General term of Taylor Series of $\sin x$ centered at $\pi/4$ What is the general term for a Taylor series of $\sin(x)$ centered at $\pi/4$?
It should be $(-1)^{[??]} \times \sqrt{2}/2 \times \frac{(x-\pi/4)^n}{n!}$
What power is $(-1)$ supposed to be raised to?
 A: Write $f(x)=\sin x$. Then your Taylor series at $\pi/4$ is
$$
\sum_{n\geq 0}\frac{f^{(n)}(\pi/4)}{n!}\left(x-\frac{\pi}{4}\right)^n
$$
Compute the first derivatives at $\pi/4$ and see the pattern. This is periodic. You can guess the sign.
Edit: So the sequence $f^{(n)}(\pi/4)$ is, from $n=0$:
$$
+\frac{\sqrt{2}}{2}, +\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2},+\frac{\sqrt{2}}{2},+\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2},\ldots
$$
This can be given by:
$$
f^{(n)}(\pi/4)=(-1)^{\lfloor n/2\rfloor}\cdot\frac{\sqrt{2}}{2}.
$$
A: You can use the result
$$ f^{(n)}(x)= \sin\left( x+\frac{n\pi}{2}\right) \implies f^{(n)}\left(\frac{\pi}{4}\right)= \sin\left( \frac{\pi}{4}+\frac{n\pi}{2}\right) $$
to construct the Taylor series at $x=\frac{\pi}{4}$. It is not hard to prove the above result. Just use the identity
$$ f(x)=\frac{1}{2i}\left( e^{ix}-e^{-ix}\right). $$
A: You could also use the exponent $\frac{n(n+1)}2$. So $-1$ raised to the power $\frac{n(n+1)}2$ will produce the sequence desired. 
