【高校数学Ⅱ】4-2-3 三角関数の合成｜問題集

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

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

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

\(\displaystyle =\sqrt{2}\left(-\frac{1}{\sqrt{2}}\sin\theta-\frac{1}{\sqrt{2}}\cos\theta\right)\)
\(\displaystyle =\sqrt{2}\sin\left(\theta-\frac{3\pi}{4}\right)\)

\(\displaystyle =\sqrt{2}\left(-\frac{1}{\sqrt{2}}\sin\theta+\frac{1}{\sqrt{2}}\cos\theta\right)\)
\(\displaystyle =\sqrt{2}\sin\left(\theta+\frac{3\pi}{4}\right)\)

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

\(\displaystyle 2\left(\frac{1}{2}\sin x+\frac{\sqrt{3}}{2}\cos x\right)=1\)
\(\displaystyle 2\sin\left(x+\frac{\pi}{3}\right)=1\)
\(\displaystyle \sin\left(x+\frac{\pi}{3}\right)=\frac{1}{2}\)
\(\displaystyle x+\frac{\pi}{3}=\frac{\pi}{6},\frac{5\pi}{6}\)
\(0\leqq x \leqq2\pi\)より、
\(\displaystyle x=\frac{\pi}{2},\frac{11\pi}{6}\)

\(\displaystyle 2\left(\frac{\sqrt{3}}{2}\sin x-\frac{1}{2}\cos x\right)=-\sqrt{2}\)
\(\displaystyle 2\sin\left(x-\frac{\pi}{6}\right)=-\sqrt{2}\)
\(\displaystyle \sin\left(x-\frac{\pi}{6}\right)=-\frac{\sqrt{2}}{2}\)
\(\displaystyle x-\frac{\pi}{6}=\frac{5\pi}{4},\frac{7\pi}{4}\)
\(0\leqq x \leqq2\pi\)より、
\(\displaystyle x=\frac{17\pi}{12},\frac{23\pi}{12}\)

\(\displaystyle 2\left(\frac{1}{2}\sin x-\frac{\sqrt{3}}{2}\cos x\right)=1\)
\(\displaystyle 2\sin\left(x-\frac{\pi}{3}\right)=1\)
\(\displaystyle \sin\left(x-\frac{\pi}{3}\right)=\frac{1}{2}\)
\(\displaystyle x-\frac{\pi}{3}=\frac{\pi}{6},\frac{5\pi}{6}\)
\(0\leqq x \leqq2\pi\)より、
\(\displaystyle x=\frac{\pi}{2},\frac{7\pi}{6}\)

\(\displaystyle \sqrt{2}\left(\frac{1}{\sqrt{2}}\sin x+\frac{1}{\sqrt{2}}\cos x\right)\geqq1\)
\(\displaystyle \sqrt{2}\sin\left(x+\frac{\pi}{4}\right)\geqq1\)
\(\displaystyle \sin\left(x+\frac{\pi}{4}\right)\geqq\frac{1}{\sqrt{2}}\)
\(\displaystyle \frac{\pi}{4}\leqq x+\frac{\pi}{4}\leqq\frac{3\pi}{4}\)
\(0\leqq x \leqq2\pi\)より、
\(\displaystyle 0\leqq x\leqq\frac{\pi}{2}\)

\(\displaystyle 2\left(\frac{\sqrt{2}}{2}\sin x+\frac{\sqrt{2}}{2}\cos x\right)\leqq-\sqrt{3}\)
\(\displaystyle 2\sin\left(x+\frac{\pi}{4}\right)\leqq-\sqrt{3}\)
\(\displaystyle \sin\left(x+\frac{\pi}{4}\right)\leqq-\frac{\sqrt{3}}{2}\)
\(\displaystyle \frac{4\pi}{3}\leqq x+\frac{\pi}{4}\leqq\frac{5\pi}{3}\)
\(0\leqq x \leqq2\pi\)より、
\(\displaystyle \frac{13\pi}{12}\leqq x\leqq\frac{17\pi}{12}\)

\(\displaystyle y=5\left(-\frac{4}{5}\sin x+\frac{3}{5}\cos x\right)\)
\(\displaystyle \ \ =5\sin\left(x+\alpha\right)\)
\(-1\leqq\sin x\leqq1\)より、
最大値は\(5\)
最小値は\(-5\)

\(\displaystyle y=2\left(-\frac{1}{2}\sin x+\frac{\sqrt{3}}{2}\cos x\right)\)
\(\displaystyle \ \ =2\sin\left(x+\frac{2\pi}{3}\right)\)
\(-1\leqq\sin x\leqq1\)より、
最大値は\(2\)(\(\displaystyle x=\frac{11\pi}{6}\)のとき)
最小値は\(-2\)(\(\displaystyle x=\frac{5\pi}{6}\)のとき)

\(\displaystyle y=2\left(\frac{\sqrt{3}}{2}\sin x+\frac{1}{2}\cos x\right)-1\)
\(\displaystyle \ \ =2\sin\left(x+\frac{\pi}{6}\right)-1\)
\(-1\leqq\sin x\leqq1\)より、
最大値は\(1\)(\(\displaystyle x=\frac{\pi}{3}\)のとき)
最小値は\(-3\)(\(\displaystyle x=\frac{4\pi}{3}\)のとき)