I have this question in my study material (an introduction to theory of real functions):
If $f(x)=\tan x$ then show that $$\sin\left(\frac{n\pi}{2}\right)=f^n(0)-{}^nC_2f^{n-1}(0)+{}^nC_4f^{n-2}(0)-\ldots$$
Let me clarify the superscript shows the derivative number and $C$ is the choose function.
I just learned about Taylor's theorem and have expanded various functions.
For this question, assuming $n\in\mathbb Z^+$, I believe the given series is finite. It stops when the binomial coefficients become zero i.e., when we have ${}^nC_k$ with $k>n$.
I am trying a proof by induction.
Base case: It's true for $n=1$. $$\sin\left(\frac{\pi}{2}\right)=\sec^20$$
Inductive step (attempt):
Suppose it's true for $n=k$ i.e.,
$$\sin\left(\frac{k\pi}{2}\right)=f^k(0)-{}^kC_2f^{k-1}(0)+{}^kC_4f^{k-2}(0)-\ldots$$
We have: $$\sin\left(\frac{(k+1)\pi}2\right)=\tan \left(\frac{k\pi}{2}\right)\cdot \sin\left(\frac{k\pi}{2}\right)$$
Can anyone give me a hint on how to proceed with this inductive step or is there any alternative approach which doesn't require to assume $n\in\mathbb Z^+$?
What could be the possible relation between $\tan x$ and $\sin{\frac{x\pi}{2}}$ which leads to such an expansion? I recognize that $f^{n}(0)$'s are related to coefficients in Maclaurin expansion of $\tan x$.