Find $\frac{dy}{dx}$ for $x=2\theta+sin2\theta$ and $y=1-cos2\theta$

Question

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.

Answer 1

$$ \frac{2 sin 2\theta }{2+2 cos 2\theta }=\\\frac{2 *2sin \theta cos \theta }{2+2 (2 cos ^2\theta -1)}=\\=\frac{4 sin \theta cos \theta }{2+4 cos^2 \theta -2}=\\\frac{4 sin \theta cos \theta }{4 cos^2 \theta }=\frac{sin \theta }{cos \theta }=tan \theta $$

Answer 2

Hint: $1+\cos2\theta = 2\cos^2\theta$, and $\sin 2\theta = 2\sin\theta \cos\theta$

Answer 3

use the double angle formulas

$$ \sin(2 \theta) = 2 \sin \theta \cos \theta \\ \cos(2 \theta) = 2 \cos^2 \theta - 1 $$ When you simplify, you'll get a $\cos \theta$ in numerator and denominator that'll cancel and you'll be done.

Answer 4

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 $\sin 2\theta = 2 \sin \theta \cos \theta$