Using Taylor Series for $\sin x$ and $\cos x$ to derive $\cos{(x-a)}$ and $\sin{(x-a)}$ My professor had this problem on our last problem set but got rid of it as it was "more involved" than he thought but I am still curious as to how it would be done (Its good that he ditched it because I had little to no idea)
Use Taylor series for $\sin x$ and $\cos x$ at $x=0$ and $x=a$ to estimate 
$$
\sin{(x-a)} ~=~ \cos a\sin x - \sin a\cos x 
$$
and 
$$
\cos{(x-a)} ~=~ \cos a\cos x + \sin a\sin x 
$$
A series for $\sin x$ and $\cos x$ isn't too tough, not quite sure what he meant by $x=0$ and $x=a$, does he mean $0$ for $\sin x$ and $a$ for $\cos x$? I don't need the problem answered completely, I just kind of want to see what I'd have had to do if it were actually due.
 A: This doesn't appear to be terribly involved.  For simplicity, let $S=\sin(a)$ and $C=\cos(a)$.
Then the Taylor series for $\sin(x-a)$ is just the linear combination of Taylor series for $\sin(x)$ and $\cos(x)$ indicated by your difference angle formula:
$$\sin(x-a) = C\; \sin(x) - S\; \cos(x)$$
That is:
$$\sin(x-a) = C \sum_0^\infty (-1)^k \frac{x^{2k+1}}{(2k+1)!} - S \sum_0^\infty (-1)^k 
\frac{x^{2k}}{(2k)!}$$
The odd powers of $x$ from the first series and even powers from the second are naturally to be alternated.  A similar expression can be given for $\cos(x-a)$.
A: I will show how to deduce that $\cos{(x+a)} = \cos a\cos x - \sin a\sin x$.
The Taylor expansion of $\cos x$ in $x=a$ is
$$
  \cos x
=
  \sum_{n=0}^\infty
  {
    \frac{\cos^{(n)}(a)}{n!}(x-a)^n
  }
$$
Since $\cos^{(2n)}(a) = (-1)^n \cos a$ and $\cos^{(2n+1)}(a)=(-1)^{n+1}\sin a$ you can split the series into two, considering pair $n$ on one and odd $n$ on the other:
$$
  \cos x
=
  \cos a
  \sum_{n=0}^\infty
  {
    \frac{(-1)^n}{(2n)!}(x-a)^{2n}
  }
-
  \sin a
  \sum_{n=0}^\infty
  {
    \frac{(-1)^n}{(2n+1)!}(x-a)^{2n+1}
  }
$$
The series multiplied by $\cos a$ is indeed that of $\cos x$ centered in $x=0$, and the other is that of $\sin x$ centered in $x=0$. Therefore
$$
\cos{(x+a)} = \cos a\cos x - \sin a\sin x
$$
In order to get the formula for $\cos{(x-a)}$ it suffices to consider $-a$ instead.
An analogous calculation applies to $\sin{(x+a)}$.
