(vii) (92,19π12)(\dfrac{9}{2},\dfrac{19\pi}{12})(29,1219π) Here r=92>0r=\dfrac{9}{2}>0r=29>0. 19π12=2π−5π12\frac{19\pi}{12}=2\pi-\frac{5\pi}{12}1219π=2π−125π So it lies in Quadrant IV. Use: cos19π12=cos5π12=6−24,sin19π12=−sin5π12=−6+24\cos\frac{19\pi}{12}=\cos\frac{5\pi}{12}=\frac{\sqrt{6}-\sqrt{2}}{4},\quad \sin\frac{19\pi}{12}=-\sin\frac{5\pi}{12}=-\frac{\sqrt{6}+\sqrt{2}}{4}cos1219π=cos125π=46−2,sin1219π=−sin125π=−46+2 x=92cos19π12=9(6−2)8y=92sin19π12=−9(6+2)8\begin{aligned} x&=\frac{9}{2}\cos\frac{19\pi}{12}=\frac{9(\sqrt{6}-\sqrt{2})}{8}\\ y&=\frac{9}{2}\sin\frac{19\pi}{12}=-\frac{9(\sqrt{6}+\sqrt{2})}{8} \end{aligned}xy=29cos1219π=89(6−2)=29sin1219π=−89(6+2)