Taylor series of $\sin(x+1)+\cos(x-1)$ I want to find the Taylor series of $f(x)=\sin(x+1)+\cos(x-1)$ for some $x_0 \in \mathbb{R}$.
We know the Taylor of $\sin(x)$ for $x_0=0$, so can I simply say that $\sin(x+1)=\sum_{n=0}^\infty \frac{(-1)^n(x+1)^{2n+1}}{(2n+1)!}~$ at $x_0=0$?
What should I do for $f$ for an unknown $x_0$?
Thanks!
 A: Hint
$$f(x_0+h) = \sin(x_0+h+1)+\cos(x_0+h-1).$$
Then use trigonometric identities like
$$\sin(x_0+h+1) = \cos (x_0+1) \sin h + \cos h \sin (x_0+1)$$
to come back to Taylor series expansion of $\sin h$ and $\cos h$ around zero that you know.
A: The successive derivatives at $x_0$ are
$$\ \ \ \sin(x_0+1)+\cos(x_0-1),
\\\ \ \ \cos(x_0+1)-\sin(x_0-1),
\\-\sin(x_0+1)-\cos(x_0-1),
\\-\cos(x_0+1)+\sin(x_0-1),$$
and so on periodically.
Hence you can express the Taylor series as four sums of powers of $(x-x_0)$,
grouping the powers with the same modulo $4$.
A: Summarizing all already said in comments, around $x=x_0$, you should have for
$$f(x)=\sin(x+1)+\cos(x-1)$$
$$f(x)=(\sin(1)+\cos(1)) \sum_{n=0}^\infty \frac{\sin \left(x_0+n\frac{\pi  }{2}\right)+\cos \left(x_0+n\frac{\pi  }{2}\right)}{n!}\,(x-x_0)^n$$
You could even make it more general for
$$g(x)=a \sin (x+b)+c \cos (x+d)$$ and obtain
$$g(x)=\sum_{n=0}^\infty \frac{a \sin \left(x_0+b+\frac{\pi  n}{2}\right)+c \cos \left(x_0+d+\frac{\pi 
   n}{2}\right)}{n!}\,(x-x_0)^n$$
A: Hint: By basic trig identities, it can be easily showed that:
$$\sin\left(x+1\right)+\cos\left(x-1\right)=\left(\sin1+\cos1\right)\left(\cos x+\sin x\right)$$
