1.次の方程式を求めなさい。ただし、\(0\leqq \theta \leqq2\pi\)とする。
(1)\(\displaystyle \sinθ=\frac{\sqrt{3}}{2}\)
\(\displaystyle \theta=\frac{\pi}{3},\frac{2\pi}{3}\)
(2)\(2\cosθ+1=0\)
\(\displaystyle \theta=\frac{2\pi}{3},\frac{4\pi}{3}\)
(3)\(\displaystyle \cosθ=-\frac{\sqrt{2}}{2}\)
\(\displaystyle \theta=\frac{3\pi}{4},\frac{5\pi}{4}\)
(4)\(\tanθ=-\sqrt{3}\)
\(\displaystyle \theta=\frac{2\pi}{3},\frac{5\pi}{3}\)
(5)\(\sinθ+1=0\)
\(\displaystyle \theta=\frac{3\pi}{2}\)
(6)\(\cosθ=0\)
\(\displaystyle \theta=\frac{\pi}{2},\frac{3\pi}{2}\)
(7)\(\displaystyle \sinθ=-\frac{1}{\sqrt{2}}\)
\(\displaystyle \theta=\frac{5\pi}{4},\frac{7\pi}{4}\)
(8)\(\displaystyle \sin\left(θ+\frac{\pi}{3}\right)=\frac{1}{\sqrt{2}}\)
\(\displaystyle \theta+\frac{\pi}{3}=\frac{\pi}{4},\frac{3\pi}{4}\)
\(0\leqq \theta \leqq2\pi\)より、
\(\displaystyle \theta=\frac{5\pi}{12},\frac{23\pi}{12}\)
(9)\(\displaystyle \sqrt{2}\cos\left(θ+\frac{\pi}{3}\right)=1\)
\(\displaystyle \theta+\frac{\pi}{3}=\frac{\pi}{4},\frac{7\pi}{4}\)
\(0\leqq \theta \leqq2\pi\)より、
\(\displaystyle \theta=\frac{17\pi}{12},\frac{23\pi}{12}\)
(10)\(\displaystyle 2\sin\left(θ-\frac{\pi}{6}\right)=-\sqrt{3}\)
\(\displaystyle \theta-\frac{\pi}{6}=\frac{4\pi}{3},\frac{5\pi}{3}\)
\(0\leqq \theta \leqq2\pi\)より、
\(\displaystyle \theta=\frac{3\pi}{2},\frac{11\pi}{6}\)
2.次の不等式を求めなさい。ただし、\(0\leqq \theta \leqq2\pi\)とする。
(1)\(\displaystyle \cosθ\leqq-\frac{1}{2}\)
\(\displaystyle \frac{2\pi}{3}\leqq\theta\leqq\frac{4\pi}{3}\)
(2)\(\displaystyle \sinθ>\frac{1}{\sqrt{2}}\)
\(\displaystyle \frac{\pi}{4}<\theta<\frac{3\pi}{4}\)
(3)\(\displaystyle \cosθ>\frac{1}{2}\)
\(\displaystyle 0\leqq\theta<\frac{\pi}{3},\frac{5\pi}{3}<\theta<2\pi\)
(4)\(\displaystyle \sinθ\leqq\frac{\sqrt{3}}{2}\)
\(\displaystyle 0\leqq\theta\leqq\frac{\pi}{3},\frac{2\pi}{3}\leqq\theta<2\pi\)
(5)\(\displaystyle \tanθ\leqq-1\)
\(\displaystyle \frac{\pi}{2}<\theta\leqq\frac{3\pi}{4},\frac{3\pi}{2}<\theta\leqq\frac{7\pi}{4}\)
(6)\(\displaystyle \tanθ<\sqrt{3}\)
\(\displaystyle 0\leqq\theta\leqq\frac{\pi}{3},\frac{\pi}{2}<\theta<\frac{4\pi}{3},\frac{3\pi}{2}<\theta<2\pi\)
(7)\(\displaystyle \sinθ>-\frac{1}{2}\)
\(\displaystyle 0\leqq\theta<\frac{7\pi}{6},\frac{11\pi}{6}<\theta<2\pi\)
3.次の方程式を求めなさい。ただし、\(0\leqq \theta \leqq2\pi\)とする。
(1)\(2\cos^2θ+3\cosθ-2=0\)
\((2\cosθ-1)(\cosθ+2)=0\)
\(2\cosθ-1=0\)より、
\(\displaystyle \theta=\frac{\pi}{3},\frac{5\pi}{3}\)
\(\cosθ+2=0\)より、解なし
よって、
\(\displaystyle \theta=\frac{\pi}{3},\frac{5\pi}{3}\)
(2)\(2\cos^2θ-\sinθ-1=0\)
\(2(1-\sin^2θ)-\sinθ-1=0\)
\(2\sin^2θ+\sinθ-1=0\)
\((2\sinθ-1)(\sinθ+1)=0\)
\(2\sinθ-1=0\)より、
\(\displaystyle \theta=\frac{\pi}{6},\frac{5\pi}{6}\)
\(\sinθ+1=0\)より、
\(\displaystyle \theta=\frac{3\pi}{2}\)
よって、
\(\displaystyle \theta=\frac{\pi}{6},\frac{5\pi}{6},\frac{3\pi}{2}\)
(3)\(2\sin^2θ+3\cosθ-3=0\)
\(2(1-\cos^2θ)+3\cosθ-3=0\)
\(2\cos^2θ-3\cosθ+1=0\)
\((2\cosθ-1)(\cosθ-1)=0\)
\(2\cosθ-1=0\)より、
\(\displaystyle \theta=\frac{\pi}{3},\frac{5\pi}{3}\)
\(\cosθ-1=0\)より、
\(\theta=0\)
よって、
\(\displaystyle \theta=0,\frac{\pi}{3},\frac{5\pi}{3}\)
4.次の不等式を求めなさい。ただし、\(0\leqq \theta \leqq2\pi\)とする。
(1)\(2\cos^2θ-\sinθ-1\geqq0\)
\(2(1-\sin^2θ)-\sinθ-1\geqq0\)
\(2\sin^2θ+\sinθ-1\leqq0\)
\((2\sinθ-1)(\sinθ+1)\leqq0\)
\(\displaystyle -1\leqq\sinθ\leqq\frac{1}{2}\)
よって、
\(\displaystyle 0\leqq\theta\leqq\frac{\pi}{6},\frac{5\pi}{6}\leqq\theta\leqq2\pi\)
(2)\(2\sin^2θ+3\cosθ-3\geqq0\)
\(2(1-\cos^2θ)+3\cosθ-3\geqq0\)
\(2\cos^2θ-3\cosθ+1\leqq0\)
\((2\cosθ-1)(\cosθ-1)\leqq0\)
\(\displaystyle \frac{1}{2}\leqq\cosθ\leqq1\)
よって、
\(\displaystyle 0\leqq\theta\leqq\frac{\pi}{3},\frac{5\pi}{3}\leqq\theta\leqq2\pi\)
5.次の関数の最大値と最小値を求めなさい。ただし、\(0\leqq \theta \leqq2\pi\)とする。
(1)\(y=-\sin^2θ-\cosθ+1\)
\(y=-(1-\cos^2θ)-\cosθ+1\)
\(\ \ =\cos^2θ-\cosθ\)
\(\displaystyle \ \ =\left(\cosθ-\frac{1}{2}\right)^2-\frac{1}{4}\)
\(-1\leqq\cosθ\leqq1\)より、
\(\displaystyle \cosθ=-1\)のとき、最大値\(2\)
\(θ=\pi\)
\(\displaystyle \cosθ=\frac{1}{2}\)のとき、最小値\(\displaystyle -\frac{1}{4}\)
\(\displaystyle θ=\frac{\pi}{3},\frac{5\pi}{3}\)
よって、
最大値は\(2\)(\(θ=\pi\)のとき)
最小値は\(\displaystyle -\frac{1}{4}\)(\(\displaystyle θ=\frac{\pi}{3},\frac{5\pi}{3}\)のとき)
(2)\(y=-\cos^2θ+\sinθ\)
\(y=-(1-\sin^2θ)+\sinθ\)
\(\ \ =\sin^2θ+\sinθ-1\)
\(\displaystyle \ \ =\left(\sinθ+\frac{1}{2}\right)^2-\frac{5}{4}\)
\(-1\leqq\sinθ\leqq1\)より、
\(\displaystyle \sinθ=1\)のとき、最大値\(1\)
\(\displaystyle θ=\frac{\pi}{2}\)
\(\displaystyle \cosθ=-\frac{1}{2}\)のとき、最小値\(\displaystyle -\frac{5}{4}\)
\(\displaystyle θ=\frac{7\pi}{6},\frac{11\pi}{6}\)
よって、
最大値は\(1\)(\(\displaystyle θ=\frac{\pi}{2}\)のとき)
最小値は\(\displaystyle -\frac{5}{4}\)(\(\displaystyle θ=\frac{7\pi}{6},\frac{11\pi}{6}\)のとき)
(3)\(y=\sin^2θ+\cosθ\)
\(y=(1-\cos^2θ)+\cosθ\)
\(\ \ =-\cos^2θ+\cosθ+1\)
\(\displaystyle \ \ =-\left(\cosθ-\frac{1}{2}\right)^2+\frac{5}{4}\)
\(-1\leqq\cosθ\leqq1\)より、
\(\displaystyle \cosθ=\frac{1}{2}\)のとき、最大値\(\displaystyle \frac{5}{4}\)
\(\displaystyle θ=\frac{\pi}{3},\frac{5\pi}{3}\)
\(\cosθ=-1\)のとき、最小値\(-1\)
\(\displaystyle θ=\pi\)
よって、
最大値は\(\displaystyle \frac{5}{4}\)(\(\displaystyle θ=\frac{\pi}{3},\frac{5\pi}{3}\)のとき)
最小値は\(-1\)(\(θ=\pi\)のとき)