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 $\sin(\theta)$?
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.
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!