Prove the identity. cos3x=4cos3x−3cosx

Question

Prove the identity.

cos3x=4cos³x−3cosx

Answer 1

cos3x = 4cos^3(x) - 3cosx

let y = cosx

cos3x = 3y^3 -3y

cos3x = cos(x+2x) = cosxcos2x -sinxsin2x = cosx(cos(x+x)) -sinxsin2x = cosx(cos^2(x) - 1) - sinxsin2x

= y(y^2-1) - sinxsin(x+x)

= y^3 -y - sinx)(2sinxcosx)

= y^3 -y - 2sin^2(x)y

= y^3 -y -2(1-cos^2(x))y

= y^3 -y -2(1-y^2)y

= y^3 -y -2y +2y^3

= 3y^3 -2y -y

= 3y^3 -2y -y

= 3y^2 -3y QED

Answer 2

cos(3x)

= cos(2x+x)

= cos(2x)cos(x) - sin(2x)sin(x) <-- Recall cos(a+b) = cos(a)cos(b) - sin(a)sin(b) by Difference Formula

= [cos²(x)-sin²(x)]cos(x) - [2cos(x)sin(x)]sin(x) <-- cos(2x) = cos²(x)-sin²(x) and sin(2x) = 2cos(x)sin(x) by Double-Angle Identities

= cos³(x) - sin²(x)cos(x) - 2cos(x)sin²(x)

= cos³(x) - 3cos(x)sin²(x)

= cos³(x) - 3cos(x)[1-cos²(x)] <-- sin²(x) = 1-cos²(x) by the Pythagorean Identity

= cos³(x) - 3cos(x) + cos³(x)

= 4cos³(x) - 3cos(x)

Hope this helped!

Answer 3

cos(3x) = cos(2x+x) = cos(2x)cos(x)-sin(2x)sin(x)

=(cos²(x)-sin²(x))cos(x)-(2cos(x)sin(x))sin(x)

=cos³(x)-sin²(x)cos(x) -2cos(x)sin²(x)

=cos³(x)-3cos(x)sin²(x)

=cos³(x)-3cos(x)(1-cos²(x))

=cos³(x)-3cos(x)+3cos³(x)

=4cos³(x)-3cos(x)