Using the half/double angle formulas to solve an equation I am completely stumped by this problem: $$\cos\theta - \sin\theta =\sqrt{2} \sin\frac{\theta}{2} $$
I know that I should start by isolating $\sin\dfrac{\theta}{2}$
and end up with $$\frac {\cos\theta - \sin\theta}{\sqrt{2}} = \sin\frac{\theta}{ 2}$$
From here on out, I have no idea what steps to take.
 A: W/O using the half/double angle formulas,
we have  $$\sin\frac\theta2=\sin\left(\frac\pi4-\theta\right)$$
$$\implies\frac\theta2=n\pi+(-1)^n\left(\frac\pi4-\theta\right)$$ where $n$ is any integer
A: $$\cos \theta . \sin \frac{\pi}{4} - \cos \frac{\pi}{4} . \sin \theta = \sin \frac{\theta}{2}$$
$$\sin (\frac{\pi}{4} - \theta) = \sin \frac{\theta}{2}$$
$$\frac{\pi}{4} - \theta = n\pi + (-1)^n(\frac{\theta}{2})$$
A: Given : $\dfrac{\cos\theta - \sin\theta}{\sqrt{2}} = \sin\dfrac{\theta}{ 2}$
using $\sin(\frac{\theta}{2})=\sqrt{\frac{1-\cos(\theta)}{2}}$ and squaring both sides,
\begin{align*}
\frac{\sin^2(\theta)+\cos^2(\theta)-2\sin(\theta)\cos(\theta)}{2} & = \frac{1-\cos(\theta)}{2}\\
1-2\sin(\theta)\cos(\theta) & = 1-\cos(\theta) && \text{because $\sin^2(\theta)+\cos^2(\theta) = 1$}\\
-2\sin(\theta)\cos(\theta) & = -\cos(\theta)\\
-\sin(\theta) & = -\frac{1}{2}\\
\sin(\theta) & = \frac{1}{2} = \sin(\frac{\pi}{3})
\end{align*}
General Solutions $\theta$ = $n\pi +(-1)^{n}\frac{\pi}{3}$ , $n\in Z$
