【数学I】例題4.1.8：三角比の式の値（One More）★★★

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% 例題I4.1.8：三角比の式の値（One More）★★★
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$\sin\theta+\cos\theta=\frac{\sqrt{2}}{2}$のとき，次の式の値を求めよ．ただし，$0^{\circ}\leqq\theta\leqq 180^{\circ}$とする．

(1) $\sin\theta\cos\theta$

(2) $\sin^3\theta+\cos^3\theta$

(3) $\sin\theta-\cos\theta$

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(1) $\sin\theta+\cos\theta=\frac{\sqrt{2}}{2}\cdots(\mathrm{i})$の両辺を2乗すると，

\[
\sin^2\theta+2\sin\theta\cos\theta+\cos^2\theta=\frac{1}{2}
\]

したがって，$1+2\sin\theta\cos\theta=\frac{1}{2}$

よって，$\sin\theta\cos\theta=-\frac{1}{4}\cdots(\mathrm{ii})$

(2) $\sin^3\theta+\cos^3\theta=(\sin\theta+\cos\theta)^3-3\sin\theta\cos\theta(\sin\theta+\cos\theta)$

(i)，(ii)を代入すると，$\sin^3\theta+\cos^3\theta=(\frac{\sqrt{2}}{2})^3-3\cdot(-\frac{1}{4})\cdot\frac{\sqrt{2}}{2}=\frac{5\sqrt{2}}{8}$

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\begin{align*}
\sin^3\theta+\cos^3\theta&=(\sin\theta+\cos\theta)(\sin^2\theta-\sin\theta\cos\theta+\cos^2\theta)\\
&={(\sin\theta+\cos\theta)(1-\sin\theta\cos\theta)}
\end{align*}

(i)，(ii)を代入すると，$\sin^3\theta+\cos^3\theta=\frac{\sqrt{2}}{2}\{1-(-\frac{1}{4})\}=\frac{5\sqrt{2}}{8}$

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(3) $(\sin\theta-\cos\theta)^2=\sin^2\theta-2\sin\theta\cos\theta+\cos^2\theta=1-2\sin\theta\cos\theta$

(ii)を代入すると，$(\sin\theta-\cos\theta)^2=1-2\cdot(-\frac{1}{4})=\frac{3}{2}$

$0^{\circ}\leqq\theta\leqq 180^{\circ}$のとき，(ii)より，$\sin\theta>0,\cos\theta<0$

したがって，$\sin\theta-\cos\theta>0$

よって，$\sin\theta-\cos\theta=\sqrt{\frac{3}{2}}=\frac{\sqrt{6}}{2}$

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% 問題I4.1.8：三角比の式の値（One More）★★★
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$\sin\theta-\cos\theta=\frac{1}{2}$のとき，次の式の値を求めよ．ただし，$0^{\circ}\leqq\theta\leqq 180^{\circ}$とする．

(1) $\sin\theta\cos\theta$

(2) $\sin^3\theta-\cos^3\theta$

(3) $\sin\theta+\cos\theta$

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(1) $\sin\theta-\cos\theta=\frac{1}{2}\cdots(\mathrm{i})$の両辺を2乗すると，$\sin^2\theta-2\sin\theta\cos\theta+\cos^2\theta=\frac{1}{4}$

したがって，$1-2\sin\theta\cos\theta=\frac{1}{4}$

よって，$\sin\theta\cos\theta=\frac{3}{8}\cdots(\mathrm{ii})$

(2) $\sin^3\theta-\cos^3\theta=(\sin\theta-\cos\theta)^3+3\sin\theta\cos\theta(\sin\theta-\cos\theta)$

(i)，(ii)を代入すると，$\sin^3\theta-\cos^3\theta=(\frac{1}{2})^3+3\cdot\frac{3}{8}\cdot\frac{1}{2}=\frac{11}{16}$

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$\sin^3\theta-\cos^3\theta=(\sin\theta-\cos\theta)(\sin^2\theta+\sin\theta\cos\theta+\cos^2\theta)$

(i)，(ii)を代入すると，$\sin^3\theta-\cos^3\theta=\frac{1}{2}\cdot(1+\frac{3}{8})=\frac{11}{16}$

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(3) $(\sin\theta+\cos\theta)^2=\sin^2\theta+2\sin\theta\cos\theta+\cos^2\theta=1+2\sin\theta\cos\theta$

(ii)を代入すると，$(\sin\theta+\cos\theta)^2=1+2\cdot\frac{3}{8}=\frac{7}{4}$

$0^{\circ}\leqq\theta\leqq 180^{\circ}$のとき，(ii)より，$\sin\theta>0,\cos\theta>0$

したがって，$\sin\theta+\cos\theta>0$

よって，$\sin\theta+\cos\theta=\sqrt{\frac{7}{4}}=\frac{\sqrt{7}}{2}$

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% 例題I4.1.8の側注：三角比の式の値（One More）★★★
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% 側注1
両辺を2乗して，$\sin^2\theta+\cos^2\theta$の形を作る．

% 側注2
$\sin^2\theta+\cos^2\theta=1$

% 側注3
\begin{align*}
&a^3+b^3\\
=&(a+b)^3-3ab(a+b)
\end{align*}

% 側注4
\begin{align*}
&a^3+b^3\\
=&(a+b)(a^2-ab+b^2)
\end{align*}

% 側注5
$\sin\theta\cos\theta=-\frac{1}{4}<0$であり，$\sin\theta>0$より，$\cos\theta<0$

よって，$-\cos\theta>0$となり，$\sin\theta-\cos\theta>0$

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% 問題I4.1.8の側注：三角比の式の値（One More）★★★
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% 側注1
両辺を2乗して，$\sin^2\theta+\cos^2\theta$の形を作る．

% 側注2
\begin{align*}
&a^3-b^3\\
=&(a-b)^3+3ab(a-b)
\end{align*}

% 側注3
\begin{align*}
&a^3-b^3\\
=&(a-b)(a^2+ab+b^2)
\end{align*}

% 側注4
$\sin\theta\cos\theta=\frac{3}{8}>0$であり，$\sin\theta>0$より，$\cos\theta>0$

よって，$\sin\theta+\cos\theta>0$

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% 例題I4.1.8の考え方・補足など：三角比の式の値（One More）★★★
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% 考え方
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$\sin^2\theta+\cos^2\theta=1$を利用できるように，式変形する．なお，(1)，(2)は$\sin\theta,\cos\theta$の対称式である．

(1) $(a+b)^2=a^2+2ab+b^2$を利用する．

(2) $a^3+b^3=(a+b)^3-3ab(a+b)$を利用する．

(3) $(\sin\theta-\cos\theta)^2$の値を利用する．このとき，$\sin\theta-\cos\theta$の符号に注意すること．

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$\sin^2\theta+\cos^2\theta=1$を利用して，式の値を求める．