Answer
$(\dfrac{\sqrt 6+\sqrt 2}{2},\dfrac{\sqrt 6 -\sqrt 2}{2})$
Work Step by Step
Conversion of polar coordinates and Cartesian coordinates are as follows:
a)$r^2=x^2+y^2 \implies r=\sqrt {x^2+y^2}$
b) $\tan \theta =\dfrac{y}{x} \implies \theta =\tan^{-1} (\dfrac{y}{x})$
c) $X=r \cos \theta$
d) $Y=r \sin \theta$
Here, we have $r=2; \theta =15^{\circ}$
$x=2 \cos 15^{\circ} \implies x=2(\dfrac{\sqrt 6+\sqrt 2}{4})=\dfrac{\sqrt 6+\sqrt 2}{2}$
$y=2 \sin 15^{\circ} \implies y=2(\dfrac{\sqrt 6-\sqrt 2}{4})=\dfrac{\sqrt 6-\sqrt 2}{2}$
Thus, our points are: $(X,Y)=(\dfrac{\sqrt 6+\sqrt 2}{2},\dfrac{\sqrt 6 -\sqrt 2}{2})$
