I'm reading a paper where this equality is claimed:
$$\cos(a + b \cos(t) ) = \cos(a) J_0(b),$$
where $a$, $b$ are constants and $J_0$ is the zeroth-order Bessel function of the first kind.
How can I show that this equality is true?
(Maybe this is related to the Fourier-Bessel series? http://mathworld.wolfram.com/Fourier-BesselSeries.html)
As you wrote it, that's obviously false. Instead, that should be an average value $$\frac1{2\pi}\int_0^{2\pi}\cos(a+b\cos t)\,dt=\cos(a)J_0(b)$$ How do we prove it? The Bessel function is defined by a differential equation, so let's define that integral as $F(b)$ and see what happens when we differentiate $F$: \begin{align*}F(b) &= \frac1{2\pi}\int_0^{2\pi}\cos(a+b\cos t)\,dt\\ F'(b) &= \frac1{2\pi}\int_0^{2\pi}-\sin(a+b\cos t)\cdot \cos t\,dt\\ F''(b) &= \frac1{2\pi}\int_0^{2\pi}-\cos(a+b\cos t)\cdot \cos^2 t\,dt\end{align*} We want to show that $b^2F(b)+bF'(b)+b^2F''(b)=0$ - the Bessel differential equation of order zero. If we just put those into the same integral now, the middle term will have a $\sin(a+b\cos t)$ where the outer terms have $\cos(a+b\cos t)$ - so let's integrate by parts in the middle term to put it into a more convenient form. \begin{align*}bF'(b) &= \frac{b}{2\pi}\int_0^{2\pi}-\sin(a+b\cos t)\cdot \cos t\,dt\\ &= \left[-\frac{b}{2\pi}\sin(a+b\cos t)\cdot \sin t\right]_{t=0}^{t=2\pi}-\frac{b}{2\pi}\int_0^{2\pi}b\sin t\cdot \cos(a+b\cos t)\cdot\sin t\,dt\\ bF'(b) &= -\frac{b^2}{2\pi}\int_0^{2\pi}\sin^2 t\cos(a+b\cos t)\,dt\end{align*} Then $$F(b)+F''(b) = \frac1{2\pi}\int_0^{2\pi}(1-\cos^2 t)\cos(a+b\cos t)\,dt=\frac1{2\pi}\int_0^{2\pi}\sin^2 t\cos(a+b\cos t)\,dt$$ $$b^2F(b)+bF'(b)+b^2F''(b) = 0$$ That's the Bessel differential equation of order zero, like we wanted. To pin down exactly which solution we get (a linear combination of $J_0$ and $Y_0$), all we need now are the initial conditions. We have $F(0)=\frac1{2\pi}\int_0^{2\pi}\cos(a)\,dt=\cos a$ and $F'(0)=\frac1{2\pi}\int_0^{2\pi}\sin(a)\cos t\,dt=0$, so from $J_0(0)=1$ and $J_0'(0)=0$, we get $F(b)=\cos a\cdot J_0(b)$. Done.
