Solve the parabolic PDE
$$
\pa_tu=\tfrac12\pa_\theta^2u,\quad
\forall\theta\neq\pm\pi:\quad u(0,x)=1,\quad
\forall t > 0:\quad u(t,\pm\pi)=0~.
$$
by Fourier analysis methods.
The normalized eigen functions of $\tfrac12\pa_\theta^2$ on $(-\pi,\pi)$ satisfying the boundary condition $f(\pm\pi)=0$ are
$$
f_n(\theta)\colon=\frac1{\sqrt{\pi}}\cos((n+\tfrac12)\theta),
g_n(\theta)\colon=\frac1{\sqrt{\pi}}\sin(n\theta)
\quad
n=0,1,2,\ldots~.
$$
Since the constant function is orthogonal to all $g_n$ and the eigen value for $f_n$ is $\l_n\colon=-\tfrac12(n+\tfrac12)^2$, we conclude that
$$
u(t,\theta)=\sum_{n=0}^\infty c_nf_n(\theta)e^{\l_nt}
\quad\mbox{where}\quad
c_n=\frac1{\sqrt{\pi}}\int_{-\pi}^{\pi}f_n(\theta)\,d\theta
=\frac{2(-1)^n}{\sqrt{\pi}(n+\frac12)}~.
$$
Hence
$$
u(t,\theta)
=\frac2{\pi}\sum_{n=0}^\infty(-1)^n\frac{\cos((n+\tfrac12)\theta)}{n+\tfrac12}
e^{-\frac12(n+\frac12)^2t}~.
$$