sierraurb
Trignometric functions help!!



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

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

cwrw238
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use a substitution by the identity
cos^2 theta = 1  sin^2 theta

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

cwrw238
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instead of cos^ theta write 1  sin^ theta
then simplify and you'll find you'll get sin^3 theta

sierraurb
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But its cos^2theta

cwrw238
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cos^2 theta = 1  sin^2 theta is an established trig identity

sierraurb
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Okay so how would i use that identity to solve?

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

cwrw238
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sin theta  sin theta * cos^ theta
= sin theta  sin theta ( 1  sin^2 theta)
now expand the brackets and simplify

cwrw238
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yes i've replaced cos^2 theta with 1  sin^2 theta

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

cwrw238
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distribute the sin theta over the parentheses is the next step

cwrw238
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* rather you distribute  sin theta

sierraurb
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so its sin theta  sin theta  sin^3 theta

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

cwrw238
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no  remember:  times  = +

cwrw238
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yes ( but note the sign of sin^3 theta is positive)

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

cwrw238
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 sin theta  sin^2 theta = + sin^3 theta

cwrw238
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yw