How does $\frac{m\sinh\phi}{\sinh\tau}\,\Re[e^{ims\cosh(\phi-\tau)}]$ become $\frac{m\sinh(\phi+\tau)}{\sinh\tau}\,\Re[e^{ims\cosh(\phi)}]$?

I'm stuck on the following steps:

$$\frac{m\sinh\phi}{\sinh\tau}\,\Re[e^{ims\cosh(\phi-\tau)}] = \frac{m\sinh(\phi+\tau)}{\sinh\tau}\,\Re[e^{ims\cosh(\phi)}]$$ I'm not quite sure how we can get the right-hand side from the left. I know $$\cosh(a-b) = \cosh a\cosh b-\sinh a\sinh b$$, but how can we take out the '$$-\tau$$' from the real part of this exponent?

• Maybe a change of variable and setting it back to the original variable, like $\phi-\tau\to\phi$? Oct 30, 2022 at 11:22
• Maybe I am wrong, but the equality does not look right to me. Maybe this is part of an integral, so that it does not matter how you change the variable Oct 30, 2022 at 11:27
• @Lorenzo Pompili Thanks, it is indeed part of integral: on both sides we are integrating with respect to $\phi$
– IGY
Oct 30, 2022 at 20:18
• This seems to be one of those questions that becomes much simpler when you consider what you actually needed to show rather than what you assumed you needed to show. If you're still confused, it could be helpful to show more of the context. At the very least, if there is an equation in the original source you are stuck on, show the equation exactly as written, complete with integral signs and boundaries. Oct 30, 2022 at 21:11
• If this is part of an integral, can you show us? This equality seems to be false. I checked it on Desmos for multiple values of $m$,$s$, and $\phi$. And is $\tau = 2\pi$ in this case? Oct 31, 2022 at 19:37

Since “We are actually integrating with respect to $$\phi$$”, let’s use $$\Phi=\phi-\tau\implies \phi=\Phi+\tau,d\Phi=d\phi$$:
$$\int_a^b\frac{m\sinh\phi}{\sinh\tau}\Re[e^{ims\cosh(\phi-\tau)}] d\phi=\int_{a-\tau}^{b-\tau} \frac{m\sinh(\Phi+\tau)}{\sinh\tau}\Re[e^{ims\cosh(\Phi)}] d\Phi$$
since it is a definite integral, we just transformed the original one into an equivalent integral meaning $$\Phi$$ is a dummy variable, so:
$$\int_a^b\frac{m\sinh\phi}{\sinh\tau}\Re[e^{ims\cosh(\phi-\tau)}] d\phi=\int_{a-\tau}^{b-\tau} \frac{m\sinh(\phi+\tau)}{\sinh\tau}\Re[e^{ims\cosh(\phi)}] d\phi$$
As for integrating the rest, $$\Re(e^{iy})=\cos(y)$$, so $$\Re(e^{i ms\cosh(\phi)})=\cos(ms\cosh(\phi));m,s,\phi\in\Bbb R$$