Answer
$$\sin\Big(\frac{3\pi}{4}-x\Big)=\frac{\sqrt2}{2}(\cos x+\sin x)$$
Work Step by Step
$$X=\sin\Big(\frac{3\pi}{4}-x\Big)$$
According to sine difference identity:
$$\sin(A-B)=\sin A\cos B-\cos A\sin B$$
That means
$$X=\sin\frac{3\pi}{4}\cos x-\cos\frac{3\pi}{4}\sin x$$
*About $\sin\frac{3\pi}{4}$ and $\cos\frac{3\pi}{4}$
$$\Big|\sin\frac{3\pi}{4}\Big|=\sin\frac{\pi}{4}\hspace{2cm}\Big|\cos\frac{3\pi}{4}\Big|=\cos\frac{\pi}{4}$$
So the values of them are not quite different. The only difference lies in the fact that while $\frac{\pi}{4}$ is in quadrant I, $\frac{3\pi}{4}$ is in quadrant II.
In quadrant II, $\sin\theta\gt0$ but $\cos\theta\lt0$.
Therefore, $$\sin\frac{3\pi}{4}=\sin\frac{\pi}{4}=\frac{\sqrt2}{2}\hspace{1.5cm}\cos\frac{3\pi}{4}=-\cos\frac{\pi}{4}=-\frac{\sqrt2}{2}$$
We replace the values just being found back into $X$:
$$X=\frac{\sqrt2}{2}\cos x-\Big(-\frac{\sqrt2}{2}\Big)\sin x$$
$$X=\frac{\sqrt2}{2}\cos x+\frac{\sqrt2}{2}\sin x$$
$$X=\frac{\sqrt2}{2}(\cos x+\sin x)$$
Overall,
$$\sin\Big(\frac{3\pi}{4}-x\Big)=\frac{\sqrt2}{2}(\cos x+\sin x)$$
