Prove that [sin x + cos x]^2 = 1 + sin 2x
$\displaystyle (\sin{x} + \cos{x})^2 = 1 + \sin(2x)$
square the left side ...
$\displaystyle \sin^2{x} + 2\sin{x}\cos{x} + \cos^2{x}$
rearrange a bit ...
$\displaystyle (\sin^2{x} + \cos^2{x}) + 2\sin{x}\cos{x}$
now, you should know what the sum of the first two terms is ... also, look up the double angle formula for sine to deal with the last term.