Prove the identity. cos3x=4cos3x−3cosx

Question

Prove the identity.

cos3x=4cos³x−3cosx

Answer 1

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 2

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)