Ask Me Help Desk - Trigonometry equation

I need to bring this equation $\sin x + \cos x - \sin x \cos x - m = 0$

to this form: $2 \cos^2 $ x - \frac{\pi}{4} $ - 2 \sqrt{2} \cos $ x - \frac{\pi}{4} $ +2m -1 = 000$

How do I get there?

Start on the second equation by using the identity cos(a-b) = cos(a)cos(b+sin(a)sin(b), where a = x and b = pi/4. The only other identity you'll need is cos^2(x) + sin^2(x) = 1. It works right out.

Thank you, Ebaines! I did work out, indeed! :)

$(\sin x + \cos x)^2 = \sin^2 x + \cos^2 x + 2 \sin x \cos x$

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

$\cos $x - \frac{\pi}{4}$ = \frac{\sqrt{2}}{2} (\cos x + \sin x)$

The first equation can be written:
$(\sin x + \cos x)^2 - 2(\sin x + \cos x) -2m -1 = 0$

$\[ \sqrt{2} \cos $x - \frac{\pi}{4} $ \]^2 - 2 \sqrt{2} \cos $ x - \frac{\pi}{4} $ - 2m -1 = 0$

... and here is the final form:
$2 \cos^2 $ x- \frac{\pi}{4} $ - 2 \sqrt{2} \cos $ x- \frac{\pi}{4} $ -2m -1 =0$