Show that $\sin2\alpha\cos\alpha+\cos2\alpha\sin\alpha = \sin4\alpha\cos\alpha – \cos4\alpha\sin\alpha$
I know that $\sin2\alpha = 2\sin\alpha\cos\alpha$
so
$$\sin2\alpha\cos\alpha=2\sin\alpha\cos^2\alpha$$
and $\cos2\alpha\sin\alpha$ can be expressed in three ways:
$$(\cos^2\alpha-\sin^2\alpha)\sin\alpha =\sin\alpha\cos^2\alpha-\sin^3\alpha$$
$$(2\cos^2\alpha -1)\sin\alpha = 2\cos^2\alpha\sin\alpha – \sin\alpha$$
$$(1-2\sin^2\alpha)\sin\alpha = \sin\alpha – 2\sin^3\alpha$$
I tried adding these, but nothing came close to the required answer.
So then I tried calculating $\sin4\alpha$ (from the required answer):
$$\sin4\alpha=2\sin(2\alpha)\cdot\cos(2\alpha)$$
$$\sin4\alpha=2\cdot2\sin\alpha\cos\alpha(\cos^2\alpha-\sin^2\alpha)$$
$$\sin4\alpha=4\sin\alpha\cos\alpha(\cos^2\alpha-\sin^2\alpha)$$
so $$\sin4\alpha\cos\alpha= 4\sin\alpha\cos^2\alpha(\cos^2\alpha-\sin^2\alpha)$$
Still looking at the answer, I calculated $\cos4\alpha$
$$\cos4\alpha = 1- \sin^2(2\alpha)$$
If $$\sin2\alpha = 2\sin\alpha\cos\alpha$$
then $$\sin^22\alpha= (2\sin\alpha\cos\alpha)^2$$
and $$\cos4\alpha = 1 – 4\sin^2\alpha\cos^2\alpha$$
and $$\cos4\alpha\sin\alpha=\sin\alpha-4\sin^3\alpha\cos^2\alpha$$
I have tried subtracting my values for $\sin4\alpha\cos\alpha$ and $\cos4\alpha\sin\alpha$, but I have not come close to a solution.
Best Answer
use $$\sin(A+B) = \sin A\cos B + \cos A\sin B $$on LHS and $$\sin(A-B) =\sin A\cos B - \cos A\sin B$$ on RHS
so $$\sin(3\alpha) = \sin(3\alpha)$$