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Is there a closed form for the function $\sin(n\arctan x)$, perhaps where $n$ is restricted to being an integer, or if not, perhaps some special integers (such as triangular numbers or some other figurate numbers)?

From playing around with a few values, it seems that

$$\sin\arctan(x)=\frac{x}{\sqrt{1+x^2}},~\sin(2\arctan x)=\frac{2x}{1+x^2},~\sin(3\arctan x)=\frac{3x-x^3}{(1+x^2)^{3/2}},$$

I can see that the denominator is $(1+x^2)^{\tfrac12n}$ but can't quite see the form of the numerator.

Motivation: This is motivated by an inconvenient but necessary change of coordinates from polar to Cartesian when a function involves not $\sin\theta$ but $\sin n\theta$ for some integer $n.$

Hobbyist
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3 Answers3

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The Chebyshev polynomials of the second kind give you $$ \sin(n \theta) = U_{n-1}(\cos \theta) \sin \theta . $$ In this formula, put $\theta = \arctan x$ and use $\cos \theta = \dfrac{1}{\sqrt{1+x^2}}$ and $\sin \theta = \dfrac{x}{\sqrt{1+x^2}}$.

Hans Lundmark
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We have that $$\tan^{-1}(x)=\frac{i}{2}\log\left(\frac{1-ix}{1+ix}\right)$$ And $$\sin(x)=\frac{i}{2}e^{-ix}-\frac{i}{2}e^{ix}$$ So $$\sin(n\tan^{-1}(x))=\frac{i}{2}\exp\left(\frac{i}{2}n\log\left(\frac{1-ix}{1+ix}\right)\right)-\frac{i}{2}\exp\left(\frac{i}{2}n\log\left(\frac{1+x}{1-ix}\right)\right)$$ $$\sin(n\tan^{-1}(x))=\frac{i}{2}\left(\frac{(1-ix)^\frac{n}{2}}{(1+ix)^\frac{n}{2}}-\frac{(1+ix)^\frac{n}{2}}{(1-ix)^\frac{n}{2}}\right)$$ $$\sin(n\tan^{-1}(x))=\frac{i}{2}\left(\frac{(1-ix)^n-(1+ix)^n}{(x^2+1)^\frac{n}{2}}\right)$$ Now we can use the Binomial theorem: $$\sin(n\tan^{-1}(x))=\frac{i}{2}\left(\frac{\sum_{k=0}^{n} \binom{n}{k}(-ix)^k-\sum_{k=0}^{n} \binom{n}{k}(ix)^k}{(x^2+1)^\frac{n}{2}}\right)$$ $$\sin(n\tan^{-1}(x))=\frac{i}{2}\left(\frac{\sum_{k=0}^{n} \binom{n}{k}x^k((-i)^k-i^k)}{(x^2+1)^\frac{n}{2}}\right)$$ $$\sin(n\tan^{-1}(x))=\frac{1}{(x^2+1)^\frac{n}{2}}\sum_{k=0}^{n} \binom{n}{k}x^k\frac{i((-i)^k-i^k)}{2}$$ $$\sin(n\tan^{-1}(x))=\frac{1}{(x^2+1)^\frac{n}{2}} \left(\binom{n}{1}x^1-\binom{n}{3}x^3+\binom{n}{5}x^5+\dots\right)$$ $$\sin(n\tan^{-1}(x))=\frac{1}{(x^2+1)^\frac{n}{2}} \sum_{k=0}^{n}\binom{n}{k}\cos\left((k-1)\frac{\pi}{2}\right)x^k$$

Botond
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  • Given that it appears the numerator is always a real-valued polynomial for any $n,$ assuming $x$ is real, is there a convenient way to factor out those $i$s? A form like that would be perfect. – Hobbyist Aug 04 '18 at 12:56
  • @Hobbyist I think I made it pure real, but please check if I didn't mess anything up. – Botond Aug 04 '18 at 13:25
  • I'm just commenting to say that it should be (k-1) instead of (n-1) in the last line. Cannot edit because it would be less than 6 characters. – Desura Feb 20 '24 at 17:14
  • @Desura thank you! – Botond Mar 12 '24 at 12:02
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Too long for a comment

Assume $x$ is real.

Let $S_n(x)=\sin(n\operatorname{arctan}(x))$.

Denote $\operatorname{arctan}$ as $g$.

Let $T_n(x)=e^{ing(x)}$.

Sloppily, $$T_n(x)=\left(e^{-ig(x)}\right)^{-n}$$

Let $f(x)=e^{-ig(x)}$.

Then, $$\frac{dT_n(x)}{df(x)}=-nT_{n+1}$$

Thus we obtain a recursive relation: $$T_{n+1}=-\frac in\cdot e^{i\operatorname{arctan}(x)}(1+x^2)\cdot T_n’(x)$$

Also, $$S_{n}(x)=\text{Im}(T_n(x))$$

Cuhrazatee
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Szeto
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  • Does $T_n(x)$ have an imaginary component for any $x \in \mathbb{R}$? Pretty sure it doesn't this would imply that $S_n(x) = 0$ for every $n$ and every $x$, I think that something went wrong in this derivation. – Cuhrazatee Nov 18 '23 at 02:06