# How do you prove (sinx + cosx)^2=1 + sin2x?

Apr 15, 2015

It depends on how far back you need to go in your proof.
If you can use:
${\sin}^{2} \left(x\right) + {\cos}^{2} \left(x\right) = 1$ (which can be derived from the Pythagorean Theorem)
and $\sin \left(2 x\right) = 2 \sin \left(x\right) \cdot \cos \left(x\right)$ (the Double Angle Formula for sin, which is considerably more complicate to prove)
then the requested proof is fairly simple:

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

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

$= \left[{\sin}^{2} \left(x\right) + {\cos}^{2} \left(x\right)\right] + 2 \sin \left(x\right) \cdot \cos \left(x\right)$

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