Q:

Why is $\frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}\cos\left[ a\xi \right]\hat{f}(\xi)d \xi = f(a)$?

Background: We are looking at the wave equation on $\mathbb{R}^n$ via the Fourier transform. If $u(x,t)$ solves $\Delta u = u_{tt}$ in $\mathbb{R}^n$, with $u(x,t) = f(x)$ at $t=0$ and $u_t(x,t) = g(x)$ at $t=0$, then I have shown that the Fourier transform of $u$ is 
$$\hat{u}(\xi,t) = \cos(|\xi|t)\hat{f}(\xi)+\frac{\sin(|\xi|t)}{|\xi|}\hat{g}(\xi).\tag{1}$$
Now my task is to show that in one dimension, this reduces to the normal solution
$$u(x,t)=\frac12 \left[ f(x+t)+f(x-t) \right] + \frac12 \int_{x-t}^{x+t}g(\rho)d\rho. \tag{2}$$
My solution so far: I am allowed to take for granted that the inverse Fourier transform of $\hat{u}$ is $u$. Now the inverse Fourier transform of $\hat{u}$ is 
$$\frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}e^{ix\xi}\left[ \cos(|\xi|t)\hat{f}(\xi)+ \frac{\sin(|\xi|t)}{|\xi|}\hat{g}(\xi) \right]d\xi.$$
Let's focus on the left term first. By evenness of cosine, I can ignore the absolute value sign in the argument.
$$\frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}\cos(\xi x)\cos(\xi t)\hat{f}(\xi)d\xi + \frac{i}{\sqrt{2\pi}}\int_{\mathbb{R}}\sin(\xi x)\cos(\xi t)\hat{f}(\xi)d\xi\tag{3}$$
Of course there should end up being no imaginary part...I have to show the right hand term is zero (or it cancels out in the end). In the left side I can use the cosine identity
$$\frac{1}{2\sqrt{2\pi}}\int_{\mathbb{R}}\cos\left[ (x+t)\xi \right]\hat{f}(\xi) + \cos\left[ (x-t)\xi \right]\hat{f}(\xi),\tag{4}$$
and this is looking strongly like the first part of (2). I could talk about the sine term in (1), but let me just ask for now:

How can I show that (4) is equal to the first part of (2)?
How can I show that the imaginary part of (3) is zero?

I would like to write (4) as $\int_{\mathbb{R}}\Re\{ e^{i(x+t)\xi}\}\dotsb$, but I'm not seeing how this would dispense with the problem.

A:

$$ \frac1{\sqrt{2\pi}} \int_{\mathbb R} \cos(a\xi) \hat f(\xi) d\xi = \frac12(f(a)+f(-a)) $$
Instead of converting complex exponentials into trig functions, go the other way and convert trig functions to complex exponentials.  You will find it so much easier to work with.

