I am almost a beginner in the topics of Fourier transform. So, I am asking this question here.
Let $n=3$ and let $\mu_t$ denote surface measure on the sphere $|x|=t$. Then how do we show that $$ \frac{\text{sin} 2\pi t|\xi|}{2\pi|\xi|}=\frac{1}{4\pi t}\hat{\mu_{t}}(\xi) $$ where $\hat{\mu_{t}}$ is the Fourier transform of $\mu_t$.
On
$\def\Rbf{\mathbf{R}}$ $\def\inta{\int_0^{\pi}}$ $\def\intb{\int_0^{2\pi}}$
This is a very nice exercise for multivariable calculus.
Denote the surface measure of the unit sphere $S^{d-1}\subset\Rbf^d$ as $d\sigma$. Then its Fourier transform is defined as $\def\ft#1{\widehat{#1}}$ $$ \ft{d\sigma}(\xi) = \int_{S^{d-1}}e^{-2\pi ix\cdot \xi}\; d\sigma(x). $$ The detailed calculation appears in Stein and Weiss's Introduction to Fourier Analysis on Euclidean Spaces (Ch5.3 page 154).
When $d=3$, for $\xi = (0,0,\xi_3)$ with $\xi_3>0$, one has $$ \ft{d\sigma}(\xi) = \inta\intb e^{-2\pi i\cos\theta_1\xi_3}\sin\theta_1\;d\theta_2d\theta_1 $$ Thus $$ \ft{d\sigma}(\xi) = 2\pi \inta e^{-2\pi i\cos\theta_1\xi_3}\sin\theta_1\;d\theta_1\;. $$ By change of variable $u = -\cos\theta_1$ and Euler's formula, one has $$ \ft{d\sigma}(\xi) = 2\pi \int_{-1}^1e^{2\pi iu\xi_3}\;du = \frac{2\pi}{2\pi i\xi_3} e^{2\pi iu\xi_3}\Big|_{-1}^1 = \frac{2\sin(2\pi\xi_3)}{\xi_3}\tag{1} $$
On the other hand, observe that $\ft{d\sigma}$ is radial: $$ \ft{d\sigma}(A\xi) = \int_{S^{d-1}}e^{-2\pi ix\cdot(A\xi)}\;d\sigma(x) = \int_{S^{d-1}}e^{-2\pi i(A^{T}x)\cdot(\xi)}\;d\sigma(x) = \ft{d\sigma}(\xi) $$ for any orthogonal matrix $A$. Consequently, (1) implies that $$ \ft{d\sigma}(\xi) = \frac{2\sin(2\pi|\xi|)}{|\xi|},\quad \xi \ne 0\;. $$
The case when the sphere is not of radius $1$ can be done by change of variable.
For other references, see also this MO question: Fourier transform of the unit sphere.
See also these two sets of notes:
It is easy if we use Kirchhoff's formula for the solution to the following wave equation: \begin{equation} \begin{cases} (\partial^2_t-\Delta)u=0, & x\in \mathbb{R}^3,\ t\in \mathbb{R}\\ u(0, x)=0 \\ u_t(0, x)=f \end{cases} \end{equation} Kirchhoff's formula says that \begin{equation} u(t,x)=\frac{1}{4\pi t} \int_{\lvert y-x\rvert=t} f(y)\, dS(y) =\frac{1}{4\pi t}d\sigma_t\ast f, \end{equation} so taking the spatial Fourier transform one gets \begin{equation} \frac{1}{4\pi t} \widehat{d\sigma}_t \hat{f}= \hat{u}(t, \xi). \end{equation} On the other hand, for fixed $\xi$ the function $\hat{u}(\cdot, \xi)$ solves the following Cauchy problem, which is a harmonic oscillator with respect to the time variable $t$: \begin{equation} \begin{cases} (\partial^2_t +4\pi^2\lvert \xi\rvert^2)\hat{u}(t,\xi)= 0,& t\in \mathbb{R} \\ \hat{u}(0, \xi)=0 \\ \hat{u}_t(0, \xi)=\hat{f}(\xi). \end{cases} \end{equation} Therefore \begin{equation} \hat{u}(t, \xi)=\frac{\sin 2\pi t\lvert \xi\rvert}{2\pi \lvert \xi\rvert}\hat{f}(\xi). \end{equation} Equating the two expressions for $\hat{u}$ yields the desired formula.
One might think that this answer is a cheat, since one usually derives Kirchhoff's formula by first computing that Fourier transform. However, that's not the only way to derive Kirchhoff's formula. Another popular approach uses spherical means and is completely devoid of any Fourier transform methods. So the computation presented in this answer is not circular.
I'll leave a reference for a more direct computation just in case: you can find it in exercises to the chapter "Topics in Fourier Analysis" of the book "Real Analysis" by G. Folland (both editions).