General term of Taylor Series of $\sin(2z-1)$ centered at $\dfrac{\pi}{4}$.

Question

I started with $\sin(z)$ expansion but couldn't move to powers of $z-\frac{\pi}{4}$ from $(2z-1)^{2n+1} ..$ is there another way to reach it?

Answer 1

To make life simpler, as Lord Shark the Unknown commented, start with $2z=t+\frac \pi 2$.

So $$\sin(2z-1)=\cos (1-t)=\cos (1) \cos (t)-\sin (1) \sin (t)$$ Now, use the series expansions of $\sin(t)$ and $\cos(t)$ around $t=0$ and, when finished, make $t=2z-\frac \pi 2$