Prove: $\sin(\theta) – \sin(\theta)\cos^2(\theta) = \sin^3(\theta)$.
Can someone show me how to prove this? Do I use the Pythagorean Identity to prove it?
I will be forever grateful. Thanks!
- Do I factor $\sin(\theta)$ out of both terms on the left side of the equal sign
and then divide both sides by $\sin(\theta)$?
Best Answer
Rewrite $$\sin\theta-\sin\theta\cos^2\theta=\sin\theta(1-\cos^2\theta)$$ Remember that $1-\cos^2\theta=\sin^2\theta$ and then plug it in: $$ \sin\theta-\sin\theta\cos^2\theta=\sin\theta(1-\cos^2\theta)= \sin\theta\cdot\sin^2\theta=\sin^3\theta. $$
Note: there's no need to divide by $\sin\theta$ which is a very dangerous thing to do, because you don't know whether it's zero or not. In this particular case it makes no real difference, because if $\sin\theta=0$ the identity clearly holds. It could be in different situations.