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prove \[\sin \Theta -\sin \Theta \times \cos ^{2}\Theta =\sin ^{3}\Theta \] show all work

I get that you'd start by multiplying sin(theta) and cos^2(theta) but idk what that comes out to

use a substitution by the identity
cos^2 theta = 1 - sin^2 theta

I have no idea what that means :( this is the one question i have no clue how to do

instead of cos^ theta write 1 - sin^ theta
then simplify and you'll find you'll get sin^3 theta

But its cos^2theta

cos^2 theta = 1 - sin^2 theta is an established trig identity

Okay so how would i use that identity to solve?

should i replace cos^2 theta with 1-sin^2

yes i've replaced cos^2 theta with 1 - sin^2 theta

do i subctract the two sin theta's or distribute the one?

distribute the sin theta over the parentheses is the next step

* rather you distribute - sin theta

so its sin theta - sin theta - sin^3 theta

and the two sin theta's cancel out and leave me with sin^3 theta?

no - remember:- - times - = +

yes ( but note the sign of sin^3 theta is positive)

I understand now!! Thank you so so much I appreciate it a lot :)

- sin theta - sin^2 theta = + sin^3 theta

yw