Question

Prove Cos³ x-sin²x = 1/16 ( 2cos x- cos 3x - cos 5x)​

Answer 1

Answer:

Step-by-step explanation:

cos2x = cos^2x - sin^2x  

sin2x = 2sinxcosx

 

cos3x = cosxcos2x - sinxsin2x = cosx(cos^2x-sin^2x) - sinx(2sinxcosx)  

 

cos3x = cos^3x - cosxsin^2x - 2sin^2xcosx = cos^3x - 3sin^2xcosx  

 

cos3x = cos^3x - 3(1-cos^2x)cosx = cos^3x - 3cosx + 3cos^3x  

 

cos3x = 4cos^3x - 3cosx  

 

sin3x = sin2xcosx + cos2xsinx = 2cosxsinxcosx + (2cos^2x - 1)sinx  

 

sin3x = 2cos^2xsinx + 2cos^2xsinx - sinx...