Finding zeroeth coefficient of a Chebyshev polynomial expansion Let $v_\theta = (\cos\theta,\sin\theta)$ be a unit vector in the plane. I have a kernel $p(\theta,\theta') = p(v_\theta\cdot v_{\theta'})$ that satisfies
$$\int_0^{2\pi} p(v_\theta\cdot v_{\theta'})\,d\theta' = 1\;\;\;(*)$$
for all $\theta\in [0,2\pi]$. I also have Chebyshev polynomials $T_0,T_1,\dots$ such that $T_k(cos\theta) = \cos(k\theta)$, normalized such that
$$\{T_0/\sqrt{\pi}\}\cup\{\sqrt{2/\pi}T_k\}_{k=1}^\infty$$
form an orthonormal basis of $L^2(-1,1)$ with weight $1/\sqrt{1-t^2}$.
Now I write the Chebyshev expansion of my kernel:
$$p(t) = \sum_{k=0}^\infty p_kT_k(t), \;\;\;\;\; t\in(-1,1)$$
and I want to show that $p_0 = \frac{1}{2\pi}$.
My progress so far: by orthonormality, we have
\begin{align*}
\int_0^{2\pi} &\frac{1}{\sqrt{\pi}}T_0(v_\theta\cdot v_{\theta'})\frac{\sqrt{2}}{\sqrt{\pi}}p(v_\theta\cdot v_{\theta'})\frac{1}{\sqrt{1 - (v_\theta\cdot v_{\theta'})^2)}}\,d\theta'\\
 &= \sum_{k=0}^\infty\int_0^{2\pi} T_0(v_\theta\cdot v_{\theta'})\frac{\sqrt{2}}{\sqrt{\pi}}p_kT_k(v_\theta\cdot v_{\theta'})\frac{1}{\sqrt{1-(v_\theta\cdot v_{\theta'})^2}}\,d\theta'\\
&= \sqrt{2}p_0.
\end{align*}
Also, noting that $T_0\equiv 1$, I know
\begin{align*}
\int_0^{2\pi} &\frac{1}{\sqrt{\pi}}T_0(v_\theta\cdot v_{\theta'})\frac{\sqrt{2}}{\sqrt{\pi}}p(v_\theta\cdot v_{\theta'})\frac{1}{\sqrt{1 - (v_\theta\cdot v_{\theta'})^2)}}\,d\theta'\\
&=\int_0^{2\pi} \frac{\sqrt{2}}{\pi} p(v_\theta\cdot v_{\theta'})\frac{1}{\sqrt{1 - (v_\theta\cdot v_{\theta'})^2}}\,d\theta'.
\end{align*}
Then it would suffice to show
$$\int_0^{2\pi}  p(v_\theta\cdot v_{\theta'})\frac{1}{\sqrt{1 - (v_\theta\cdot v_{\theta'})^2}}\,d\theta' = \frac{1}{2}.$$
This is where I'm stuck: I'm not sure how to use (*) in the expression above. Indeed since the expression above is constant in $v_\theta$ as we showed earlier, we are free to pick a particular value, say, $v_\theta = (0,1)$, to make this
$$\int_0^{2\pi} \frac{p(\cos\theta')}{\sin\theta'}\,d\theta',$$
but still I am not sure what to do with this.

As it turns out, using orthonormality was a a red herring, and the solution is actually quite simple. Choosing $v_\theta = (1,0)$, we compute
\begin{align*}
1 &= \int_0^{2\pi} p(v_\theta\cdot v_{\theta'})\,d\theta'\\
&= \sum_{k=0}^\infty \int_0^{2\pi} p_kT_k(v_\theta\cdot v_{\theta'})\,d\theta'\\
&= \sum_{k=0}^\infty \int_0^{2\pi} p_kT_k(\cos\theta')\,d\theta'\\
&= \sum_{k=0}^\infty \int_0^{2\pi} p_k\cos(k\theta')\,d\theta'\\
&= 2\pi p_0 + \sum_{k=1}^\infty \underbrace{\int_0^{2\pi} p_k\cos(k\theta')\,d\theta}_{= 0},
\end{align*}
and so $p_0 = 1/2\pi$.
