# Trig multiple angle identity

I have a quick problem that I can't solve. I know it's simple, but I can't seem to crack it.

Determine whether the equation is an identity. Answer true or false.

$$\sin4\theta = \cos\theta(4\sin\theta-8\sin^3\theta)$$

The answer is true. Any help would be appreciated.

• All you need for this is the double angle identities. It's like a 3 line proof. – user137731 Oct 29 '15 at 19:24

$$\sin(4x)= 2\sin(2x)\cos(2x) = 4\sin(x)\cos(x)(1 - 2\sin^2(x))$$ Therefore, $$\sin(4x) = \cos(x)(4\sin(x) - 8\sin^3(x))$$

• Neat, simply explained and clean. +1 – Shailesh Oct 30 '15 at 1:33

You may write \begin{align} \sin4\theta &=2\sin(2\theta)\cos(2\theta)\\\\ &=\cos \theta \:(4\sin\theta\cos(2\theta))\\\\ &=\cos \theta \:(4\sin\theta(\cos^2\theta-\sin^2 \theta))\\\\ &=\cos \theta \:(4\sin\theta(1-2\sin^2 \theta))\\\\ &=\cos \theta \:(4\sin\theta-8\sin^3 \theta)\\\\ \end{align} as announced.

Notice, $$RHS=\cos\theta(4\sin\theta-8\sin^3\theta)$$ $$=2\cos\theta(2\sin\theta-4\sin^3\theta)$$ $$=2\cos\theta((3\sin\theta-4\sin^3\theta)-\sin\theta)$$ $$=2\cos\theta(\sin3\theta-\sin\theta)$$ $$=2\sin3\theta\cos\theta-2\sin\theta\cos\theta$$ Applying $\color{red}{2\sin A\cos B=\sin(A+B)+\sin(A-B)}$

$$=\sin(3\theta+\theta)+\sin(3\theta-\theta)-\sin2\theta$$ $$=\sin4\theta+\sin2\theta-\sin2\theta$$ $$=\sin4\theta=LHS$$ Hence, the given equation: $\color{blue}{\sin4\theta=\cos\theta(4\sin\theta-8\sin^3\theta)}$ is an identity.

HINT: Use the double angle identities:

$\sin{2\theta} = 2 \sin{\theta} \cos{\theta}$ and $\cos{2\theta} = 1-2 \sin^2{\theta}$.

From this point, I'm confident you can expand the left hand side of your problem to get the right hand side, but I'll leave you to it.