【数学I】例題4.1.7：三角比の相互関係2（One More）★★

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% 例題I4.1.7：三角比の相互関係2（One More）★★
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(1) $\sin\alpha=\frac{5}{13}$のとき，$\cos\alpha,\tan\alpha$の値を求めよ．ただし，$90^{\circ}<\alpha<180^{\circ}$とする．

(2) $\tan\beta=-\frac{3}{4}$のとき，$\sin\beta,\cos\beta$の値を求めよ．ただし，$0^{\circ}\leqq\beta\leqq 180^{\circ}$とする．

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% 例題I4.1.7の解答（One More）★★
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(1) $\sin^2\alpha+\cos^2\alpha=1$より，

\[
\cos^2\alpha=1-(\frac{5}{13})^2=\frac{144}{169}
\]

$90^{\circ}<\alpha<180^{\circ}$において，$\cos\alpha<0$

よって，$\cos\alpha=-\sqrt{\frac{144}{169}}=-\frac{12}{13}$

また，$\tan\alpha=\frac{\sin\alpha}{\cos\alpha}=\frac{5}{13}\div(-\frac{12}{13})=\frac{5}{13}\times(-\frac{13}{12})=-\frac{5}{12}$

(2) $1+\tan^2\beta=\frac{1}{\cos^2\beta}$より，

\[
\frac{1}{\cos^2\beta}=1+\tan^2\beta=1+(-\frac{3}{4})^2=\frac{25}{16}
\]

したがって，$\cos^2\beta=\frac{16}{25}$

$\tan\beta=-\frac{3}{4}<0$より，$90^{\circ}<\beta<180^{\circ}$であるから，$\cos\beta<0$

よって，$\cos\beta=-\sqrt{\frac{16}{25}}=-\frac{4}{5}$

また，$\tan\beta=\frac{\sin\beta}{\cos\beta}$より，$\sin\beta=\tan\beta\cdot\cos\beta=-\frac{3}{4}\cdot(-\frac{4}{5})=\frac{3}{5}$

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% 問題I4.1.7：三角比の相互関係2（One More）★★
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(1) $\sin\alpha=\frac{2}{3}$のとき，$\cos\alpha,\tan\alpha$の値を求めよ．ただし，$90^{\circ}<\alpha<180^{\circ}$とする．

(2) $\tan\beta=-2$のとき，$\sin\beta,\cos\beta$の値を求めよ．ただし，$0^{\circ}\leqq\beta\leqq 180^{\circ}$とする．

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% 問題I4.1.7の解答（One More）★★
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(1) $\sin^2\alpha+\cos^2\alpha=1$より，$\cos^2\alpha=1-(\frac{2}{3})^2=\frac{5}{9}$

$90^{\circ}<\alpha<180^{\circ}$において，$\cos\alpha<0$

よって，$\cos\alpha=-\sqrt{\frac{5}{9}}=-\frac{\sqrt{5}}{3}$

また，$\tan\alpha=\frac{\sin\alpha}{\cos\alpha}=\frac{2}{3}\div(-\frac{\sqrt{5}}{3})=\frac{2}{3}\times(-\frac{3}{\sqrt{5}})=-\frac{2}{\sqrt{5}}$

(2) $1+\tan^2\beta=\frac{1}{\cos^2\beta}$より，$\frac{1}{\cos^2\beta}=1+\tan^2\beta=1+(-2)^2=5$

したがって，$\cos^2\beta=\frac{1}{5}$

$\tan\beta=-2<0$より，$90^{\circ}<\beta<180^{\circ}$であるから，$\cos\beta<0$

よって，$\cos\beta=-\sqrt{\frac{1}{5}}=-\frac{1}{\sqrt{5}}$

また，$\tan\beta=\frac{\sin\beta}{\cos\beta}$より，$\sin\beta=\tan\beta\cdot\cos\beta=-2\cdot(-\frac{1}{\sqrt{5}})=\frac{2}{\sqrt{5}}$

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% 例題I4.1.7の側注：三角比の相互関係2（One More）★★
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% 側注1
符号に注意すること．

% 側注2
$\tan\alpha=\frac{\sin\alpha}{\cos\alpha}=\frac{\frac{5}{13}}{-\frac{12}{13}}$

% 側注3
$\cos\beta<0$より，$\frac{4}{5}$は不適である．

% 側注4
$\tan\theta=\frac{\sin\theta}{\cos\theta}$より，$\sin\theta=\tan\theta\cos\theta$

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% 問題I4.1.7の側注：三角比の相互関係2（One More）★★
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% 側注1
符号に注意すること．

% 側注2
$\tan\alpha=\frac{\sin\alpha}{\cos\alpha}=\frac{\frac{2}{3}}{-\frac{\sqrt{5}}{3}}$

% 側注3
$\cos\beta<0$より，$\frac{1}{\sqrt{5}}$は不適である．

% 側注4
$\tan\beta=\frac{\sin\beta}{\cos\beta}$より，$\sin\beta=\tan\beta\cos\beta$

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% 例題I4.1.7の考え方・補足など：三角比の相互関係2（One More）★★
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% 考え方
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三角比の相互関係を利用する．このとき，与えられた角度の条件をもとに，三角比の値の符号に注意すること．

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% 三角比の相互関係
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\[
\tan\theta=\frac{\sin\theta}{\cos\theta},\sin^2\theta+\cos^2\theta=1,1+\tan^2\theta=\frac{1}{\cos^2\theta}
\]