Simplification of polar equation

Question

How was this answer simplified? How come the $4$ turn into an $8$, and the $\sin θ \cos θ$ turn into $\sin2θ$?

$9(r \cos θ)(r \sin θ) = 4,\quad 9r^2 \sin θ \cos θ = 4,\quad r^2 \sin 2θ = 8/9$

Answer 1

(First time trying the align function...)

\begin{align} 9r^2 \cos \theta \sin \theta & = 4 \\[4px] \frac{9}{2} r^2 \sin 2 \theta & = 4 && (\text {because} \cos \theta \sin \theta = \tfrac{1}{2} \sin 2 \theta) \\[4px] {9} r^2 \sin 2 \theta & = 8 && (\text {multiplying by 2})\\[4px] r^2 \sin 2 \theta & = \!\frac{8}{9} && (\text {dividing by 9})\\ \end{align}

Answer 2

note that $$9r\cos(\theta)r\sin(\theta)=9r^2\sin(\theta)\cos(\theta)=\frac{9}{2}r^22\sin(\theta)\cos(\theta)=\frac{9}{2}r^2\sin(2\theta)$$ since $$\sin(2x)=2\sin(x)\cos(x)$$

Answer 3

$$9r^2\sin θ \cos θ = 4,\quad $$

multiply both sides by $2$

$$9r^2 2 \sin θ \cos θ = 8,\quad $$

Since

$$2 \sin θ \cos θ = \sin 2θ, \quad $$

use it to simplify and shorten it further

$$ 9\,r^2\sin 2θ = 8$$

Transpose $9$ to the RHS

$$ r^2\sin 2θ = 8/9.$$