Giải thích các bước giải:
$a)VT=2sin(\frac{\pi}{4}+x)sin(\frac{\pi}{4}-x)\\ =2.\frac{1}{2}.\left [ cos(\frac{\pi}{4}+x-\frac{\pi}{4}+x)-cos(\dfrac{\pi}{4}+x+\frac{\pi}{4}-x) \right ]\\
=cos2x-cos\frac{\pi}{2}=cos2x=VP\Rightarrow ĐPCM\\
b) VT=sinx(1+cos2x)=sinx(1+2cos^2x-1)=2sinx.cos^2x\\
=2sinxcosx.cosx=sin2x.cosx=VP\\
c)VT=\dfrac{sinx+sin3x+sin5x}{cosx+cos3x+cos5x}=\dfrac{(sinx+sin5x)+sin3x}{(cosx+cos5x)+cos3x}\\ \\
=\dfrac{2sin\frac{x+5x}{2}cos\frac{x-5x}{2}+sin3x}{2cos\frac{x+5x}{2}cos\frac{x-5x}{2}+cos3x}\\ \\
=\dfrac{2sin3xcos2x+sin3x}{2cos3xcos2x+cos3x}\\ \\
=\dfrac{sin3x(2cos2x+1)}{cos3x(2cos2x+1)}\\
=\dfrac{sin3x}{cos3x}=tan3x\Rightarrow ĐPCM\\
d) VT=\dfrac{1+sin2x-cos2x}{1+sin2x+cos2x}=\dfrac{sin^2x+cos^2x+sin2x-cos^2x+sin^2x}{sin^2x+cos^2x+sin2x+cos^2x-sin^2x}\\ \\
=\dfrac{2sin^2x+sin2x}{2cos^2x+sin2x}\\ \\
=\dfrac{2sin^2x+2sinxcosx}{2cos^2x+2sinxcosx}\\ \\
=\dfrac{2sinx(sinx+cosx)}{2cosx(sinx+cosx)}\\ \\
=\dfrac{sinx}{cosx}=tanx=VP\Rightarrow ĐPCM$
