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Let $$f(x) = 1 + \left(\frac12\cdot x\right)^2+\left(\frac12\cdot\frac34 \cdot x^2\right)^2+\left(\frac12\cdot\frac34\cdot\frac56\cdot x^3\right)^2+\cdots$$ Prove that $$\sin x\;f(\sin x)\;f^\prime(\cos x)+\cos x\;f(\cos x)\;f^\prime(\sin x)=\frac{2}{\pi\sin x\cos x}$$

No hints are available. Any help would be appreciated. Thanks.

I have been trying to solve this problem by finding explicit form of $f(x)$. I have noticed that the series is similar to the power series of $1/\sqrt{1-x}$, thus I have spent a lot of time connecting it to $f(x)$. Apparently, it did not work.

I also have tried to transform the claimed equality into well-known identity of trigonometric functions, such as $\sin^2x+\cos^2x =1$.

(Edited) Thanks Parcly Taxel for editing the article. I did not know how to ask properly.

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    $\begingroup$ What's the source of this problem? Is the expectation that the problem requires some sophisticated theorems, or that it becomes trivial by invoking "one cool trick"? $\endgroup$ – Blue Oct 28 '18 at 11:43
  • $\begingroup$ It is fairly easy to see that $LHS=f(\sin x)f(\cos x) \frac{d}{dx} \log \frac{f(\sin x)}{f(\cos x)}$. Just not sure what's next. $\endgroup$ – AdditIdent Oct 28 '18 at 12:09
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    $\begingroup$ As requested by @Blue you should mention the source of the problem. I also saw a comment giving link to the source of the problem which was later deleted. Anyway the link is now a part of my answer and should help add context to your problem. $\endgroup$ – Paramanand Singh Oct 28 '18 at 14:21
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The problem is taken from this website. This is Legendre's identity for elliptic integrals in disguise.


We have $$f(x) = \frac{2K(x)}{\pi}\tag{1}$$ where $$K(x)=\int_{0}^{\pi/2}\frac{dt}{\sqrt{1-x^2\sin^2t}},E(x)=\int_{0}^{\pi/2}\sqrt{1-x^2\sin^2t}\,dt\tag {2}$$ Now Legendre's identity states that $$K(x) E(\sqrt {1-x^2})+K(\sqrt{1-x^2})E(x)-K(x)K(\sqrt{1-x^2})=\frac{\pi}{2}\tag{3}$$ We also need the formula $$E(x) =x(1-x^2) \frac{dK(x)} {dx} +(1-x^2) K(x)\tag{4}$$ Using $(1)$ and $(4)$ the identity $(3)$ can be transformed into an identity connecting $f$ and its derivative $f'$ and the finally replacing $x$ by $\sin x$ completes the job.

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  • $\begingroup$ Thanks Singh! It really helped a lot! $\endgroup$ – Euduardo Oct 28 '18 at 13:28
  • $\begingroup$ @Euduardo: if you like the answer you may upvote/accept it. $\endgroup$ – Paramanand Singh Oct 28 '18 at 13:51

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