Proving $ \sin4\theta = 4\cos^3\theta \sin\theta - 4\cos\theta \sin^3 \theta$

Question

I am trying to prove this formula:

$ \sin4\theta = 4\cos^3\theta \sin\theta - 4\cos\theta \sin^3 \theta$

Using De Moivre's formula:

$(\cos\theta + i\sin\theta)^4 = \cos4\theta + i\sin4\theta$

How can I do this?

Answer 1

We can also do this without the de-Moivres formula, only using the double angle identities for $\sin$ and $\cos$. Using $\cos(2\theta) = \cos^2 (\theta) - \sin^2 (\theta)$ and $\sin(2\theta) = 2\sin(\theta) \cos(\theta)$, we may calculate $$4\cos^3 (\theta) \sin(\theta)-4\cos(\theta) \sin^3 (\theta) = 4\cos(\theta) \sin(\theta) (\cos^2 (\theta) - \sin^2 (\theta))$$ $$ = 2 (2\sin(\theta) \cos(\theta)) \cos(2\theta) = 2\sin(2\theta) \cos(2\theta) = \sin(4\theta)$$ as desired.

Answer 2

Use de-Moivres formula, as you say, and expand the left hand side using binomial theorem. Then look at the imaginary part of this expansion to get your formula.

Answer 3

Use the fact that $2\sin x \cos x = \sin 2x$ and $\cos^2x-\sin ^2x = \cos 2x$:

\begin{eqnarray} 4\cos^3\theta \sin\theta - 4\cos\theta \sin^3 \theta &=& 4\cos \theta \sin \theta (\cos ^2 \theta -\sin ^2 \theta )\\ &=& 2\sin 2\theta \cos 2\theta \\ &=& \sin 4\theta \end{eqnarray}

Answer 4

Because $$4\cos^3\theta\sin\theta-\sin^3\theta\cos\theta=4\sin\theta\cos\theta(\cos^2\theta-\sin^2\theta)=2\sin2\theta\cos2\theta=\sin4\theta$$