The n-Angle Trigonometric with Pascal's Triangles

I'm back at it again with another fun math inquiry. We learn double angle formula for $\sin$ and $\cos$ but we usually do not go any further. Of course, we can build triple and quadruple angle formula with Angle Sum Identities, but why do we not discuss patterns of $\sin(n\theta)$ or $\cos(n\theta)$?
Today, we will discuss intriguing patterns that arrises from n-Angle Trigs.

Basics

Let's first go over the basic identities:
$$\begin{align*}
\sin(\theta + \phi) &= \sin(\theta)\cos(\phi) + \cos(\theta)\sin(\phi) \\
\sin(\theta - \phi) &= \sin(\theta)\cos(\phi) - \cos(\theta)\sin(\phi) \\
\cos(\theta + \phi) &= \cos(\theta)\cos(\phi) - \sin(\theta)\sin(\phi) \\
\cos(\theta - \phi) &= \cos(\theta)\cos(\phi) + \sin(\theta)\sin(\phi) \\
 \sin(2\theta) &= 2\sin(\theta)\cos(\theta) \\
 \cos(2\theta) &= \cos^2 (\theta) - \sin^2(\theta) \end{align*}$$
From these simple identities we can build all n-angled trigs. But why aren't we taught them?
Well, for one thing, solving for the equations become quite complicated and messy. So students are given only the essential identities above and can solve for more complex identities when absolutely necessary (which usually is almost never).

But for sake of fun, and for spirit of Christmas, we will try to expand all n-Angles to their simplest forms.

Expansions

After many, many pages of working-out, getting confused, and reworking-out the expansion, I learned the essential importance of consistent organization. We will order all our terms in descending order of $\cos$.
$$\begin{align*}
\cos(3\theta) & = \cos(\theta + 2\theta)\\
& = \cos(\theta)\cos(2\theta) - \sin(\theta)\sin(2\theta) \\
& = \cos(\theta)( \cos^2(\theta) - \sin^2(\theta) ) - \sin(\theta)(2\sin(\theta)\cos(\theta)) \\
& = \cos^3(\theta) - \cos(\theta)\sin^2(\theta) - 2\cos(\theta)\sin^2(\theta) \\
& = \cos^3(\theta) - 3\cos(\theta)\sin^2(\theta) \\ \\
\sin(3\theta) &= \sin(\theta + 2\theta)\\
& =\sin(\theta)\cos(2\theta) + \cos(\theta)\sin(2\theta) \\
& = \sin(\theta)(\cos^2(\theta) -\sin^2(\theta) ) + \cos(\theta)(2\sin(\theta)\cos(\theta)) \\
& = \cos^2(\theta)\sin(\theta) - \sin^3(\theta) + 2\cos^2(\theta)\sin(\theta) \\
& = 3\cos^2(\theta)\sin(\theta)-\sin^3(\theta)
\end{align*}
$$
Do you see the pattern already?

With these, solving for Quadruple-Angle becomes more manageable.
$$\begin{align*}
\cos(4\theta) & = \cos(\theta + 3\theta) \\
&= \cos(\theta)\cos(3\theta) - \sin(\theta)\sin(3\theta) \\
& = \cos( \theta)( \cos^3(\theta) - 3\cos(\theta)\sin^2(\theta) ) - \sin(\theta)( 3\cos^2(\theta)\sin(\theta) - \sin^3(\theta) ) \\
& = \cos^4(\theta) - 3\cos^2(\theta)\sin^2(\theta) - 3\cos^2(\theta)\sin^2(\theta) + \sin^4(\theta) \\
& = \cos^4(\theta) - 6\cos^2(\theta)\sin^2(\theta) + \sin^4(\theta) \\ \\
\sin(4\theta) & = \sin(\theta + 3\theta) \\
& = \sin(\theta)\cos(3\theta) + \cos(\theta)\sin(3\theta) \\
& = \sin(\theta)( \cos^3(\theta) - 3\cos(\theta)\sin^2(\theta) ) + \cos(\theta)( 3\cos^2(\theta)\sin(\theta) - \sin^3(\theta) ) \\
&= \cos^3(\theta)\sin(\theta) - 3\cos(\theta)\sin^3(\theta) + 3\cos^3(\theta)\sin(\theta) - \cos(\theta)\sin^3(\theta) \\
& = 4\cos^3(\theta)\sin(\theta) - 4\cos(\theta)\sin^3(\theta)
\end{align*}$$
Do you see the pattern now?

Patterns in Coefficients 

The coefficients are every other values from the Pascal's Triangle with alternating signs!

With this, we can predict the next Identities to be
$$\begin{align*}
\cos(5\theta) & = \cos^5(\theta) - 10\cos^3(\theta)\sin^2(\theta) + 5\cos(\theta)\sin^4(\theta) \\ \sin(5\theta) & = 5\cos^4(\theta)\sin(\theta) - 10\cos^2(\theta)\sin^3(\theta) + \sin^5(\theta) \\ \\
\cos(6\theta) & = \cos^6(\theta) - 15\cos^4(\theta)\sin^2(\theta) + 15\cos^2(\theta)\sin^4(\theta) - \sin^6(\theta) \\
\sin(6\theta) & = 6\cos^5(\theta)\sin(\theta) - 20\cos^3(\theta)\sin^3(\theta) + 6\cos(\theta)\sin^5(\theta) \\ \\
\cos(7\theta) & = \cos^7(\theta) - 21\cos^5(\theta)\sin^2(\theta) + 35\cos^3(\theta)\sin^4(\theta) -7\cos(\theta)\sin^6(\theta) \\
\sin(7\theta) & = 7\cos^6(\theta)\sin(\theta) - 35\cos^4(\theta)\sin^3(\theta) + 21\cos^2(\theta)\sin^5(\theta) - \sin^7(\theta)\\
\vdots
\end{align*}$$
and indeed, they are true. And in fact, even the Double-Angle Formulas fits this description.
You can find n-Angle Trigonometry using Pascal'sTriangle!

$$\begin{align*}
\cos(n\theta) & = \binom{n}{0}\cos^n(\theta) - \binom{n}{2}\cos^{n-2}(\theta)\sin^2(\theta) + \binom{n}{4}\cos^{n-4}(\theta)\sin^4(\theta) - \dots + (-1)^x \binom{n}{2x} \cos^{n-2x}(\theta) \sin^{2x}(\theta) - \dots \\
\sin(n\theta) &= \binom{n}{1}cos^{n-1}(\theta)\sin(\theta) - \binom{n}{3}\cos^{n-3}(\theta)\sin^3(\theta) + \binom{n}{5}\cos^{n-5}(\theta)\sin^5(\theta) - \dots + (-1)^x\binom{n}{2x + 1}\cos^{n-2x-1}(\theta)\sin^{2x+1}(\theta) - \dots
\end{align*}$$
So why does this turn out this way?

Euler's Formula

First, if we add the two $\cos$ and $\sin$ together and order them, we get the full Binomials. Using $n=5$ as an example, we get
$$\cos(5\theta)+\sin(5\theta) = \\
\cos^5(\theta) + 5\cos^4(\theta)\sin(\theta) - 10\cos^3(\theta)\sin^2(\theta) - 10\cos^2(\theta)\sin^3(\theta) + 5\cos(\theta)\sin^4(\theta) + \sin^5(\theta)
$$
This has a striking resemblance to $(\cos(\theta)-\sin(\theta))^5$ with one crucial difference: instead of signs alternating each term, it is grouped into pairs. This alternating every second term occurs when $i$ is involved.
In fact, if we multiply $\sin(5\theta)$ by $i$, we can perfectly factor the equation into a polynomial.
$$\cos(5\theta)+i\sin(5\theta) = (\cos(\theta)+i\sin(\theta))^5$$ This generalizes to
$$\cos(n\theta)+i\sin(n\theta) = (\cos(\theta) + i\sin(\theta))^n$$
The Euler fan will now already know where this is headed. When we see a $\cos(\theta)+i\sin(\theta)$ we immediately know it equals to $e^{i\theta}$, thanks to Euler's Formula.
With help of Euler, we find a simple proof for this:

$$\begin{align*}
\cos(n\theta)+i\sin(n\theta) & = e^{i*n\theta} \\
& = (e^{i\theta})^n \\
& = (\cos(\theta) + i\sin(\theta))^n \, \blacksquare
\end{align*}$$

and for $n \in \mathbb{N}$ we can use Pascal's Triangle to generate the necessary coefficients.
$$(\cos(\theta)+i\sin(\theta))^n = \binom{n}{0}\cos^n(\theta)+i\binom{n}{1}\cos^{n-1}(\theta)\sin(\theta)-\binom{n}{2}\cos^{n-2}(\theta)\sin^2(\theta)-i\binom{n}{3}\cos^{n-3}(\theta)\sin^3(\theta)+\dots$$ Since only every second terms of expanded polynomial carry $i$, every other coefficient belongs to either $\cos(n\theta)$ or $\sin(n\theta)$.

Once again, any interesting concept we seemed to have stumbled upon, Euler had discovered centuries ago. But this doesn't make the journey any less exciting. We can keep exploring until we come across a idea truly original.