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.

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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 $\cos(x-a)$.

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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 $\sin{(x+a)}$.

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