【高校数学C】4-2-2 極座標と直交座標｜問題集

\(\displaystyle x=2\cos\frac{\pi}{6}=\sqrt{3}\)
\(\displaystyle y=2\sin\frac{\pi}{6}=1\)
よって、
\((\sqrt{3},1)\)

\(\displaystyle x=\sqrt{2}\cos\frac{\pi}{4}=1\)
\(\displaystyle y=\sqrt{2}\sin\frac{\pi}{4}=1\)
よって、
\((1,1)\)

\(\displaystyle x=2\cos\frac{\pi}{4}=\sqrt{2}\)
\(\displaystyle y=2\sin\frac{\pi}{4}=\sqrt{2}\)
よって、
\((\sqrt{2},\sqrt{2})\)

\(\displaystyle x=3\cos\left(-\frac{5}{6}\pi\right)=-\frac{3\sqrt{3}}{2}\)
\(\displaystyle y=3\sin\left(-\frac{5}{6}\pi\right)=-\frac{3}{2}\)
よって、
\(\displaystyle \left(-\frac{3\sqrt{3}}{2},-\frac{3}{2}\right)\)

\(x=3\cos\pi=-3\)
\(y=3\sin\pi=0\)
よって、
\((-3,0)\)

\(r=\sqrt{2^2+2^2}=2\sqrt{2}\)
\(\displaystyle \cos\theta=\frac{2}{2\sqrt{2}},\sin\theta=\frac{2}{2\sqrt{2}}\)より、
\(\displaystyle \theta=\frac{\pi}{4}\)
よって、
\(\displaystyle \left(2\sqrt{2},\frac{\pi}{4}\right)\)

\(r=\sqrt{(-1)^2+(\sqrt{3})^2}=2\)
\(\displaystyle \cos\theta=-\frac{\sqrt{3}}{2},\sin\theta=\frac{\sqrt{3}}{2}\)より、
\(\displaystyle \theta=\frac{2}{3}\pi\)
よって、
\(\displaystyle \left(2,\frac{2}{3}\pi\right)\)

\(r=\sqrt{(-\sqrt{3})^2+(-1)^2}=2\)
\(\displaystyle \cos\theta=-\frac{\sqrt{3}}{2},\sin\theta=-\frac{1}{2}\)より、
\(\displaystyle \theta=\frac{7}{6}\pi\)
よって、
\(\displaystyle \left(2,\frac{7}{6}\pi\right)\)

\(r=\sqrt{(-1)^2+1^2}=\sqrt{2}\)
\(\displaystyle \cos\theta=-\frac{1}{\sqrt{2}},\sin\theta=\frac{1}{\sqrt{2}}\)より、
\(\displaystyle \theta=\frac{3}{4}\pi\)
よって、
\(\displaystyle \left(\sqrt{2},\frac{3}{4}\pi\right)\)

\(r=\sqrt{(\sqrt{3})^2+3^2}=2\sqrt{3}\)
\(\displaystyle \cos\theta=\frac{\sqrt{3}}{2\sqrt{3}},\sin\theta=\frac{\sqrt{3}}{2\sqrt{3}}\)より、
\(\displaystyle \theta=\frac{\pi}{3}\)
よって、
\(\displaystyle \left(2\sqrt{3},\frac{\pi}{3}\right)\)

\(r=\sqrt{0^2+(-2)^2}=2\)
\(\displaystyle \cos\theta=\frac{0}{2},\sin\theta=\frac{-2}{2}\)より、
\(\displaystyle \theta=\frac{3}{2}\pi\)
よって、
\(\displaystyle \left(2,\frac{3}{2}\pi\right)\)