第1章 1.11

(1)

$(4,\dfrac{\pi}{4})$

$x=r\cos\theta=4\cos\dfrac{\pi}{4}=4\cdot\dfrac{\sqrt{2}}{2}=2\sqrt{2} \qquad y=r\sin\theta=4\sin\dfrac{\pi}{4}=2\sqrt{2} \qquad \therefore(2\sqrt{2},2\sqrt{2})$

(2)

$(2,\dfrac{2}{3}\pi)$

$x=r\cos\theta=2\cos\dfrac{2}{3}\pi=2\cdot(-\dfrac{1}{2})=-1 \qquad y=r\sin\theta=2\sin\dfrac{2}{3}\pi=2\cdot\dfrac{\sqrt{3}}{2}=\sqrt{3} \qquad \therefore(-1,\sqrt{3})$

(3)

$(3,\dfrac{\pi}{6})$

$x=r\cos\theta=3\cos\dfrac{\pi}{6}=3\cdot\dfrac{\sqrt{3}}{2}=\dfrac{3\sqrt{3}}{2} \qquad y=r\sin\theta=3\sin\dfrac{\pi}{6}=3\cdot\dfrac{1}{2}=\dfrac{3}{2} \qquad \therefore(\dfrac{3\sqrt{3}}{2},\dfrac{3}{2})$

(4)

$(1,-\dfrac{2}{3}\pi)$

$x=r\cos\theta=\cos(-\dfrac{2}{3}\pi)=-\dfrac{1}{2} \qquad y=r\sin\theta=\sin(-\dfrac{2}{3}\pi)=-\dfrac{\sqrt{3}}{2} \qquad \therefore(-\dfrac{1}{2},-\dfrac{\sqrt{3}}{2})$

(5)

$(3,-\dfrac{\pi}{2})$

$x=r\cos\theta=3\cos(-\dfrac{\pi}{2})=3\cdot0=0 \qquad y=r\sin\theta=3\sin(-\dfrac{\pi}{2})=3\cdot(-1)=-3 \qquad \therefore(0,-3)$

(6)

$(2,-\pi)$

$x=r\cos\theta=2\cos(-\pi)=2\cdot(-1)=-2 \qquad y=r\sin\theta=2\sin(-\pi)=2\cdot0=0 \qquad \therefore(-2,0)$