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

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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]
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