Question

Корень из 2cosx-2cos (45°+x)/2sin (45°+x)-корень из 2sinx

Answer 1

$\displaystyle \frac{\sqrt{2}\cos x-2\cos\left(45^\circ+x\right)}{2\sin\left(45^\circ+x\right)-\sqrt{2}\sin x}=\frac{\sqrt{2}\cos x-2(\cos45^\circ\cos x-\sin45^\circ\sin x)}{2(\sin 45^\circ\cos x+\cos 45^\circ\sin x)-\sqrt{2}\sin x}=\\ \\ \\ =\frac{\sqrt{2}\cos x-2\left(\frac{\sqrt{2}}{2}\cos x-\frac{\sqrt{2}}{2}\sin x\right)}{2\left(\frac{\sqrt{2}}{2}\cos x+\frac{\sqrt{2}}{2}\sin x\right)-\sqrt{2}\sin x}=\frac{\sqrt{2}\cos x-\sqrt{2}\cos x+\sqrt{2}\sin x}{\sqrt{2}\cos x+\sqrt{2}\sin x-\sqrt{2}\sin x}=$

$=\displaystyle \frac{\sqrt{2}\sin x}{\sqrt{2}\cos x}=\frac{\sin x}{\cos x}={\rm tg}\, x$