<s>[INST] Prove $\sin8\theta-\sin10\theta=\cot9\theta(\cos10\theta-\cos8\theta)$ $$\sin8\theta-\sin10\theta=\cot9\theta(\cos10\theta-\cos8\theta)$$
So far I've done
$$2\cos\left(\dfrac {18}2\right)\theta\sin\left(\dfrac {8-10}2\right)\theta$$
I am stuck, can anyone help?
Am I supposed to multiply them by $2$ to get rid of the $2$'s and then divide all by $\sin$?
 [/INST] From the angle addition identities 
$$\begin{align*}
\sin (a \pm b) &= \sin a \cos b \pm \sin b \cos a \\
\cos (a \pm b) &= \cos a \cos b \mp \sin a \sin b
\end{align*}$$
we find 
$$\begin{align*}
\sin (a+b) - \sin(a-b) &= 2\sin b \cos a \\
\cos (a+b) - \cos(a-b) &= -2 \sin a \sin b.
\end{align*}$$
Then choosing $a = 9\theta$, $b = \theta$, we get
$$\begin{align*}
\sin 10\theta - \sin 8\theta &= 2 \sin \theta \cos 9\theta \\
\cos 10\theta - \cos 8\theta &= -2 \sin 9\theta \sin \theta.
\end{align*}$$
Consequently, $$\frac{\sin 8\theta - \sin 10\theta}{\cos 10\theta - \cos 8\theta} = \frac{-2\sin \theta \cos 9\theta}{-2 \sin 9\theta \sin \theta} = \cot 9\theta.$$
</s>