Expansion of Sine and Cosine

With the Pythagoras derived by rotational dynamics from the last post, it is natural to think of sines and cosines as they are both related to rotation more so directly.  

1 Definitions

A big difference about sines and cosines is that they depend on angles rather than side lengths. So what is an angle? Well it is nothing but the length of the arc swiped counterclockwise by a line segment that equals to 1, 

here $OA$ rotates about the origin $O$ then become $OC$, and the length of arc $AC$ is the definition of an angle(in radian), let us call this angle $\alpha$. Now in a 2-D coordinate system, an angle is defined w.r.t. the positive x-axis. For example, let us put this model in a 2-D coordinate, where $O$ is the origin, $A$ is at $(1,0)$, then $OA$ would be at a angle 0 as it overlaps with the positive x-axis and $OC$ would be at an angle $\alpha$. Now to define sine and cosine, $\sin \left(\alpha \right)$ is defined to be the y-coordinate of $C$(or vertical component of $OC$) and $\cos \left(\alpha \right)$ is defined to be the x-coordinate of $C$(or horizontal component of $OC$).  

2 Setup

Now sine and cosine of an angle are defined, but they are only definitions, we want to actually calculate them. We will look at a situation where $OC$ is already at an angle $\alpha$ w.r.t. the horizon, and from there it rotates an angle $\beta$ to become $OE$. 

And of course the radius is always 1, that is $OA=OC=OE=1$, and remember we say angles but angles are really just arc lengths. $\alpha$ is really just the length of arc $\mathrm{AC}$ and the angle $\mathrm{\beta }$ is just the length of the arc $EC$. Our goal is to find the coordinate of the end point $E$, whose y-coordinate(or vertical components of $OE$) correspond to $\sin \left(\mathrm{\alpha }+\mathrm{\beta }\right)$, and x-coordinate(or horizontal and components of $OE$) correspond to $\cos \left(\mathrm{\alpha }+\mathrm{\beta }\right)$. The reason not let the rotation start from the horizon but rather from a non-zero angle $\alpha$ will become clear at the end of this article. 

3 Discrete rotation 

The approach we use is very similar to the one we used in the derivation of Pythagoras, a rotation can be approximated by discrete linear movements, 

i.e. $OC$ is the starting point and $OC=1$, it takes 4 linear segments(or call them tangent movements) $CP$, $PQ$, $QR$, $RE$ to become $OE$, and of course the direction of motion is always perpendicular to the "radius" at that point so

 $PC\perp OC QP\perp OP RQ\perp OQ ER\perp OR$

and every linear movement is equal in length,

 $PC=QP=RQ=ER$

so this time for each step it moves the same distance, rather than moving the same height as in the Pythagoras case. Now let us get into action and calculate how the horizontal component and the vertical component evolve in this 4-step-rotation. For simplicity, we will use HC and VC as short hands for horizontal component and vertical component.

The first step,

the starting position is at $C$, it has coordinate $(a,b)$ and $OC=1$. The first tangent movement is $C$ to $P$ where $PC\perp OC$ and $PC=\Delta$. $PM$ and $MC$ represent the changes in HC and VC, by similar triangles, 

$PM=\Delta b$ 

$MC=\Delta a$

so after the first step at the new point $P$, its coordinate becomes 

$(a-\Delta b,b+\Delta a)$

let us call it $\left(a^{\text{'}},b^{\text{'}}\right)$ for simplicity. Now the second step,

again, $QP\perp PO$ and $QP=\Delta$, this time $OP$ does not equal to 1 anymore, but in the limiting case when each tangent movement or $\Delta$ is infinitesimally small, then it becomes a proper rotation and radius is preserved. So let us just pretend $\Delta$ is small and radius is still 1 in this case. Again by similar triangle 

$QN=\Delta b^{\text{'}}$

$NP=\Delta a^{\text{'}}$

so after the second step at the new point $Q$, its coordinate becomes 

$(a^{\text{'}}-\Delta b^{\text{'}},b^{\text{'}}+\Delta a^{\text{'}})$ 

call it $\left(a^{\text{'}\text{'}},b^{\text{'}\text{'}}\right)$. The third step,

again, $RQ\perp QO$ and $RQ=\Delta$, and the new point $R$ would have coordinate $(a^{\text{'}\text{'}}-\Delta b^{\text{'}\text{'}},b^{\text{'}\text{'}}+\Delta a^{\text{'}\text{'}})$. 

Needless to do more steps, the pattern is quite neat. HC is always reduced by $\Delta *VC$ at that point, and VC is always increased by $\Delta *HC$ at that point, if we write the ith step of HC and VC as $H{C}_i$ and $V{C}_i$, then, 

$H{C}_{i+1}=H{C}_i-\Delta ·V{C}_i$.

$V{C}_{i+1}=V{C}_i+\Delta ·H{C}_i$.

 Now let us write the expressions out in terms of the original $a$ and $b$ and try to find some pattern. Spoiler alert, the terms will look exactly like binomial coefficients but with interchanging signs. The trick is to look at terms of different orders one by one, namely coefficients of ${\Delta }^0, {\Delta }^1, {\Delta }^2$... First the terms without $\Delta$, so they are the original $a$ and $b$, at each step, they remain unchanged as whatever is added or subtracted is multiplied by a factor of $\Delta$, so the added or subtracted terms are at least of order ${\Delta }^1$, so $a$ and $b$ remain unchanged for any steps. Now the ${\Delta }^1$ term, because $a$ and $b$ are always in every step, when they are multiplied by Delta and added or substracted to VC or HC, there is always an $\Delta ·a$ or $\Delta ·b$ picked up by VC or HC in every step, from the 1st step onward, namely:

so at ith step we always have a plus or minus $i·\Delta$ term in VC or HC. By the same logic, these ${\Delta }^1$ terms will also be multiplied by a $\Delta$ then added or subtracted to VC or HC, in every step from the second step onward, so there will always be a $(i-1)·{\Delta }^2$ term added or subtracted at the ith step:

now we find the ${\Delta }^2$ term for the ith step would be $[1+2+3+...(\mathrm{i}-1\left)\right]$. By the same logic, this ${\Delta }^2$ term will be multiplied by a $\Delta$ then added or subtracted back. So from the 3rd step onward, there will be a ${\Delta }^3$ term with coefficient $[1+(1+2)+(1+2+3)+...+(1+2+3+...+\mathrm{i}-2\left)\right]$ added or subtracted, for any ith step. Now the pattern should become clear, needless to do the ${\Delta }^4$ terms in specific. The result of first 6 steps look like,

And again they look exactly like binomial expansion but with interchanging signs, so all of the coefficients are binomial coefficients as well. Now instead of just using the binomial coefficients and proceed, let us invent a new set of summation expressions, where it is equivalent to binomial coefficients.

a general formula would be:

${\sum }_n^m={\sum }_1^{m-1}+{\sum }_2^{m-1}+{\sum }_3^{m-1}+...+{\sum }_n^{m-1}$

with some integers $m$ and $n$. To calculate any ${\sum }_n^m$, one can use the definition above add its way up to get a result, but one can also use the multiplication form to get the result quicker,

${\sum }_n^m=\frac{n(n+1)(n+2)...(n+m-1)}{m!}$

and this addition formula is equivalent to the binomial coefficient, related by:

${\sum }_n^m=\left(\begin{array}{c}n+m-1\\ m\end{array}\right)$ 

hopefully it is obvious to see what the summation expression represent, and of course, as for why, like how it is related to binomial coefficient, how summing up numbers have an equivalent multiplication form, if it has generalizations to non-integers and so on, these questions deserve another blog purely for summation form and binomial coefficients, our main goal here is about sine and cosine. One more thing about the reason we introduce this summation form rather than just using the binomial coefficients, is because the summation form is defined from summing up numbers which is exactly what is going on in the evolution of HC and VC in our case, while binomial coefficient is defined from combinations and permutations, not as directly related. 

Now, using the summation form as coefficients, HC and VC after 6 steps would look like:

$HC=a-{\sum }_6^1\Delta b-{\sum }_5^2{\Delta }^{2}a+{\sum }_4^3{\Delta }^{3}b+{\sum }_3^4{\Delta }^{4}a-{\sum }_2^5{\Delta }^{5}b+{\sum }_1^6{\Delta }^{6}a$

$VC=b+{\sum }_6^1\Delta a-{\sum }_5^2{\Delta }^{2}b-{\sum }_4^3{\Delta }^{3}a+{\sum }_3^4{\Delta }^{4}b+{\sum }_2^5{\Delta }^{5}a-{\sum }_1^6{\Delta }^{6}b$

and the result after n steps would be:

$HC=a-{\sum }_n^1\Delta b-{\sum }_{n-1}^2{\Delta }^{2}a+{\sum }_{n-2}^3{\Delta }^{3}b+{\sum }_{n-3}^4{\Delta }^{4}a-···{\sum }_1^n{\Delta }^{n}a$

$VC=b+{\sum }_n^1\Delta a-{\sum }_{n-1}^2{\Delta }^{2}b-{\sum }_{n-2}^3{\Delta }^{3}a+{\sum }_{n-3}^4{\Delta }^{4}b+···{\sum }_1^n{\Delta }^{n}b$

use the equivalent multiplication form as coefficients:

$HC=a-\frac{n}{1!}\Delta b-\frac{(n-1)n}{2!}{\Delta }^{2}a+\frac{\left(n-2\right)\left(n-1\right)n}{3!}{\Delta }^{3}b+\frac{\left(n-3\right)\left(n-2\right)\left(n-1\right)n}{4!}{\Delta }^{4}a-...\frac{1\times 2\times 3···n}{n!}{\Delta }^{n}a$

$VC=b+\frac{n}{1!}\Delta a-\frac{\left(n-1\right)n}{2!}{\Delta }^{2}b-\frac{\left(n-2\right)\left(n-1\right)n}{3!}{\Delta }^{3}a+\frac{\left(n-3\right)\left(n-2\right)\left(n-1\right)n}{4!}{\Delta }^{4}b+...\frac{1\times 2\times 3···n}{n!}{\Delta }^{n}b$

Going back to our setup, $OC$ rotates an angle $\beta$ to become $OE$, on the other hand, in the limiting case of our discrete rotation model, it would take infinite steps to complete the discrete rotation, let us call the number of steps $N$(even though $N$ is infinite). Now in the limiting case, the rotation becomes perfect, the path that the tangent movements swiped through is exactly equal to the angle $\beta$, or the arc length $\beta$, so we have:

$\beta =N\Delta$

$\Delta =\frac{\beta }{N}$

now try to substitute $N$ and $\beta$ into HC:

 

$HC=a-\frac{N}{1!}\frac{\beta }{N}b-\frac{\left(N-1\right)N}{2!}{\left(\frac{\beta }{N}\right)}^{2}a+\frac{\left(N-2\right)\left(N-1\right)N}{3!}{\left(\frac{\beta }{N}\right)}^{3}b+\frac{\left(N-3\right)\left(N-2\right)\left(N-1\right)N}{4!}{\left(\frac{\beta }{N}\right)}^{4}a-···$ 

but as $N$ is infinite, the $N\mathit{-}\mathit{1}$, $N\mathit{-}\mathit{2}$ or $N$ minus any finite number term would cancel out with an $N$ on the denominator, so we have

$HC=a-\beta b-{\beta }^{2}a+{\beta }^{3}b+{\beta }^{4}a-···$

$VC=b+\beta a-{\beta }^{2}b-{\beta }^{3}a+{\beta }^{4}b+···$

now all of the terms have an $a$ or $b$, rearrange the equations by $a$ and $b$ we get

$HC=a(1-{\beta }^2+{\beta }^4-{\beta }^6+···)-b(\beta -{\beta }^3+{\beta }^5-{\beta }^7+···)$

$VC=b(1-{\beta }^2+{\beta }^4-{\beta }^6+···)+a(\beta -{\beta }^3+{\beta }^5-{\beta }^7+···)$

remember in our setup, HC is nothing but $\cos \left(\alpha +\beta \right)$ and VC is just $\sin \left(\alpha +\beta \right)$, so

$\cos \left(\alpha +\beta \right)=a\left(1-{\beta }^2+{\beta }^4-{\beta }^6+···\right)-b\left(\beta -{\beta }^3+{\beta }^5-{\beta }^7+···\right)$

$\sin \left(\alpha +\beta \right)=b\left(1-{\beta }^2+{\beta }^4-{\beta }^6+···\right)+a\left(\beta -{\beta }^3+{\beta }^5-{\beta }^7+···\right)$

now let the angle $\alpha$ be 0, such that the rotation actually starts from the horizon or the positive x-axis, then we have $b=0$ and $a=1$,

$\cos \left(\beta \right)=1-{\beta }^2+{\beta }^4-{\beta }^6+···$

$\sin \left(\beta \right)=\beta -{\beta }^3+{\beta }^5-{\beta }^7+···$

there we have it, the sine and cosine expansion in terms of angle(or arc length). 

Now coming back to the equations where $\alpha \ne 0$, one might have noticed that we accidentally derived the angle addition formulae for sine and cosine on the way of deriving sine and cosine expansion, as $a$ is just $\cos \left(\alpha \right)$ and $b$ is just $\sin \left(\alpha \right)$ according to definition also using the sine and cosine expansion as we discovered,

$\cos \left(\alpha +\beta \right)=\cos \left(\alpha \right)\cos \left(\beta \right)-\sin \left(\alpha \right)\sin \left(\beta \right)$

$\sin \left(\alpha +\beta \right)=\sin \left(\alpha \right)\cos \left(\beta \right)+\cos \left(\alpha \right)\sin \left(\beta \right)$