Prove: $\sin(\theta) - \sin(\theta)\cos^2(\theta) = \sin^3(\theta)$. Help please? I'm lost. 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 $\sin(\theta)$? 

 A: 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.
A: HINT: $$\sin(\theta)(1 - \cos^2(\theta)) = \sin^3(\theta)$$
A: Let's simplify:
$$\sin{\theta}-\sin{\theta}\cos^{2}{\theta}=\sin^{3}{\theta}$$
$$1-\cos^{2}{\theta}=\sin^{2}{\theta}$$
$$1=\sin^{2}{\theta}+\cos^{2}{\theta}$$
And that being a trigonometric identity makes your theorem proved (The case where $\sin{\theta}=0$ is trivial to solve).
A: sin(θ)−sin(θ)cos2(θ)=sin3(θ)
First, subtract sin(θ) from both sides.
-sin(θ)*cos2(θ)=sin3(θ)-sin(θ)
Now, divide both sides by -sin(θ). Mind the - sign.
Cos2(θ)= [sin3(θ)-sin(θ)]/-sin(θ)
Factor the top by grouping. Pull out a -sin(θ) from both sides. Mind the signs.
Cos2(θ)=-sin(θ)(-sin2(θ)+1)-sin(θ)
Cancel out -sin(θ) from the top and bottom
cos2(θ)=-sin2(θ)+1
Move the 1
cos2(θ)=1-sin2(θ)
Take the square root of both sides
Cos(θ)= +-√1-sin2(θ)
Bam. Trig identity!
