Verifying Trigonomic Identities

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The discussion revolves around verifying the trigonometric identity cos(3θ) = cos3θ - 3sin²θcosθ. Participants suggest starting with the left side of the equation, using the cosine addition formula cos(u + v) = cosu cosv - sinu sinv to simplify cos(3θ). There is a focus on breaking down cos(3θ) into cos(2θ + θ) for easier manipulation. The conversation emphasizes the use of known trigonometric identities, such as sin²u = 2sinu cosu, to aid in solving the equation. Overall, the thread seeks collaborative input to clarify and solve the identity.
lucy1234
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Hey,

Can anyone help me solve this equation please?

Thanks

Homework Statement



cos(3θ)=cos3θ-3sin2θcosθ


Homework Equations



cos(u+v) = cosu cosv - sinu sinv
sin2u = 2sinu cosu

The Attempt at a Solution


 
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I started with taking cosθ out of the equation, but got stuck from there
 
lucy1234 said:
Hey,

Can anyone help me solve this equation please?

Thanks

Homework Statement



cos(3θ)=cos3θ-3sin2θcosθ


Homework Equations



cos(u+v) = cosu cosv - sinu sinv
sin2u = 2sinu cosu
I would start in on the left side.
cos(3θ) = cos(2θ + θ) = ?
 
cos3x = cos(2x+x) if that helps.
 
Mark44 said:
I would start in on the left side.
cos(3θ) = cos(2θ + θ) = ?

iRaid said:
cos3x = cos(2x+x) if that helps.
I think that's pretty much what I said.
 
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