# How do you verify (1+sinx+cosx)^2 = 2(1+sinx)(1+cosx)?

Jun 15, 2015

You will be squaring a quantity that contains both $\sin x$ and $\cos x$, so you will need the identity ${\sin}^{2} x + {\cos}^{2} x = 1$. I would start by multiplying both of these out.

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

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

while

$2 \left(1 + \sin x\right) \left(1 + \cos x\right) = 2 \left(1 + \cos x + \sin x + \sin x \cos x\right) = 2 + 2 \cos x + 2 \sin x + 2 \sin x \cos x$

Compare:
$1 + {\sin}^{2} x + {\cos}^{2} x + \cancel{2 \sin x + 2 \cos x + 2 \sin x \cos x} = 2 + \cancel{2 \cos x + 2 \sin x + 2 \sin x \cos x}$

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

$1 + 1 = 2$

$2 = 2$

They're equal.