Shift in origin for polar coordinates Consider a set of polar coordinates, $(r, \theta)$ for a plane. Let's say there is a point, at $\left(1,\frac{\pi}{4}\right)$ for which I want to define a translated and rotated coordinate system, $(R, \Theta)$ for which that point has the coordinates $(0,0)$.
I just want to verify here, the transformation is given by $R = r - 1$ and $\Theta = \theta - \frac{\pi}{4}$, correct?
 A: Hint Suppose we want to define new coordinates $(R, \Theta)$ using a reference point $(\rho, \alpha)$ given in the original polar coordinates $(r, \theta)$. Then, the old polar coordinates $(r, \theta)$ and new polar coordinates $(R, \Theta)$ of the arbitrary (red) point are related as in the diagram below.
By the Law of Cosines, $$R^2 = r^2 + \rho^2 - 2 \rho r \cos (\theta - \alpha) .$$
Likewise, the Law of Sines gives
$$\sin \Theta = \frac{r}{R} \sin (\theta - \alpha) .$$

A: 
Hint

We can consider polar coordinates as vectors and then use vector subtraction

A: 


*

*$rsin\theta=\rho.sin\alpha+Rcos(\phi-(\frac\pi2-\alpha))$




*$rsin(\theta-\alpha)=Rsin(\pi-\phi)$


*

*$rsin\theta.cos\alpha-rcos\theta.sin\alpha=Rsin\phi$

*$[\rho.sin\alpha+Rcos(\phi-(\frac\pi2-\alpha))]cos\alpha-rcos\theta.sin\alpha=Rsin\phi$

*$rcos\theta=\frac{[\rho.sin\alpha+Rcos(\phi-(\frac\pi2-\alpha))]cos\alpha-Rsin\phi}{sin\alpha}$
$r=\sqrt((rcos\theta)^2+(rsin\theta)^2)$
$tan\theta=\frac{rsin\theta}{rcos\theta}$
Substitute r, $\theta$ wherever you need
