Trigonometric Functions The question is to show that $A\sin(x + B)$ can be written as $a\sin x + b\cos x$ for suitable a and b.
Also, could somebody please show me how $f(x)=A\sin(x+B)$ satisfies $f + f ''=0$?

If
$$
f(x) = A\sin(x+B)
$$
then
$$
f'(x) = A\cos(x+B)\cdot\frac{d}{dx}(x+B) = A\cos(x+B)\cdot1,
$$
and
$$
f''(x) = -A\sin(x+B)\cdot\frac{d}{dx}(x+B) = -A\sin(x+B).
$$
So
$$
f''(x)+f(x) = -A\sin(x+B)+A\sin(x+B) = 0.
$$
For the initial question, the standard trigonometric identity
$$
\sin(x+B) = \sin x\cos B+ \cos x\sin B
$$
is most of what you need to know.  Then you have
$$
A\sin(x+B) = A\Big( \sin x\cos B+ \cos x\sin B \Big)
$$
$$
= \Big(A\cos B\Big) \sin x + \Big( A\sin B\Big) \cos x
$$
$$
=a\sin x+b\cos x.
$$
