<s>[INST] Find $\frac{dy}{dx}$ for $x=2\theta+sin2\theta$ and $y=1-cos2\theta$ 
The parametric equations of a curve are
  $$x=2\theta+\sin2\theta,\:y=1-\cos2\theta.$$
  Show that $\frac{dy}{dx}=\tan\theta$.

I can use the chain rule to get
$$\frac{dx}{d\theta}=2+2\cos2\theta$$
$$\frac{dy}{d\theta}=2\sin2\theta$$
$$\frac{dy}{dx}=\frac{dy}{d\theta}\div\frac{dx}{d\theta}$$
$$=\frac{2\sin2\theta}{2+2\cos2\theta},$$
but I'm not sure how to get to a final proof.
 [/INST] You can see that $$\frac{2\sin 2 \theta}{2+2 \cos 2 \theta} = \frac{\sin 2 \theta}{1+ \cos 2 \theta} \\ = \frac{2 \cos \theta \sin \theta}{2 \cos^2 \theta} = \tan \theta$$
I used the identities $\cos 2\theta = 2 \cos^2 \theta - 1$ and $\sin 2\theta = 2 \sin \theta \cos \theta$
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