Question

j[ppo]po]

Answer 1

Uh cos3x = 4cos^3x - 3cosx
Formulas used:
cos(a + b) = cosacosb - sinasinb
sin(a + b) = sinacosb + cosasinb
sin^2a + cos^2a = 1
Consider the left hand side:
cos3x
= cos(2x + x)
= cos2xcosx - sin2xsinx
= cos(x + x)cosx - sin(x + x)sinx
= (cosxcosx - sinxsinx)cosx - (sinxcosx + cosxsinx)sinx
= (cos^2x - sin^2x)cosx - (2sinxcosx)sinx
= (cos^2x - (1 - cos^2x))cosx - 2sin^2xcosx
= (cos^2x - 1 + cos^2x)cosx - 2(1 - cos^2x)cosx
= (2cos^2x - 1)cosx - 2(cosx - cos^3x)
= 2cos^3x - cosx - 2cosx + 2cos^3x
= 4cos^3x - 3cosx
= right hand side

Answer 2

thanks so much

Answer 3

You're welcome friend. ;u;