# How do you verify sin x + cos x = (2sin^2 x - 1) /(sin x -cos x)?

Jul 9, 2016

$\sin x + \cos x = \frac{2 \left(1 - {\cos}^{2} x\right) - 1}{\sin x - \cos x}$

sinx + cosx= (2 - 2cos^2x - 1)/(sinx - cosx

$\sin x + \cos x = \frac{1 - 2 {\cos}^{2} x}{\sin x - \cos x}$

Here it looks like we're at a dead end, but in fact we're not.

$\left(\sin x + \cos x\right) \left(\sin x - \cos x\right) = 1 - 2 {\cos}^{2} x$

${\sin}^{2} x - {\cos}^{2} x = 1 - 2 {\cos}^{2} x$

$1 - {\cos}^{2} x - {\cos}^{2} x = 1 - 2 {\cos}^{2} x$

$1 - 2 {\cos}^{2} x = 1 - 2 {\cos}^{2} x \to \text{Identity proved!}$

Hopefully this helps!

Jul 10, 2016

$R H S = \frac{2 {\sin}^{2} x - 1}{\sin x - \cos x}$

$= \frac{2 {\sin}^{2} x - \left({\cos}^{2} x + {\sin}^{2} x\right)}{\sin x - \cos x}$

=(2sin^2x-cos^2x -sin^2x)/(sinx-cosx)

$= \frac{{\sin}^{2} x - {\cos}^{2} x}{\sin x - \cos x}$

$= \frac{\left(\cancel{\sin x - \cos x}\right) \left(\sin x + \cos x\right)}{\cancel{\sin x - \cos x}}$

$= \left(\sin x + \cos x\right) = L H S$