Sequence Curves: Essence of Sine and Cosine

Last time, we covered Geometric Series and before that we covered Arithmetic Series.
Now we will try to find formulas for more unusual Series such as $\sum\sin$ and $\sum\cos$.
$$\sum_{i=n}^m \sin(i) = \int_{n-1}^mȢ\{\sin(x)\}dx, \\
\sum_{i=n}^m \cos(i) = \int_{n-1}^mȢ\{\cos(x)\}dx $$

Essence Identity

$$Ȣ\{f(x+1) - f(x)\} = \frac{d}{dx}f(x+1) $$ Last time we were able to calculate Essence of Exponentials with help of this Identity (also called Essence Formula). This will also help us calculate Essence for Sine and Cosine, though through a more complicated process. Unlike Exponentials, the difference cannot be grouped into a single term. To compute $Ȣ\{\sin(x)\}$ and $Ȣ\{\cos(x)\}$, we will have to solve them simultaneously side-by-side.

$Ȣ\{\sin(x)\} \text{ and }Ȣ\{\cos(x)\}$

Firstly, let us use Essence Identity to find that
$$\begin{align*}
Ȣ\{\sin(x+1)-\sin(x)\}&=\frac{d}{dx}\sin(x+1)\\
Ȣ\{\sin(1)\cos(x)+\cos(1)\sin(x) - \sin(x)\}&=\cos(x+1)\\
Ȣ\{\sin(1)\cos(x) + (\cos(1)-1)\sin(x) \}&=\\
\sin(1)Ȣ\{\cos(x)\}+(\cos(1)-1)Ȣ\{\sin(x)\}&=\\
\therefore Ȣ\{\sin(x)\} &= \frac{\cos(x+1) - \sin(1)Ȣ\{\cos(x)\}}{\cos(1)-1}
\end{align*}$$ and $$\begin{align*}
Ȣ\{\cos(x+1)-\cos(x)\}&= \frac{d}{dx}\cos(x+1)\\
Ȣ\{\cos(1)\cos(x)-\sin(1)\sin(x)-\cos(x)\}& = -\sin(x+1) \\
Ȣ\{-\sin(1)\sin(x) + (\cos(1)-1)\cos(x)\}&= \\
-\sin(1)Ȣ\{\sin(x)\}+(\cos(1)-1)Ȣ\{\cos(x)\}&=\\
\therefore Ȣ\{\sin(x)\}&=\frac{(\cos(1)-1)Ȣ\{\cos(x)\}+\sin(x+1)}{\sin(1)} \end{align*}$$ We thus find two definitions of Essence of Sine each in terms of Essence of Cosine. We can use substitution to solve for Essence of Cosine.
$$\begin{align*}\frac{(\cos(1)-1)Ȣ\{\cos(x)\}+\sin(x+1)}{\sin(1)} &=  \frac{\cos(x+1) - \sin(1)Ȣ\{\cos(x)\}}{\cos(1)-1} \\
(\cos(1)-1)^2Ȣ\{\cos(x)\} + (\cos(1)-1)\sin(x+1) &= \sin(1)\cos(x+1) - \sin^2(1)Ȣ\{\cos(x)\} \\
( (\cos(1)-1)^2+\sin^2(1) )Ȣ\{\cos(x)\} & = \sin(1)\cos(x+1) - (\cos(1)-1)\sin(x+1) \\
( \cos^2(1) + 1 - 2\cos(1) + \sin^2(1)  )Ȣ\{\cos(x)\} & = \sin(1)\cos(x+1) - \cos(1)\sin(x+1) + \sin(x+1) \\
(2 - 2\cos(1))Ȣ\{\cos(x)\} &= \sin(1 - (x+1) ) + \sin(x+1) \\
2(1-\cos(1))Ȣ\{\cos(x)\} &= \sin(-x) + \sin(x+1) \\
\therefore Ȣ\{\cos(x) \} = \frac{\sin(x+1)-\sin(x)}{2(1-\cos(1))}
\end{align*}$$ Now knowing Essence of Cosine, we can substitute this back into one of the definitions for Essence of Sine. We will use the second definition in order to easily cancel out the $1-\cos(1)$ in the denominator. $$\begin{align*}
\because Ȣ\{\sin(x)\}&=\frac{(\cos(1)-1)Ȣ\{\cos(x)\}+\sin(x+1)}{\sin(1)} \\
&= \frac{-(1-\cos(1))\frac{\sin(x+1)-\sin(x)}{2(1-\cos(1))}+\sin(x+1)}{\sin(1)} \\
& = \frac{-\frac{1}{2}\sin(x+1) + \frac{1}{2}\sin(x) + \sin(x+1)}{\sin(1)} \\
& = \frac{\frac{1}{2}\sin(x+1) + \frac{1}{2}\sin(x)}{\sin(1)} \\
\therefore Ȣ\{\sin(x)\} & = \frac{\sin(x+1) + \sin(x)}{2\sin(1)}
\end{align*}$$ We arrived at simple equations for both Essences, but these are not the only possible definitions.
Using the Interchangeability of Essence and Differentiation, $$\frac{d}{dx}Ȣ\{f(x)\} = Ȣ\{\frac{d}{dx}f(x)\}$$ we can find that $$\begin{align*}
Ȣ\{\frac{d}{dx}\sin(x)\} & = \frac{d}{dx}\frac{\sin(x+1) + \sin(x)}{2\sin(1)} \\
\thereforeȢ\{\cos(x)\} & = \frac{\cos(x+1) + \cos(x)}{2\sin(1)} \\ \\
Ȣ\{\frac{d}{dx}\cos(x)\} & = \frac{d}{dx}\frac{\sin(x+1)-\sin(x)}{2(1-\cos(1))} \\
-Ȣ\{\sin(x)\} &= \frac{\cos(x+1) - \cos(x)}{2(1-\cos(1))} \\
\thereforeȢ\{\sin(x)\} &= \frac{-\cos(x+1) + \cos(x)}{2(1-\cos(1))} \end{align*}$$ With this, we conclude the Essence formulas for both:
$$\begin{align*}
Ȣ\{\sin(x)\} & = \frac{\sin(x+1) + \sin(x)}{2\sin(1)}  & =\frac{-\cos(x+1) + \cos(x)}{2(1-\cos(1))} \\
Ȣ\{\cos(x)\} & =  \frac{\cos(x+1) + \cos(x)}{2\sin(1)} & = \frac{\sin(x+1)-\sin(x)}{2(1-\cos(1))} \\
&&\blacksquare
\end{align*}$$

We note an interesting symmetry with the two definitions of each Sine and Cosine: when $\sin(1)$ is in the denominator, it is as if it is taking the average of the current and the next term of the sequence; when $1-\cos(1)$ is at the denominator, it is taking the difference of the current and next term of the other trigonometric function.
These simple pattern lends itself to be easily memorized for future uses. Personally, the first definitions with $\sin(1)$ seems to be more easily memorized and used. $$\sum_{i=n}^mf(i) = \int_{n-1}^mȢ\{f(x)\}dx$$ By definition of Essences, we find that $$\sum_{i=n}^m \sin(i) = \int_{n-1}^m\frac{\sin(x+1)+\sin(x)}{2\sin(1)}dx = \frac{\cos(n)+\cos(n-1)-\cos(m+1)-\cos(m)}{2\sin(1)} \\
\sum_{i=n}^m\cos(i) = \int_{n-1}^m \frac{\cos(x+1)+\cos(x)}{2\sin(1)}dx = \frac{\sin(m+1)+\sin(m) - \sin(n) - \sin(n-1)}{2\sin(1)}
$$ or if you wish to use alternate definition, $$\sum_{i=n}^m\sin(i) = \int_{n-1}^m\frac{-\cos(x+1)+\cos(x)}{2(1-\cos(1))}dx = \frac{-\sin(m+1) + \sin(m) + \sin(n) - \sin(n-1)}{2(1-\cos(1))} \\
\sum_{i=n}^m\cos(i) = \int_{n-1}^m\frac{\sin(x+1)-\sin(x)}{2(1-\cos(1))}dx = \frac{\cos(n)-\cos(n-1) - \cos(m+1) - \cos(m)}{2(1-\cos(1))}
$$
By similar process we can calculate $Ȣ_x\{\sin(nx)\}$ and $Ȣ_x\{\cos(nx)\}$ where $Ȣ_x$ denotes Essence with respect to $x$.

$$\begin{align*}
\because \frac{d}{dx} f(x+1) &= Ȣ\{ f(x+1)-f(x) \} \\ \frac{d}{dx}\sin(n(x+1)) & = Ȣ\{\sin(n(x+1)) - \sin(nx) \} \\
n\cos(n(x+1)) & = Ȣ\{ \sin(nx+n) - \sin(nx) \} \\
n\cos(nx+n) &= Ȣ\{ \sin(n)\cos(nx) + \cos(n)\sin(nx) - \sin(nx) \} \\
&= \sin(n)Ȣ\{\cos(nx)\} + (\cos(n)-1)Ȣ\{\sin(nx)\} \\
\therefore Ȣ\{\sin(nx)\} &= \frac{n\cos(nx+n) - \sin(n)Ȣ\{\cos(nx)\}}{\cos(n) - 1} \\ \\
\frac{d}{dx}\cos(n(x+1)) &= Ȣ\{\cos(n(x+1)) - \cos(nx) \} \\
-n\sin(n(x+1)) &= Ȣ\{\cos(nx+n) - \cos(nx) \} \\
-n\sin(nx+n) &= Ȣ\{\cos(n)\cos(nx) - \sin(n)\sin(nx) - \cos(nx) \} \\
-\sin(nx + n) &= (\cos(n)-1)Ȣ\{\cos(nx)\} - \sin(n)Ȣ\{\sin(nx)\} \\
\therefore Ȣ\{\sin(nx)\} &= \frac{n\sin(nx+n) + (\cos(n)-1)Ȣ\{\cos(nx)\}}{\sin(n)}
\end{align*} \\ $$ $$
\begin{align*}
\therefore \frac{n\cos(nx+n) - \sin(n)Ȣ\{\cos(nx)\}}{\cos(n) - 1} &= \frac{n\sin(nx+n) + (\cos(n)-1)Ȣ\{\cos(nx)\}}{\sin(n)} \\ n\sin(n)\cos(nx+n) - \sin^2(n)Ȣ\{\cos(nx)\} & = n(\cos(n)-1)\sin(nx+n) + (\cos(n)-1)^2Ȣ\{\cos(nx)\} \\
n(\sin(n)\cos(nx+n) - (\cos(n)-1)\sin(nx+n) ) &= (\cos(n)-1)^2 + \sin^2(n))Ȣ\{\cos(nx)\} \\
\because (\cos(\theta)-1)^2 = 2(1-\cos(\theta))-\sin^2(\theta) \\
n(\sin(n)\cos(nx+n) - \cos(n)\sin(nx+n) + \sin(nx+n)  ) &= 2(1-\cos(n))Ȣ\{\cos(nx)\} \\
n(\sin(n-(nx+n)) + \sin(nx+n)) &= \\
n(\sin(nx+n) - \sin(nx)) & =\\ \\
\therefore Ȣ_x\{\cos(nx)\} = \frac{n(\sin(nx+n) - \sin(nx))}{2(1-\cos(n))} \\
\end{align*} $$ $$
\begin{align*}
\because Ȣ\{\sin(nx)\} & = \frac{n\sin(nx+n) + (\cos(n)-1)Ȣ\{\cos(nx)\}}{\sin(n)} \\
& = \frac{n\sin(nx+n) - (1-\cos(n))\frac{n(\sin(nx+n) - \sin(nx))}{2(1-\cos(n))}}{\sin(n)} \\
& = \frac{n( 2\sin(nx+n) - \sin(nx+n) + \sin(nx) )}{2\sin(n)} \\ \\
\therefore Ȣ_x\{\sin(nx)\} &= \frac{n(\sin(nx+n) + \sin(nx) )}{2\sin(n)}
\end{align*} $$ $$
\begin{align*}
\because Ȣ\{\frac{d}{dx}f(x) \} & = \frac{d}{dx}Ȣ\{f(x)\} \\
\therefore Ȣ\{ \frac{d}{dx} \sin(nx) \} &= \frac{d}{dx}Ȣ\{\sin(nx) \} \\
Ȣ\{ n\cos(nx) \} & = \frac{d}{dx} frac{n(\sin(nx+n) + \sin(nx) )}{2\sin(n)} \\
nȢ\{\cos(nx)\} &= \frac{n^2(\cos(nx+n) + \cos(nx) )}{2\sin(n)} \\
\therefore Ȣ\{\cos(nx) \} &= \frac{n(\cos(nx+n) + \cos(nx) )}{2\sin(n)} \\ \\
\therefore Ȣ\{\frac{d}{dx} \cos(nx)\} &= \frac{d}{dx}Ȣ\{\cos(nx)\} \\
Ȣ\{-n\sin(nx)\} & = \frac{d}{dx}\frac{n(\sin(nx+n) - \sin(nx))}{2(1-\cos(n))} \\
-nȢ\{\sin(nx) \} &= \frac{n^2(\cos(nx+n) - \cos(nx))}{2(1-\cos(n))} \\
\therefore Ȣ\{\sin(nx)\} &= \frac{n(-\cos(nx+n) + \cos(nx))}{2(1-\cos(n))}
\end{align*}$$ $$
\begin{align*}
Ȣ_x\{\sin(nx)\} &=  \frac{n(\sin(nx+n) + \sin(nx) )}{2\sin(n)} &= \frac{n(-\cos(nx+n) + \cos(nx))}{2(1-\cos(n))}  \\
Ȣ_x\{\cos(nx) \} &= \frac{n(\cos(nx+n) + \cos(nx) )}{2\sin(n)} &= \frac{n(\sin(nx+n) - \sin(nx))}{2(1-\cos(n))}  \\
&&\blacksquare \end{align*} $$

Notice that this is consistent with original definitions above when $n=1$.`
With these we can find that
$$
\begin{align*}
\sum_{i=n}^m\sin(r*i) & = \int_{n-1}^m\frac{r(\sin(rx+r) + \sin(rx) )}{2\sin(r)} dx &&= \int_{n-1}^m \frac{r(-\cos(rx+r) + \cos(rx))}{2(1-\cos(r))} dx \\
& = \frac{\cos(rn)+\cos(rn-r) - \cos(rm+r) - \cos(rm)}{2\sin(r)} &&= \frac{ -\sin(rm+r) + \sin(rm) + \sin(rn) - \sin(rn-r) }{2(1-\cos(r))} \\
\sum_{i=n}^m \cos(r*i) &= \int_{n-1}^m \frac{ r(\cos(rx+r) +\cos(rx) ) }{2\sin(r)} dx &&= \int_{n-1}^m \frac{ r( \sin(rx+r) - \sin(rx) ) }{2(1-\cos(r))}dx \\
&= \frac{ \sin(rm+r)+\sin(rm) - \sin(rn) - \sin(rn-r) }{2\sin(r)} &&= \frac{ \cos(rn) -\cos(rn-r) -\cos(rm+r) + \cos(rm) }{2(1-\cos(r))}
\end{align*} $$ What fascinates me the most is that with these formulas we are able to condense a long sum of sine and cosine, which has no obvious reduction rules by themselves, into a concise single equation. Also, the added benefit of being able to calculate non-integer boundary series.
In addition to those, $\cos(\pi x)$ is a popular continuous substitute for $(-1)^x$ and so knowing its Essence will help us interpolate between more discontinuous functions later on.

Endnote

This has been a particularly equation-heavy article, but I hope you enjoyed exploring this concept with me nonetheless. I will try to post more frequently, with clearer explanations, and for more interesting topics in this new decade and I hope you will follow me through that journey.

Happy New Year and Happy New Decade!