Integral involving trigonometric functions I am learning a theory involving the Bessel functions and I encountered an integral equality that I could not prove.
Let $z$ be a fixed complex number and $n$ an odd number.
The author wrote that, $$\frac{1}{\pi} \int_0^\pi \sin n\theta \sin (z\sin \theta)\,d\theta=\frac{2}{\pi} \int_0^{\pi/2} \sin n\theta \sin (z\sin \theta)\,d\theta .$$
Therefore I wanted to prove that $$\int_0^{\pi/2} \sin n\theta \sin (z\sin \theta)\,d\theta=\int_{\pi/2}^{\pi} \sin n\theta \sin (z\sin \theta)\,d\theta.$$
But I failed doing so.
My attempt:
By change of variables, $\theta \mapsto \theta + \pi/2,$ we have
$$\int_{\pi/2}^{\pi} \sin n\theta \sin (z\sin \theta)\,d\theta=\int_{0}^{\pi/2} \sin (n\theta+\frac{n\pi}{2}) \sin (z\sin (\theta+\frac{\pi}{2}))\,d\theta= \int_{0}^{\pi/2} \cos n\theta \sin\frac{n\pi}{2} \sin (z\cos \theta)\,d\theta.$$
Note that $\sin\frac{n\pi}{2}$ can be either $1$ or $-1$ and it depends on an integer $m$ which depends on $n$, where $m\equiv n$ (mod $4$).
However, the result is still much different from the desired equality that I wanted. Did I miss something simple or what?
Any help will be appreciated.
 A: Well, I managed a way to prove the desired integral equality. The proof is all about elementary but tedious trigonometric identities.
Let $n$ be an odd integer.
A direct calculation with appealing to elementary trigonometric identities gives $$\sin (n\theta +\frac{n\pi}{2})=-\sin (n\theta - \frac{n\pi}{2}).$$
Therefore by the equality above and the change of variables, $\theta \mapsto \theta+\frac{\pi}{2},$ we have
\begin{align*}
 & \int_0^{\pi/2 } \sin (n\theta +\frac{n\pi}{2}) \sin(z\sin(\theta+\frac{\pi}{2}))d\theta\\  & =- \int_0^{\pi/2 } \sin (n\theta -\frac{n\pi}{2}) \sin(z\sin(\theta+\frac{\pi}{2}))d\theta\\  & =- \int_{-\pi/2 }^0 \sin n\theta\sin(z\sin(\theta+\pi))d\theta\\ & = \int_{-\pi/2 }^0 \sin n\theta\sin(z\sin\theta)d\theta.
\end{align*}
Applying change of variables $\theta \mapsto -\theta$, then we derive $$ \int_{-\pi/2 }^0 \sin n\theta\sin(z\sin\theta)d\theta= \int_0^{\pi/2 } \sin n\theta\sin(z\sin\theta)d\theta.$$
Hence, we have proved $$\int_0^{\pi/2 } \sin (n\theta +\frac{n\pi}{2}) \sin(z\sin(\theta+\frac{\pi}{2}))d\theta= \int_0^{\pi/2 } \sin n\theta\sin(z\sin\theta)d\theta,$$
where is the desired integral equality.
