Eahidn
Answered

Join IDNLearn.com and start getting the answers you've been searching for. Ask any question and get a detailed, reliable answer from our community of experts.

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

Sagot :

Luanaidn
[tex]cosx*cos2x+\frac{(sin2x)^2}{2cosx}=cosx\\\\ cosx*(cos^2x-sin^2x)+\frac{(2sinxcosx)^2}{2cosx}=cosx\\\\ cosx*(cos^2x-sin^2x)+\frac{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[/tex]
[tex]cosxcos2x+\frac{sin^22x}{2cosx}=cosx\\\\L=cosx(cos^2x-sin^2x)+\frac{(2sinxcosx)^2}{2cosx}\\\\=cos^3x-sin^2xcosx+\frac{(2sinxcosx)^2}{2cosx}=\frac{2cosx(cos^3x-sin^2xcosx)+(2sinxcosx)^2}\\\\\\=\frac{2cos^4x-2sin^2xcos^2x+4sin^2xcos^2x}{2cosx}=\frac{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}[/tex]


[tex]Other\ theorems:\\\\sin2x=2sinxcosx\\\\cos2x=cos^2x-sin^2x[/tex]
Thank you for using this platform to share and learn. Keep asking and answering. We appreciate every contribution you make. Find the answers you need at IDNLearn.com. Thanks for stopping by, and come back soon for more valuable insights.