Relation between the Cartesian and the Polar Co-ordinates

To connect polar co-ordinates to the familiar Cartesian system, take the polar axis $OX$ as the X-axis and the line through $O$ perpendicular to $OX$ as the Y-axis, with the pole $O$ as the origin. Let $P$ be any point other than the origin, with Cartesian co-ordinates $(x,y)$ and polar co-ordinates $(r,\theta)$. By the definition of the sine and cosine of an angle applied to the right triangle formed by dropping a perpendicular from $P$ to the X-axis,

$\sin\theta=\dfrac{y}{r}$ and $\cos\theta=\dfrac{x}{r}$, so

This is the relation between Cartesian and polar co-ordinates: given $(r,\theta)$ these formulas give $(x,y)$ directly; given $(x,y)$, one first finds $r=\sqrt{x^2+y^2}$ and then solves $\cos\theta=x/r$, $\sin\theta=y/r$ simultaneously for $\theta$ in $[0,2\pi)$, taking care to pick the value of $\theta$ in the correct quadrant (matching both the sign of $\cos\theta$ and the sign of $\sin\theta$, not just one of them).

Ex.(1) Find the Cartesian co-ordinates of the point whose polar co-ordinates are $\left(2,\dfrac{\pi}{4}\right)$. Here $r=2$, $\theta=\dfrac{\pi}{4}$. $x=r\cos\theta=2\cos\dfrac{\pi}{4}=2\times\dfrac{1}{\sqrt2}=\sqrt2$; $y=r\sin\theta=2\sin\dfrac{\pi}{4}=2\times\dfrac{1}{\sqrt2}=\sqrt2$. The required Cartesian co-ordinates are $(\sqrt2,\sqrt2)$. …