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Find $x$ with $0<|x|<2$ such that $$\tan^{-1}(x^2-\frac{x^4}{2}+\frac{x^6}{4}-\cdots)+ \cot^{-1}(x^4-\frac{x^8}{2}+\frac{x^{12}}{4}-\cdots)=\pi/ 2$$

My try:

$(x^2-\frac{x^4}{2}+\frac{x^6}{4}-\cdots)=\alpha$

$(x^4-\frac{x^8}{2}+\frac{x^{12}}{4}-\cdots)=\beta$

$\alpha=\frac{\pi }{2}-\beta \to \tan \alpha=\tan(\frac{\pi }{2}-\beta)=\cot \beta$

Now?

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    "α=π/2−β" Is that so? – 2017-02-12

1 Answers 1

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What I can come up with is as below

$\tan^{-1}(\alpha) =A \implies \tan(A)=\alpha$

$\cot^{-1}(\beta) =B \implies \cot(B)=\beta$

Also, we have

$A+B=\frac{\pi}{2}$

Add $\frac{\pi}{2}$ to both sides

$A+B+\frac{\pi}{2}=\pi$

Take $\tan$ from both sides and use the famous formula of $\tan(x+y)=\frac{\tan(x)+\tan(y)}{1-\tan(x)\tan(y)}$

$0=\tan(\pi)=\frac{\tan(A)+\tan(B+\frac{\pi}{2})}{1-\tan(A)\tan(B+\frac{\pi}{2})}$

Knowing $\tan(x+\frac{\pi}{2})=\cot(x)$, we get

$0=\tan(\pi)=\frac{\tan(A)+\cot(B)}{1-\tan(A)\cot(B)}=\frac{\alpha+\beta}{1-\alpha \beta}$

So, we need to have $\alpha+\beta=0$. Of course, we cannot have $1-\alpha \beta=0$, at the same time. Because, the n we have $\alpha=\frac{1}{\beta}$ and as we want to have $\alpha = -\beta$, by multiplying them we get the contradiction $\alpha^2=-1$.

Then, we need to find a closed formula for $(x^2-\frac{x^4}{2}+\frac{x^6}{4}-\cdots)=\alpha$ and $(x^4-\frac{x^8}{2}+\frac{x^{12}}{4}-\cdots)=\beta$, which can be found by separating each of them into positive and negative terms and using geometric series summation formula. Note that the limit on $0<|x|<2$ comes useful here, as we need the geometric summations to be convergent. Finally, if I have not made any mistakes, you need to solve a polynomial of order $10$ to find $x$.

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    There is no need to solve a polynomial equation: $\alpha = x^2\sum_{k=0}^\infty (-x^2/2)^k = \frac{2x^2}{2+x^2} \geq 0$ and $\beta$ is the same formula with $x\to x^2$ so both are $\geq 0$ and there is only one way they can sum to zero. – 2017-02-12
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    @Winther, Thanks for giving more detail. – 2017-02-12
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    There is one issue here though: the claim that $\alpha+\beta = 0$ does not fit with $x = \pm 1$ being solutions. If for example $x=1$ then $\alpha = \beta = \frac{2}{3}$ and $\tan^{-1}(2/3) + \cot^{-1}(2/3) = \frac{\pi}{2}$. – 2017-02-12
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    There is just a simple sign-mistake: $\tan(x+\pi/2) = -\cot(x)$ so we end up with $\frac{\alpha-\beta}{1+\alpha\beta} = 0$. – 2017-02-12
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    @Winther, I usually come to this website, because I have no good idea to work on. What brings you here? You seem to have serious stuff going on for you. – 2017-02-12
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    Procrastination I guess. It's something to do to take my mind of work. And solving problems is always nice. – 2017-02-12
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    Fair enough, I would say. Thanks for responding. – 2017-02-13