How do I calculate the phase shift between sine and cosine? I know that $\sin(\alpha + x)=\cos(\alpha)$. How do I find $x$ ?
I'd start by using the angle sum identity for sine:
$\cos(\alpha)*\sin(x)+\sin(\alpha)*\cos(x)=\cos(\alpha)$
I had some ideas about what to do next but they didn't get me anywhere.
 A: I'll try to expand a bit on joriki's answer. Since we want the identity 
$$\cos(\alpha)\sin(x)+\sin(\alpha)\cos(x)=\cos(\alpha)$$
to be true for all $\alpha$, it has to be true in particular for $\alpha=0$ and $\alpha=\frac{\pi}{2}$. Thus, the $x$ we are looking for must satisfy both
$$\cos(0)\sin(x)+\sin(0)\cos(x)=\cos(0)$$
$$1\cdot\sin(x)+0\cdot\cos(x)=1$$
$$\sin(x)=1$$
and
$$\cos(\tfrac{\pi}{2})\sin(x)+\sin(\tfrac{\pi}{2})\cos(x)=\cos(\tfrac{\pi}{2})$$
$$0\cdot\sin(x)+1\cdot\cos(x)=0$$
$$\cos(x)=0$$
Which value of $x$ satisfies both $\sin(x)=1$ and $\cos(x)=0$?
A: Good start. Now you want the left-hand side to be $\cos\alpha$, so you want the coefficient of $\cos\alpha$, which is $\sin x$, to be $1$ and the coefficient of $\sin\alpha$, which is $\cos x$, to be $0$. Which $x$ yields those values?
A: Somewhat similar to Zev's method: if you say that $\sin(x+\alpha)=\cos(\alpha)$, then it is also true that $\sin(x-\alpha)=\cos(-\alpha)=\cos(\alpha)$. If you apply the usual sum and difference formulae for the trigonometric functions, you should obtain a system of two equations in the two unknowns $\sin(x)$ and $\cos(x)$. Solving those equations will yield $\sin(x)=1$ and $\cos(x)=0$
As already mentioned, $x=\pi/2$ is one such value of $x$; in general, due to periodicity, any number of the form $\pi/2+2k\pi$, $k$ an integer, is a possible value of $x$.
