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I am trying to find an identity for $\arctan(ab)$ in terms of any trigonometric function involving $a$ and $b$ being separate.

I have attempted doing so using $(\arctan(a)+\arctan(b))^2$ but that got me nowhere. I have not tried any other ideas because I can't think of anything that is worth putting down on paper and seeing where it leads.

Can anybody help me out here? Do note that I am asking because this is required for a question that is part of a question set I am submitting to UKMT for their Senior Mathematical Challenge (will not disclose the question).

Edit 1: Looks like there is no identity for $\arctan(ab)$, but I am now curious as to whether an identity exists for $\arctan(a^2)$ since that also helps at the stage of this question I am fleshing out.

Yajat Shamji
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  • I don't think it's going to be easy to express $\arctan(ab)$ in terms of functions of $a$ and $b$. See the comments on this question for example. – Sahaj Dec 21 '23 at 18:33
  • There is no known formula to simplify this expression even for some very simple cases. For example, there's no known formula for $\arctan(2x)$ (in terms of trigonometric functions or its inverses evaluated at $x$) – jjagmath Dec 21 '23 at 18:44
  • Having the product $ab$ as the argument makes this particularly difficult because you essentially are solving $$a b = \tan \theta$$ and due to symmetry $a$ and $b$ could be anything. – John Alexiou Dec 21 '23 at 21:37
  • $$ 2 \arctan(z) = \arctan \left( \frac{2}{1-z^2} z \right) $$ – John Alexiou Dec 21 '23 at 21:55
  • OK. Then what about $\arctan(a^2)$? Because that also helps for me. – Yajat Shamji Dec 22 '23 at 11:03

3 Answers3

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I have found an identity for $\arctan(a^2)$ on my own. It is:

$$\fbox{$\arctan(a^2)=\frac{1}{2}[\arctan(\frac{2a}{a^4-2a^3+1})+\arctan(\frac{2a(1-a)}{a^4-2a^3-1})]$}$$

This was derived from the $\arctan(a+b)$ identity from https://math.stackexchange.com/a/4294539/789034

($b=-a+a^2$ was the substitution used to get the above identity.).

As for $\arctan(ab)$, the identity produced has a mixture of $a$'s and $b$'s, which doesn't satisfy my question's initial conditions.

Also, you can generalise for $\arctan(a^n)$ (using the $\arctan(a+b)$ identity and $b=-a+a^n$ as the substitution), which is:

$$\fbox{$\arctan(a^n)=\frac{1}{2}[\arctan(\frac{2a}{a^{2n}-2a^{n+1}+1})+\arctan(\frac{2a(1-a^{n-1})}{a^{n}-2a^{n+1}-1})]$}$$

Yajat Shamji
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$\arctan(a^2)$ can be written as $$\arctan\left(\sum_{n=1}^a(2n+1)\right)$$ implying that we can repeatedly use the $\arctan(a+b)$ identity on this. But this is not a good, clean identity at all, and although I cannot prove that no identity exists, I'd bet money that no identity does exist, for $\arctan(a^2)$ (with a finite number of terms, generalized, clean) etc. For your information, $$\arctan(a+b)=\frac{1}{2}\arctan\left(\frac{2a}{1-a^2+b^2}\right)+\frac{1}{2}\arctan\left(\frac{2b}{1+a^2-b^2}\right)$$

wolfqz
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According to Wolfram functions:

$$\tan^{-1}(z^n)=\sum_{k=1}^n(-1)^{k-1}\tan^{-1}\left(\frac{z\sin\left(\frac{(2k-1)\pi}{2n}\right)}{1-z\cos\left(\frac{(2k-1)\pi}{2n}\right)}\right);|z|<1,n\in\Bbb N$$

For $n=2$:

$$\tan^{-1}(a^2)=\cot^{-1}\left(\frac{\sqrt2}a-1\right)-\cot^{-1}\left(\frac{\sqrt2}a+1\right)+\begin{cases}0,&-\sqrt2<a\le\sqrt2\\\pi,&\text{otherwise}\end{cases}$$

which may be tested here. We may define the right hand side to be $0$ at $a=0$

Тyma Gaidash
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