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Trignometric functions help!!

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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

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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
sin theta - sin theta * cos^ theta = sin theta - sin theta ( 1 - sin^2 theta) now expand the brackets and simplify
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

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