Question

Prove: Sin Theta-sin theta(cos^2 theta)=sin^3 theta. I have to show all work. I already know I have to factor out the sin theta from the left side and that I have to use the identity : sin^2 θ + cos^2 θ = 1but I'm not sure where to go from there

Answer 1

\(\bf sin(\theta)-sin(\theta)cos^2(\theta)=sin^3(\theta)?\)

Answer 2

hmmm

Answer 3

yes. When I factor out sin theta i get

sin theta(1-cos^2 theta)=sin^3 theta

I have no idea where to go from there

Answer 4

wait... how did you get that?

Answer 5

hmmm ok so \(\bf {\color{brown}{ sin^2(\theta)+cos^2(\theta)=1\to cos^2(\theta)=1-sin^2(\theta)}}
\\ \quad \\ \quad \\

sin(\theta)-sin(\theta)cos^2(\theta)=sin^3(\theta)
\\ \quad \\
sin(\theta)-sin(\theta)[{\color{brown}{ 1-sin^2(\theta)}}]\implies ?\)

Answer 6

do i distribute \[-\sin \theta [1-\sin^2 \theta]\]

Answer 7

yeap

Answer 8

\[\sin \theta-\sin \theta +\sin^3 \theta\]

Answer 9

Can you subtract \[\sin^3 \theta \] from both sides? and make both sides =0?