# Using sum-to-product formula to solve $\sin(2\theta)+\sin(4\theta)=0$

Trying to use the sum-to-product formula to solve $$\sin(2\theta)+\sin(4\theta)=0$$ over the interval $$[0,2\pi)$$, but I'm missing solutions.

$$\sin(2\theta)+\sin(4\theta)=0$$

Apply sum-to-product formula:

$$2\sin\left(\frac{2\theta+4\theta}{2}\right)\cos\left(\frac{2\theta-4\theta}{2}\right)=0$$

$$2\sin(3\theta)\cos(-\theta)=0$$

By odd-even identities: $$\cos(-\theta)=\cos(\theta)$$

$$2\sin(3\theta)\cos(\theta)=0$$

$$\sin(3\theta)\cos(\theta)=0$$

By the zero-product property

$$\sin(3\theta)=0$$ or $$\cos(\theta)=0$$

Then solving for theta gives: $$\theta=0, \frac{\pi}{2}, \frac{3\pi}{2}, \pi$$.

However, there are missing solutions $$\frac{\pi}{3}, \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{5\pi}{3}$$.

A solution online used double angle identities instead:

$$\sin(2\theta)+\sin(4\theta)=0$$

$$\sin(2\theta)+\sin(2*2\theta)=0$$

Apply double angle identity for: $$\sin(2*2\theta)$$

$$\sin(2\theta)+2\sin(2\theta)\cos(2\theta)=0$$

Factor out $$\sin(2\theta)$$

$$\sin(2\theta)*[1+2\cos(2\theta)]=0$$

Apply double angle identities:

$$\cos(2\theta)= 1-2\sin^2(\theta)$$

$$\sin(2\theta)= 2\sin(\theta)\cos(\theta)$$

$$2\sin(\theta)\cos(\theta)*[1+2(1-2\sin^2(\theta))]=0$$

$$2\sin(\theta)\cos(\theta)*[-4\sin^2(\theta)+3]=0$$

By the zero-product property

$$2\sin(\theta)\cos(\theta)=0$$ or $$-4\sin^2(\theta)+3=0$$

Which further simplifies to

$$\sin(\theta)=0$$, $$\cos(\theta)=0$$, or $$-4\sin^2(\theta)+3=0$$

Solving for theta now gives all possible solutions over $$[0, 2\pi)$$.

My questions are: (1) Can the sum-to-product formula be used to solve this equation?

(2) If so, why were solutions missing when using the sum-to-product formula but not the double angle identities? What was I doing incorrectly?

• No solutions were missing! – Andrew Chin Jul 2 '20 at 23:05
• Wouldn't it be easier to use $$\sin(2\theta)+\sin(4\theta)= \sin(2\theta)+2\sin(2\theta)\cos(2\theta)=\sin(2\theta)\cdot(2+\cos(2\theta)?$$ – Michael Hoppe Jul 3 '20 at 11:45

## 1 Answer

This is an excellent way to proceed with this problem, and the reduction to $$\sin(3\theta)\cos(\theta)=0$$ is great; this implies that $$\sin(3\theta)=0$$ or $$\cos(\theta)=0$$.

• The solutions to $$\cos(\theta)=0$$ are $$\theta = \dots,\frac\pi2,\frac{3\pi}2,\dots$$.
• The solutions to $$\sin(\alpha)=0$$ are $$\alpha = \dots, 0, \pi, 2\pi, \dots$$. But we have $$\sin(3\theta)=0$$, and so the solutions are $$3\theta = \dots, 0, \pi, 2\pi, \dots$$, which is the same as $$\theta=\dots,0,\frac\pi3,\frac{2\pi}3,\pi,\frac{4\pi}3,\frac{5\pi}3,\dots$$.
• Ah yes, I see my mistake now. Thank you. – Slecker Jul 2 '20 at 23:18