Is the following curve a circle? Let $\gamma(\theta)=(\sin(\theta+\alpha)\cos\theta,\sin(\theta+\alpha)\sin\theta),\theta\in[0,2\pi]$. Is $\gamma(\theta)$ a circle? What is the radius of it?
 A: Let's recall two trigonometric identities:
$$
\begin{align}
\sin\gamma\cos\delta & = \frac{\sin(\gamma+\delta)+\sin(\gamma-\delta)}{2} \\  \\
\sin\gamma\sin\delta & = \frac{\cos(\gamma-\delta)-\cos(\gamma+\delta)}{2}
\end{align}
$$
So put $\theta+\alpha$ in the role of $\gamma$ and $\theta$ in the role of $\delta$.  Then we have
$$
\begin{align}
\sin(\theta+\alpha)\cos\theta & = \frac{\sin\alpha+\sin(2\theta+\alpha)}{2} \\  \\
\sin(\theta+\alpha)\sin\theta & = \frac{\cos\alpha-\cos(2\theta+\alpha)}{2}
\end{align}
$$
As $\theta$ goes from $0$ to $2\pi$, this point goes twice around a circle centered at $(\sin\alpha,\cos\alpha)/2$, with radius $1/2$ (and diameter $1$).
A: Using polar coordinates $(r,\theta)$, the equation of this curve is 
$$
r=\sin(\theta+\alpha)=\sin(\theta)\cos(\alpha)+\cos(\theta)\sin(\alpha).
$$ 
One sees that $r^2=r\sin(\theta)\cos(\alpha)+r\cos(\theta)\sin(\alpha)=y\cos(\alpha)+x\sin(\alpha)$, hence
$$
x^2+y^2-x\sin(\alpha)-y\cos(\alpha)=0.
$$
This is indeed the equation of a circle. Let me indicate its radius is $
\frac12$ and leave you the joy to find its center.
