問題の解答
検索用コード（LaTeX）
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% 例題I4.1.4：三角比の相互関係1（One More）★
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$A$は鋭角とする．このとき，次の問いに答えよ．
(1) $\sin A=\frac{5}{7}$のとき，$\cos A$と$\tan A$の値を求めよ．
(2) $\tan A=2$のとき，$\sin A$と$\cos A$の値を求めよ．
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% 例題I4.1.4の解答（One More）★
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(1) $\sin^2A+\cos^2A=1$より，
\[
\cos^2A=1-\sin^2A=1-(\frac{5}{7})^2=\frac{24}{49}
\]
$A$は鋭角であるから，$\cos A>0$
よって，$\cos A=\frac{\sqrt{24}}{7}=\frac{2\sqrt{6}}{7}$
また，$\tan A=\frac{\sin A}{\cos A}=\frac{5}{7}\div\frac{2\sqrt{6}}{7}=\frac{5}{2\sqrt{6}}$
(2) $1+\tan^2A=\frac{1}{\cos^2A}$より，$\frac{1}{\cos^2A}=1+2^2=5$
したがって，$\cos^2A=\frac{1}{5}$
$A$は鋭角であるから，$\cos A>0$
よって，$\cos A=\frac{1}{\sqrt{5}}$
また，$\sin A=\tan A\cos A=2\cdot\frac{1}{\sqrt{5}}=\frac{2}{\sqrt{5}}$
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% 例題I4.1.4の別解
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$A$が鋭角，$\sin A=\frac{5}{7}$より，右の図のような直角三角形ABCがかける．
三平方の定理より，
\[
\mathrm{AC}=\sqrt{\mathrm{AB}^2-\mathrm{BC}^2}=\sqrt{7^2-5^2}=\sqrt{24}=2\sqrt{6}
\]
よって，$\cos A=\frac{2\sqrt{6}}{7},\tan A=\frac{5}{2\sqrt{6}}$
(2) も同様に，$\tan A=\frac{2}{1}$と考えて，直角三角形をかくことで解くこともできる．
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% 問題I4.1.4：三角比の相互関係1（One More）★
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$A$は鋭角とする．このとき，次の問いに答えよ．
(1) $\sin A=\frac{4}{5}$のとき，$\cos A$と$\tan A$の値を求めよ．
(2) $\tan A=\frac{\sqrt{2}}{4}$のとき，$\sin A$と$\cos A$の値を求めよ．
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% 問題I4.1.4の解答（One More）★
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(1) $\sin^2A+\cos^2A=1$より，$\cos^2A=1-\sin^2A=1-(\frac{4}{5})^2=\frac{9}{25}$
$A$は鋭角であるから，$\cos A>0$
よって，$\cos A=\frac{\sqrt{9}}{5}=\frac{3}{5}$
また，$\tan A=\frac{\sin A}{\cos A}=\frac{4}{5}\div\frac{3}{5}=\frac{4}{3}$
(2) $1+\tan^2A=\frac{1}{\cos^2A}$より，$\frac{1}{\cos^2A}=1+(\frac{\sqrt{2}}{4})^2=\frac{18}{16}$
したがって，$\cos^2A=\frac{16}{18}$
$A$は鋭角であるから，$\cos A>0$
よって，$\cos A=\frac{2\sqrt{2}}{3}$
また，$\sin A=\tan A\cos A=\frac{\sqrt{2}}{4}\cdot\frac{2\sqrt{2}}{3}=\frac{1}{3}$
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