Answer
$$\cos\Big(\frac{3\pi}{4}- x\Big)=\frac{\sqrt2(\sin x-\cos x)}{2}$$
Work Step by Step
$$X=\cos\Big(\frac{3\pi}{4}-x\Big)$$
To expand the formula, cosine difference identity would be used:
$$\cos(A-B)=\cos A\cos B+\sin A\sin B$$
That means
$$X=\cos\frac{3\pi}{4}\cos x+\sin\frac{3\pi}{4}\sin x$$
1) Now we need to calculate $\cos\frac{3\pi}{4}$ and $\sin\frac{3\pi}{4}$.
$\cos\frac{3\pi}{4}$ and $\sin\frac{3\pi}{4}$ are in fact like $\cos\frac{\pi}{4}$ and $\sin\frac{\pi}{4}$. The only difference is that they are in quadrant II.
In quadrant II, $\sin\theta\gt0$ but $\cos\theta\lt0$.
Therefore, $$\sin\frac{3\pi}{4}=\sin\frac{\pi}{4}=\frac{\sqrt2}{2}$$ while $$\cos\frac{3\pi}{4}=-\cos\frac{\pi}{4}=-\frac{\sqrt2}{2}$$
2) Now apply back to $X$:
$$X=-\frac{\sqrt2}{2}\cos x+\frac{\sqrt2}{2}\sin x$$
$$X=\frac{-\sqrt2\cos x+\sqrt2\sin x}{2}$$
$$X=\frac{\sqrt2(\sin x-\cos x)}{2}$$
Therefore, $$\cos\Big(\frac{3\pi}{4}- x\Big)=\frac{\sqrt2(\sin x-\cos x)}{2}$$
