# $f(x)=\sum_{k=0}^{n}\binom{n}{k}\sin^{k}x\sec^{k-n}{x}.$

Let $$n$$ be a non-negative integer and let $$f(x)=\sum_{k=0}^{n}\dbinom{n}{k}\sin^{k}x\sec^{k-n}{x}.$$ Prove that $$f(x)$$ is periodic and find its amplitude.

I don't really know how to start, and all I did was write $$\sin^{k}x\sec^{k-n}{x}$$ as $$\sin^k x\cos^{n-k}x$$. I'm not sure how to proceed from here, and any guidance would be appreciated!

• Hint: Use the binomial theorem: $$(x+y)^n = \sum_{k=0}^n \binom n k x^k y^{n-k}$$ Jun 10 at 21:22

According to @EeveeTrainer's comment, you can express the proposed sum as \begin{align*} f(x) & = \sum_{k=0}^{n}{n \choose k}\sin^{k}(x)\sec^{k-n}(x)\\ & = \sum_{k=0}^{n}{n \choose k}\sin^{k}(x)\cos^{n-k}(x)\\\\ & = (\sin(x) + \cos(x))^{n}\\\\ & = 2^{n/2}\left(\frac{1}{\sqrt{2}}\sin(x) + \frac{1}{\sqrt{2}}\cos(x)\right)^{n}\\\\ & = 2^{n/2}\sin^{n}\left(x + \frac{\pi}{4}\right) \end{align*}
• You changed $\sec^{k-n}(x)$ to $\sec^{n-k}(x)$. Jun 10 at 22:04