Check out some similar questions!

hannahls · Nov 20, 2011, 11:22 AM

I can't verify this identity! I'm so stuck. Please help and show full work. Thanks!

I'll try to make it look like it does in my book...
cos(x) sin(x) - cos(x) - sin(x) = 2tan(2x)
cos(x) - sin(x) cos(x) sin(x)

Aurora2000 · Nov 20, 2011, 12:25 PM

By direct computation:

$\frac{\cos x+\sin x}{\cos x-\sin x}-\frac{\cos x-\sin x}{\cos x+\sin x}=$
$=\frac{(\cos x+\sin x)^2 - (\cos x-\sin x)^2 }{\cos^2 x-\sin^2 x}=$
$= \frac{(\cos^2 x+\sin^2 x +2\cos x\sin x) - (\cos^2 x+\sin^2 x-2\cos x\sin x) }{\cos^2 x-\sin^2 x}=$
$\frac{4\cos x\sin x}{\cos^2 x-\sin^2 x}.$

Using sum formula for sin and cos,
$\sin(2x)=2\sin x\cos x$
$\cos(2x)=\cos^2 x-\sin ^2 x$
$\frac{4\cos x\sin x}{\cos^2 x-\sin^2 x}=\frac{2\sin(2x)}{\cos(2x)}=2\tan(2x)$

as the tan is sin/cos.