Compound angle formula confusion I'm working through my book, on the section about compound angle formulae. I've been made aware of the identity $\sin(A + B) \equiv \sin A\cos B + \cos A\sin B$. Next task was to replace B with -B to show $\sin(A - B) \equiv \sin A\cos B - \cos A \sin B$ which was fairly easy. I'm struggling with the following though:
"In the identity $\sin(A - B) \equiv \sin A\cos B - \cos A\sin B$, replace A by $(\frac{1}{2}\pi - A)$ to show that $\cos(A + B) \equiv \cos A\cos B - \sin A\sin B$."
I've got $\sin((\frac{\pi}{2} - A) - B) \equiv \cos A\cos B - \sin A\sin B$ by replacing $\sin(\frac{\pi}{2} - A)$ with $\cos A$ and $\cos(\frac{\pi}{2} - A)$ with $\sin A$ on the RHS of the identity. It's just the LHS I'm stuck with and don't know how to manipulate to make it $\cos(A + B)$.
P.S. I know I'm asking assistance on extremely trivial stuff, but I've been staring at this for a while and don't have a tutor so hope someone will help!

Update: Since $\sin (\frac{\pi }{2}-x)\equiv \cos x$, you have
$$\sin \left(\frac{\pi }{2}-A-B\right)\equiv \sin \left(\frac{\pi }{2}-(A+B)\right)\equiv \cos (A+B).$$

Replacing $A$ by $\frac{\pi }{2}-A$ in
$$\sin (A-B)\equiv \sin A\cos B-\cos A\sin B,$$
gives
$$\sin \left(\frac{\pi }{2}-A-B\right)\equiv \sin \left( \frac{\pi }{2}-A\right)
\cos B-\cos \left( \frac{\pi }{2}-A\right) \sin B.$$
Since $\sin (\frac{\pi }{2}-x)\equiv \cos x$ and $\cos (\frac{\pi }{2}-x)\equiv \sin x$, you have
$$\sin \left(\frac{\pi }{2}-A-B\right)\equiv \sin \left(\frac{\pi }{2}-(A+B)\right)\equiv \cos (A+B),$$
$$\sin \left( \frac{\pi }{2}-A\right) \equiv \cos A,$$
and
$$\cos \left( \frac{\pi }{2}-A\right) \equiv \sin A.$$
Hence
$$\cos (A+B)\equiv \cos A\cos B-\sin A\sin B.$$
