Show that $2\sin{\phi}\sin{\theta}+\frac{2\sin{\phi}\cos{(2\theta)}}{\sin{\theta}}-2\cot{\theta}\sin{\phi}\cos{\theta}=0$ as simply as possible.

Question

Using the fewest possible number of trig identities, how do you show that

$\quad2\sin{\phi}\sin{\theta}+\frac{2\sin{\phi}\cos{(2\theta)}}{\sin{\theta}}-2\cot{\theta}\sin{\phi}\cos{\theta}=0$?

Answer 1

Hint: Use $$\cos(2\theta) = \cos^2\theta - \sin^2 \theta$$

Answer 2

You see that the only trigonometric function there that doesn't have a single angle is the middle term. In fact, you have a double-angle. Consider expanding that guy with the double angle identity for cosine and see your result magically vanish.

Answer 3

Hint: It is $$2\sin(\phi)\sin(\theta)+\frac{2\sin(\phi)\cos(2\theta)}{\sin(\theta)}=\frac{2\sin(\phi)\cos(\theta)^2}{\sin(\theta)}$$

Answer 4

Multiply by $\dfrac{\sin\theta}{2\sin\phi}$, and get

$$\sin^2{\theta}+\cos{(2\theta)}-\cos^2{\theta}=0.$$

Answer 5

$$LHS$$ $=\quad2\sin{\phi}\sin{\theta}+\frac{2\sin{\phi}\cos{(2\theta)}}{\sin{\theta}}-2\cot{\theta}\sin{\phi}\cos{\theta}=\frac{2\sin{\phi}\sin^2{\theta}-2\sin{\phi}\sin^2{\theta}+2\sin{\phi}\cos^2{\theta}-2\sin{\phi}\cos^2{\theta}}{sin{\theta}}=\frac{0}{sin{\theta}}=0=RHS$