$$~~~~~\cos{\left(-{ \frac{2}{\,3\,}}\pi\right)}=-{ \frac{1}{\,2\,}}$$$$~~~~~\tan{\left(-{ \frac{2}{\,3\,}}\pi\right)}=\sqrt{3}$$$${\small (4)}~\sin{5\pi}=0$$$$~~~~~\cos{5\pi}=-1$$$$~~~~~\tan{5\pi}=0$$

p.144 Training 2$$~~~\cos{\theta}=-{ \frac{\,3\sqrt{10}\,}{10}}~,~\sin{\theta}=-{ \frac{\,\sqrt{10}\,}{10}}$$

p.144 Training 3$${\small (1)}~{ \frac{7}{\,18\,}}$$$${\small (2)}~{ \frac{\,4\,}{3}}$$$${\small (3)}~{ \frac{\,25\sqrt{2}\,}{54}}$$

p.144 Training 4[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.144 Training 5$${\small (1)}~b$$$${\small (2)}~-b$$$${\small (3)}~-{ \frac{\,b\,}{a}}$$

p.144 Training 6\({\small (1)}~\)周期 \(\pi\)

\({\small (2)}~\)周期 \(2\pi\)

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

p.144 Training 7$${\small (1)}~\theta={ \frac{\,7\,}{6}}\pi~,~{ \frac{\,11\,}{6}}\pi$$$${\small (2)}~\theta={ \frac{5}{\,6\,}}\pi~,~{ \frac{\,7\,}{6}}\pi$$$${\small (3)}~\theta={ \frac{\,\pi\,}{6}}~,~{ \frac{\,7\,}{6}}\pi$$

p.144 Training 8$${\small (1)}~\theta={ \frac{\,\pi\,}{12}}~,~{ \frac{5}{\,12\,}}\pi$$$${\small (2)}~\theta={ \frac{\,\pi\,}{2}}~,~{ \frac{\,3\,}{2}}\pi$$

p.144 Training 9$${\small (1)}~{ \frac{\,\pi\,}{3}}≦\theta≦{ \frac{2}{\,3\,}}\pi$$$${\small (2)}~0≦\theta<{ \frac{\,\pi\,}{4}}~,~{ \frac{\,7\,}{4}}\pi<\theta<2\pi$$$${\small (3)}~0≦\theta<{ \frac{\,\pi\,}{2}}~,~{ \frac{5}{\,6\,}}\pi≦\theta<{ \frac{\,3\,}{2}}\pi$$$$~~~~~~{ \frac{\,11\,}{6}}\pi≦\theta<2\pi$$

p.144 Training 10$$~~~y=\sin{\theta}+2$$$$~~~y=\cos{2\left( \theta-\frac{\,\pi\,}{\,4\,} \right)}$$

p.147 問1$$~~~{ \frac{\,\sqrt{6}-\sqrt{2}\,}{4}}~,~{ \frac{\,\sqrt{6}+\sqrt{2}\,}{4}}$$

p.147 問2$$~~~{ \frac{\,\sqrt{2}+\sqrt{6}\,}{4}}~,~{ \frac{\,\sqrt{2}-\sqrt{6}\,}{4}}$$→ 加法定理

p.147 問3$${\small (1)}~-{ \frac{56}{\,65\,}}$$$${\small (2)}~-{ \frac{\,63\,}{65}}$$$${\small (3)}~{ \frac{33}{\,65\,}}$$→ 加法定理と式の値

p.148 問4$$~~~2+\sqrt{3}~,~2-\sqrt{3}$$

p.149 問5$$~~~{ \frac{\,\pi\,}{4}}$$→ ２直線のなす角

p.149 問6$$~~~{ \frac{24}{\,25\,}}~,~{ \frac{7}{\,25\,}}$$→ ２倍角の公式

p.150 問7$$~~~\theta={ \frac{\,\pi\,}{\,3\,}}~,~{ \frac{\,\pi\,}{\,2\,}}~,~{ \frac{\,3\,}{\,2\,}}\pi~,~{ \frac{\,5\,}{\,3\,}}\pi$$

p.150 問8$$~~~{ \frac{\,\sqrt{2+\sqrt{2}}\,}{2}}~,~{ \frac{\,\sqrt{2+\sqrt{2}}\,}{2}}$$→ 半角の公式

p.153 問9$${\small (1)}~2\sqrt{3}\sin{\left(\theta+{ \frac{\,\pi\,}{3}}\right)}$$$${\small (2)}~\sqrt{2}\sin{\left(\theta-{ \frac{3}{\,4\,}}\pi\right)}$$→ 三角関数の合成

p.153 問10\({\small (1)}~\)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.154 Challenge 問1$${\small (1)}~\theta={ \frac{\pi}{\,12\,}}~,~{ \frac{\,17\,}{12}}\pi$$$${\small (2)}~\theta=\pi~,~{ \frac{\,5\,}{3}}\pi$$→ 合成を用いる方程式と不等式

p.155 Training 11$${\small (1)}~-{ \frac{63}{\,65\,}}$$$${\small (2)}~-{ \frac{16}{\,65\,}}$$

p.155 Training 12$${\small (1)}~\sqrt{3}\cos{\theta}$$$${\small (2)}~\sqrt{3}\cos{\theta}$$$${\small (3)}~-\sin{\theta}$$$${\small (4)}~1$$

p.155 Training 13$${\small (1)}~3$$$${\small (2)}~{ \frac{1}{\,2\,}}$$

p.155 Training 14$${\small (1)}~-{ \frac{\,4\sqrt{2}\,}{9}}$$$${\small (2)}~{ \frac{7}{\,9\,}}$$$${\small (3)}~-{ \frac{\,4\sqrt{2}\,}{7}}$$

p.155 Training 15$${\small (1)}~\theta={ \frac{\,\pi\,}{6}}~,~{ \frac{\,\pi\,}{2}}~,~{ \frac{5}{\,6\,}}\pi~,~{ \frac{\,3\,}{2}}\pi$$$${\small (2)}~\theta=0~,~{ \frac{\,\pi\,}{3}}~,~\pi~,~{ \frac{\,5\,}{3}}\pi$$

p.155 Training 16$${\small (1)}~\sqrt{2}\sin{\left(\theta+{ \frac{3}{\,4\,}}\pi\right)}$$$${\small (2)}~2\sqrt{3}\sin{\left(\theta-{ \frac{\,\pi\,}{6}}\right)}$$

p.155 Training 17　最大値 \(5\)、最小値 \(-5\)
→ 三角関数の最大値・最小値

p.155 Training 18　②

p.156 Level Up 1$${\small (1)}~-{ \frac{\,\sqrt{6}\,}{2}}$$$${\small (2)}~-{\large \frac{\,3\sqrt{6}\,}{8}}$$

p.156 Level Up 2$${\small (1)}~2\sin{\theta}$$$${\small (2)}~0$$

p.156 Level Up 3\({\small (1)}~\)周期 \(\pi\)

\({\small (2)}~\)周期 \(\pi\)

\({\small (3)}~\)周期 \(4\pi\)

p.156 Level Up 4$$~~~r=2~,~a=3~,~b={ \frac{\,\pi\,}{2}}$$

p.156 Level Up 5$${\small (1)}~\theta={ \frac{\,\pi\,}{3}}~,~{ \frac{\,\pi\,}{2}}~,~{ \frac{\,4\,}{3}}\pi~,~{ \frac{\,3\,}{2}}\pi$$$${\small (2)}~0≦\theta≦{ \frac{\,\pi\,}{2}}~,~{ \frac{\,7\,}{6}}\pi≦\theta<2\pi$$

p.156 Level Up 6\({\small (1)}~\)$$~~~{ \frac{\,\pi\,}{2}}<\theta<{ \frac{3}{\,4\,}}\pi~,~{ \frac{\,5\,}{4}}\pi<\theta<{ \frac{\,3\,}{2}}\pi$$\({\small (2)}~\)$$~~~0≦\theta≦{ \frac{\,\pi\,}{6}}~,~{ \frac{5}{\,6\,}}\pi≦\theta≦{ \frac{\,7\,}{6}}\pi$$$$~~~~,~{ \frac{\,11\,}{6}}\pi≦\theta<2\pi$$

p.156 Level Up 7$$~~~{ \frac{4}{3}}$$

p.157 Level Up 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.157 Level Up 9$${\small (1)}~\theta={ \frac{\,\pi\,}{2}}~,~{ \frac{2}{\,3\,}}\pi~,~{ \frac{\,4\,}{3}}\pi~,~{ \frac{\,3\,}{2}}\pi$$$${\small (2)}~\theta={ \frac{\,\pi\,}{6}}~,~{ \frac{\,5\,}{6}}\pi$$

p.157 Level Up 10$${\small (1)}~{ \frac{1}{\,2\,}}\sin{2\theta}$$$${\small (2)}~-{ \frac{1}{\,2\,}}\cos{2\theta}+{ \frac{1}{\,2\,}}$$

p.157 Level Up 11　\(\theta={\large \frac{\,\pi\,}{3}}\) で最大値 \(2\sqrt{3}\)
　\(\theta=\pi\) で最小値 \(-\sqrt{3}\)

p.157 Level Up 12$${\small (1)}~y={ \frac{1}{\,2\,}}t^2+t-{ \frac{1}{\,2\,}}$$$${\small (2)}~-1≦y≦{ \frac{1}{\,2\,}}+\sqrt{2}$$

p.157 Level Up 13$${\small (1)}~\sqrt{2}\sin{\left(2\theta-{ \frac{\,\pi\,}{4}}\right)}$$$${\small (2)}~{ \frac{\,\pi\,}{4}}~,~{ \frac{\,\pi\,}{2}}$$

p.157 Level Up 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 \(-\sqrt{2}+3\)