I searched quite a bit online but could only find the MACLAURIN SERIES of $\sin x$ and $\cos x$:
$$\sin x = x – \frac{x^3}{3!} + \frac{x^5}{5!} – \frac{x^7}{7!}+\cdots$$
$$\cos x = 1 – \frac{x^2}{2!} + \frac{x^4}{4!} – \frac{x^6}{6!}+\cdots$$
Can anyone explain how we can express the TAYLOR SERIES of $\sin x$ and $\cos x$, and also show me the derivation?
Thanks in advance!
Best Answer
Hint/Partial Solution:
Recall the definition of the Taylor Series formula at a point $a$: $$\sum_{k=0}^\infty \frac{f^{(k)}(a)}{k!}(x-a)^k$$
(Note that $f^{(k)}(a)$ is a short-hand for the $k$th derivative of $f(x)$ evaluated at $a$)
Letting $f(x)=\sin x$, we note that the numerator of the fraction is the following cycle: $$\sin(c), \cos(c), -\sin(c), -\cos(c), \cdots$$
The rest is just going to be expanding the series. It won't condense well if $a$ is not a rational multiple of $\pi$ though. The same logic follows for the cosine expansion. As an example of this, let $x=1$. Then we get the following series: $$\sum_{k=0}^\infty \frac{\sin^{(k)}(1)}{k!}(x-1)^k$$ Expanded, this becomes
$$\sin(1)+(x-1) \cos(1)-\frac{1}{2}(x-1)^2 \sin(1)-\frac 16 (x-1)^3 \cos(1)+\frac{1}{24} (x-1)^4 \sin(1)+\cdots$$
After seeing this, you might think to yourself that this could be broken up into a sum without any derivative symbols. That hypothetical you is right! We can rewrite this as
$$\sum_{k=0}^{\infty} \left(\frac{\sin(1)(-1)^k}{(2k)!}(x-1)^{2k}+\frac{\cos(1)(-1)^{k+1}}{(2k+1)!}(x-1)^{2k+1}\right)$$ A similar formula emerges for the cosine expansion, but I leave that up to the OP!