【数学I】例題4.1.14：三角比を含む2次関数の最大・最小（One More）★★★

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% 例題I4.1.14：三角比を含む2次関数の最大・最小（One More）★★★
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関数$y=\sin^2\theta+\cos\theta-1$の最大値と最小値を求め，そのときの$\theta$の値を求めよ．ただし，$0^{\circ}\leqq\theta\leqq 180^{\circ}$とする．

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% 例題I4.1.14の解答（One More）★★★
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$y=\sin^2\theta+\cos\theta-1=(1-\cos^2\theta)+\cos\theta-1=-\cos^2\theta+\cos\theta\cdots(\mathrm{i})$ $\cos\theta=t$とおくと，$0^{\circ}\leqq\theta\leqq 180^{\circ}$より，
\[
-1\leqq t\leqq 1
\]

このとき，(i)に$t$を代入すると，

\[
y=-t^2+t=-(t-\frac{1}{2})^2+\frac{1}{4}
\]

したがって，グラフは右の図のようになる．

ゆえに，$y$は$t=\frac{1}{2}$,すなわち，$\cos\theta=\frac{1}{2}$のとき，最大値$\frac{1}{4}$をとり，$t=-1$，すなわち，$\cos\theta=-1$のとき，最小値$-2$をとる．

$0^{\circ}\leqq\theta\leqq 180^{\circ}$より，$\cos\theta=\frac{1}{2}$のとき，$\theta=60^{\circ}$，$\cos\theta=-1$のとき，$\theta=180^{\circ}$

よって，$\theta=60^{\circ}$のとき，最大値$\frac{1}{4}$，$\theta=180^{\circ}$のとき，最小値$-2$

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% 問題I4.1.14：三角比を含む2次関数の最大・最小（One More）★★★
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関数$y=\sin^2\theta-\cos\theta$の最大値と最小値を求め，そのときの$\theta$の値を求めよ．ただし，$0^{\circ}\leqq\theta\leqq 180^{\circ}$とする．

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% 問題I4.1.14の解答（One More）★★★
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$y=\sin^2\theta-\cos\theta=(1-\cos^2\theta)-\cos\theta=-\cos^2\theta-\cos\theta+1\cdots(\mathrm{i})$ $\cos\theta=t$とおくと，$0^{\circ}\leqq\theta\leqq 180^{\circ}$より，
\[
-1\leqq t\leqq 1
\]

このとき，(i)に$t$を代入すると，

\[
y=-t^2-t+1=-(t+\frac{1}{2})^2+\frac{5}{4}
\]

したがって，グラフは右の図のようになる．

ゆえに，$y$は$t=-\frac{1}{2}$,すなわち，$\cos\theta=-\frac{1}{2}$のとき，最大値$\frac{5}{4}$をとり，$t=1$，すなわち，$\cos\theta=1$のとき，最小値$-1$をとる．

$0^{\circ}\leqq\theta\leqq 180^{\circ}$より，$\cos\theta=-\frac{1}{2}$のとき，$\theta=120^{\circ}$，$\cos\theta=1$のとき，$\theta=0^{\circ}$

よって，$\theta=120^{\circ}$のとき，最大値$\frac{5}{4}$，$\theta=0^{\circ}$のとき，最小値$-1$

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% 例題I4.1.14の側注：三角比を含む2次関数の最大・最小（One More）★★★
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% 側注1
$t$の範囲を求める．

% 側注2
\begin{align*}
&-t^2+t\\
=&-(t^2-t)\\
=&-\{(t-\frac{1}{2})^2-(\frac{1}{2})^2\}
\end{align*}

% 側注3
$\sin^2\theta+\cos^2\theta=1$

% 側注4
$t=-1$の方が軸から遠いので，$t=-1$で最小となる．

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% 問題I4.1.14の側注：三角比を含む2次関数の最大・最小（One More）★★★
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% 側注1
$t$の範囲を求める．

% 側注2
$\sin^2\theta+\cos^2\theta=1$

% 側注3
$t=1$の方が軸から遠いので，$t=1$で最小となる．

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% 例題I4.1.14の考え方・補足など：三角比を含む2次関数の最大・最小（One More）★★★
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% 考え方
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$\sin^2\theta+\cos^2\theta=1$を利用して，三角比の種類を減らす．扱いやすさから，$\cos\theta=t$などと文字を用いておくとよい．$t$の値の範囲を求め，2次関数のグラフをかいて最大値と最小値を求める．

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三角比の種類を減らし，2次関数のグラフを考える．