# Analogous “Dark Sector” Trigonometric (and Hyperbolic) Functions

Existing Definitions: $$\zeta(n)=\sum_{k=1}^\infty \frac{ 1 }{k^n}$$ $$\lambda(n)=\sum_{k=1}^\infty \frac{ 1 }{(2k-1)^n}=\frac{\left(2^n-1\right)}{2^n}\zeta (n)$$ $$\eta(n)=\sum_{k=1}^\infty \frac{(-1)^{k-1}}{k^n}=\left(1 -2^{1-n} \right) \zeta (n)$$ $$\beta(n)=\sum_{k=1}^\infty \frac{(-1)^{k-1}}{(2k-1)^n}$$

The existing well known trigonometric functions $$\csc(x)$$, $$\sec(x)$$, $$\tan(x)$$ and $$\cot(x)$$ in infinite series form are:

$$\csc(x)=\frac{1}{x}+2 \sum _{k=1}^{\infty } \frac{\eta(2 k) }{\pi ^{2 k}}\;x^{2 k-1}$$

$$\sec(x)=2\sum _{k=1}^{\infty } \frac{\ 2^{2 k-1} \beta(2 k-1) }{\pi ^{2 k-1}}\,x^{2 k-2}$$

$$\tan(x)=2 \sum _{k=1}^{\infty } \frac{2^{2 k} \lambda(2 k) }{\pi ^{2 k}}\,x^{2 k-1}$$

$$\cot(x)=\frac{1}{x}-2 \sum _{k=1}^{\infty } \frac{ \zeta (2 k)}{\pi ^{2 k}}\,x^{2 k-1}$$

I am now going to define some analogous new definitions, using the odd Zeta constants and the even Beta constants, and postfix them with an "i":

$$\text{csci}(x) =2 \sum _{k=1}^{\infty } \frac{\eta(2 k+1) }{\pi ^{2 k+1}}\,x^{2 k}-\frac{1}{x}+\frac{2 \log (2)}{\pi }$$

$$\text{seci}(x)=2\sum _{k=1}^{\infty } \frac{\ 2^{2 k}\; \beta(2 k) }{\pi ^{2 k}}\,x^{2 k-1}$$ $$\text{tani}(x)=2 \sum _{k=1}^{\infty } \frac{2^{2 k+1} \;\lambda(2 k+1) }{\pi ^{2 k+1}}\,x^{2 k}+\frac{2 \log (2)}{\pi }$$ $$\text{coti}(x)=-2 \sum _{k=1}^{\infty } \frac{ \zeta (2 k+1)}{\pi ^{2 k+1}}\,x^{2 k}-\frac{1}{x}+\frac{2 \log (2)}{\pi }$$

You could try the same analogy with "dark sector" hyperbolic functions. e.g. $$\text{sechi}(x)=2\sum _{k=1}^{\infty } \frac{(-1)^{k-1} 2^{2 k}\; \beta(2 k) }{\pi ^{2 k}}\,x^{2 k-1}$$

It is immediately apparent that there are certain similarities between normal trigonometry and "dark sector" trigonometry

For example $$\text{seci}(x)=\text{csci}(x+\frac{\pi}{2})$$, and $$\text{csci}(x)=\text{coti}(x/2)-\text{coti}(x)$$.

But also differences for example $$\text{tani}(x)$$ is not equal to the inverse of $$\text{coti}(x)$$. However both the inverses appear to lead to the same new function, that differs only by an inversion and a phase shift of $$\pi/2$$.

Graph for $$1/\text{tani}(x)$$

Graph for $$1/\text{coti}(x)$$

Similar thing happens with the inverse of $$1/\text{csci}(x)$$ and $$1/\text{seci}(x)$$ to give $$\text{sini}(x)$$ and $$\text{cosi}(x)$$

Graph for $$\text{sini}(x)$$

Graph for $$\text{cosi}(x)$$

Examples of combining "dark sector" functions with normal trigonometric functions look quite interesting:

Graph for $$\text{cosi}(x)+\cos(x)$$

Graph for $$\text{cosi}(x)-\cos(x)$$

I've just sketched this structure using Mathematica before I waste too much time on it. There may be a better way of defining the four starting analogous functions: $$\text{csci}(x)$$,$$\text{seci}(x)$$,$$\text{tani}(x)$$ and $$\text{coti}(x)$$.

Does anyone know of attempts to develop what I call here "dark sector" trig or hyperbolic functions?

Does anyone recognise any of these functions and where they might have an application?

• This is really interesting (and beautiful) ! $+1$ fur sure and I shall spend time working this post. Thanks for posting such a nice work. Cheers. – Claude Leibovici Jan 27 at 14:44
• @ClaudeLeibovici: Thanks. By the way the constant $\frac{2 \log 2}{\pi}$ is actually $\frac{2\, \eta(1)}{\pi}$ which can't be calculated using the Mathematica friendly definition for $\eta(n)$ in terms of the Zeta Function I used at the top of the page, where $n>1$. – James Arathoon Jan 28 at 16:33
• Could you explain why you call these functions "dark sector" functions? Also: are you looking for closed forms? I'm not exactly sure what you're asking – clathratus Jan 29 at 1:13
• @clathratus: I am interested in the overall structure, which you don't see the possibility of unless you approach infinite series using zeta and beta constants. I'm calling it the "dark sector" because I know some sort of limited analogical structure to trigonometry can be formulated, but I don't see more than a rough outline. As an example of my lack of understanding, differentiating tani[x] produces a function that is similar to but not the same as seci[x]^2 calculated in Mathematica directly. This may mean something or nothing I have no idea. – James Arathoon Jan 29 at 1:49
• The identity $$\operatorname{tani}(x)=-\operatorname{coti}\left(x+\frac{\pi}{2}\right)$$ does not yet seem to be in your list. This should establish your observation about the reciprocals. – J. M. is a poor mathematician Mar 27 at 12:08