Question

Prove: cosx.cos2x+(sin2x)^ 2/2cosx=cosx

Answer 1

`cosxcos2x+(sin^22x)/(2cosx)=cosx\\\\L=cosx(cos^2x-sin^2x)+((2sinxcosx)^2)/(2cosx)\\\\=cos^3x-sin^2xcosx+((2sinxcosx)^2)/(2cosx)=\frac{2cosx(cos^3x-sin^2xcosx)+(2sinxcosx)^2}\\\\\\=(2cos^4x-2sin^2xcos^2x+4sin^2xcos^2x)/(2cosx)=(2cos^2x(cos^2x+sin^2x))/(2cosx)\\\\=cosx\underbrace{(sin^2x+cos^2x)}_(use:sin^2x+cos^2x=1)=cosx=R\\\center\boxed{L=R}`



`Other\ theorems:\\\\sin2x=2sinxcosx\\\\cos2x=cos^2x-sin^2x`

Answer 2

`cosx*cos2x+((sin2x)^2)/(2cosx)=cosx\\\\ cosx*(cos^2x-sin^2x)+((2sinxcosx)^2)/(2cosx)=cosx\\\\ cosx*(cos^2x-sin^2x)+(4sin^2xcos^2x)/(2cosx)=cosx\\\\ cosx*(cos^2x-sin^2x)+2sin^2xcosx=cosx\ \ \ |Divide\ by\ cosx\\\\ (cos^2x-sin^2x)+2sin^2x=1\\\\ cos^2x-sin^2x+2sin^2x=1\\\\ cos^2x+sin^2x=1\\\\ \boxed{1=1}\ \ TRUE`