Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
add comment

2 Answers

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 $\cos(x-a)$.

share|improve this answer
add comment

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

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.