I know that
$$ \frac{1}{2\pi}\int_0^{2\pi}e^{i\,z\,\cos\theta}d\theta=J_0(z) $$
where $J_n(z)$ denotes the Bessel function of the first kind of integral order. My question is - how do I show that
$$ \frac{1}{2\pi}\int_0^{2\pi}e^{i\,z(a\cos\theta+b\sin\theta)}d\theta=J_0(z\sqrt{a^2+b^2}) $$
It is probably something trivial, but I can't seem to figure it out. Any tips or hints will be appreciated.
Since: $$ a\cos\theta + b\sin\theta = \sqrt{a^2+b^2}\cos\left(\theta-\arctan\frac{b}{a}\right)$$ the claim follows.