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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?

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Hint… – David Speyer Jan 25 '12 at 17:13
Plot it (for some suitably chosen values of $\alpha$, such as $0$ and $\pi/2$). Does is look like a circle? If so, what does its center and radius appear to be? If you write down the naive parameterization of a circle with that center and radius, can you prove that it equals your $\gamma$? – Henning Makholm Jan 25 '12 at 17:15
Pedantic note: $\gamma$ might or might not be (the equation of) a circle (here, it is) but $\gamma(\theta)$ is a point in the plane. – Did Jan 25 '12 at 17:20
up vote 3 down vote accepted

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 $1$).

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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.

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