【高校数学Ⅱ】4-2-3 三角関数の合成｜要点まとめ

\(\displaystyle =\sqrt{2}\left(\frac{1}{\sqrt{2}}\sin\theta+\frac{1}{\sqrt{2}}\cos\theta\right)\)
\(\displaystyle =\sqrt{2}\sin\left(\theta+\frac{\pi}{4}\right)\)

\(\displaystyle =2\left(-\frac{\sqrt{3}}{2}\sin\theta-\frac{1}{2}\cos\theta\right)\)
\(\displaystyle =2\sin\left(\theta-\frac{5\pi}{6}\right)\)

\(\displaystyle 2\left(\frac{1}{2}\sin x+\frac{\sqrt{3}}{2}\cos x\right)=\sqrt{3}\)
\(\displaystyle 2\sin\left(x+\frac{\pi}{3}\right)=\sqrt{3}\)
\(\displaystyle \sin\left(x+\frac{\pi}{3}\right)=\frac{\sqrt{3}}{2}\)
\(\displaystyle x+\frac{\pi}{3}=\frac{\pi}{3},\frac{2\pi}{3}\)
\(0\leqq x \leqq2\pi\)より、
\(\displaystyle x=0,\frac{\pi}{3}\)

\(\displaystyle 2\sqrt{2}\left(\frac{1}{2}\sin x-\frac{\sqrt{3}}{2}\cos x\right)\leqq2\)
\(\displaystyle 2\sqrt{2}\sin\left(x-\frac{\pi}{3}\right)\leqq2\)
\(\displaystyle \sin\left(x-\frac{\pi}{3}\right)\leqq\frac{1}{\sqrt{2}}\)
\(\displaystyle 0\leqq x-\frac{\pi}{3}\leqq\frac{\pi}{4},\frac{3\pi}{4}\leqq x-\frac{\pi}{3}<2\pi\)
\(0\leqq x \leqq2\pi\)より、
\(\displaystyle 0\leqq x\leqq\frac{7\pi}{12},\frac{13\pi}{12}\leqq x<2\pi\)

\(y=-(1-2\sin^2x)+2\sin x-1\)
\(\ \ =2\sin^2x+2\sin x-1\)
\(\displaystyle \ \ =2\left(\sin x+\frac{1}{2}\right)^2-\frac{5}{2}\)
\(-1\leqq\sin x\leqq1\)より、
\(\displaystyle \sin x=1\)のとき、最大値\(2\)
\(\displaystyle x=\frac{\pi}{2}\)
\(\displaystyle \sin x=-\frac{1}{2}\)のとき、最小値\(\displaystyle -\frac{5}{2}\)
\(\displaystyle x=\frac{7\pi}{6},\frac{11\pi}{6}\)
よって、
最大値は\(2\)(\(\displaystyle x=\frac{\pi}{2}\)のとき)
最小値は\(\displaystyle -\frac{5}{2}\)(\(\displaystyle x=\frac{7\pi}{6},\frac{11\pi}{6}\)のとき)

\(\displaystyle y=2\sqrt{2}\left(\frac{1}{\sqrt{2}}\sin x+\frac{1}{\sqrt{2}}\cos x\right)\)
\(\displaystyle \ \ =2\sqrt{2}\sin\left(x+\frac{\pi}{4}\right)\)
\(\displaystyle -1\leqq\sin\left(x+\frac{\pi}{4}\right)\leqq1\)より、
最大値は\(2\sqrt{2}\)(\(\displaystyle x=\frac{\pi}{4}\)のとき)
最小値は\(-2\sqrt{2}\)(\(\displaystyle x=\frac{5\pi}{4}\)のとき)