Question

Take r=4sin(theta) into rectangular (x,y) form

Answer 1

\[\begin{cases}r=\sqrt{x^2+y^2}\\\theta=\tan^{-1}\dfrac{y}{x}\end{cases}\]
\[\sqrt{x^2+y^2}=4\sin\left(\tan^{-1}\frac{y}{x}\right)\]

\[\sin\theta=\sin\left(\tan^{-1}\frac{y}{x}\right)=\frac{y}{\sqrt{x^2+y^2}}\]
\[\begin{align*}\sqrt{x^2+y^2}&=\frac{4y}{\sqrt{x^2+y^2}}\\
x^2+y^2&=4y&\text{this form is fine, but if you want to continue...}\\
y^2-4y&=-x^2\\
y^2-4y+4&=4-x^2\\
(y-2)^2&=4-x^2\\
y&=2\pm\sqrt{4-x^2}
\end{align*}\]