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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!

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    $\begingroup$ Hint: Use the binomial theorem: $$(x+y)^n = \sum_{k=0}^n \binom n k x^k y^{n-k}$$ $\endgroup$
    – PrincessEev
    Jun 10, 2022 at 21:22

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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*}

Can you take it from here?

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    $\begingroup$ You changed $\sec^{k-n}(x)$ to $\sec^{n-k}(x)$. $\endgroup$
    – aschepler
    Jun 10, 2022 at 22:04
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    $\begingroup$ @aschepler Thanks for pointing that out! I am going to fix that typo. $\endgroup$
    – Átila Correia
    Jun 10, 2022 at 22:05

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