I know that there is an answer here Is there a quicker way to write $\cos (n\theta)$ in terms of $\cos \theta$? but it uses the Chebychev polynomials which aren't present in my curriculum (so I can't use this method in exams). I was just wondering is there a fast way that I could obtain an expression for $\cos{6\theta}$ or higher that only involves complex numbers and trig identities? Faster than expanding $(z+\frac{1}{z})^6$ and then having to convert all the $\sin$ terms to cos which involves just a tedious amount of expansions.
Fastest way to find $\cos{6\theta}$ or higher in terms of powers of $\cos$
complex numberstrigonometry
Related Solutions
Exact answers may not have closed forms for sufficiently small angles but the general method is as follows
Say we know the exact answer to $\sin(u)$ and now want to calculate $\sin(\frac{u}{k})$
For an integer $k$
$$\sin(k\theta)$$
In general by using the rule derived from the angle sum formula:
One can expand this out into a large sum of sines and cosines
Now for each of the cosines one can substitute
$$\cos(x) = \sqrt{1 - \sin(x)^2}$$
To get an expression of the form
$$\sin(k \theta) = SOME \ HAIRY \ ALGEBRAIC \ MESS \ FOR \ SUFFICIENTLY \ LARGE \ K$$
Nevertheless we can now "solve" our equation above for $\sin(\theta)$ which allows us to expressed
$$\sin(\theta) = F(\sin(k\theta))$$
Where F is the, generally more complex inverse to the algebraic expression earlier.
Now substitute $u = k \theta, \rightarrow \theta = \frac{u}{k}$
And you've got yourself a formula for integer divisions of angles
So you can actually get exact formulas for ANY and ALL Rational Numbers.
There is one slight crux to this all. The answers can no longer be expressed in radicals when consider rational numbers whose denominators are a multiple of a prime greater than or equal to 5 because in order to express such rational numbers one EVENTUALLY is forced to solve a polynomial of degree 5 or greater (in the attempt to invert the $\sin(x)$ expression) and in general this is not solvable.
But at least you can always get an exact algebraic answer if you extend your set of tools to include ultra-radicals
To see a worked example: consider
$$\sin(2x) = 2\sin(x)\cos(x)$$
(I am cheating and using a prepared identity instead of attempting to trudge through with the angle sum identity, i assure you both methods work)
now some algebraic hocus-pocus
$$\sin(2x) = 2 \sin(x)\sqrt{1 - \sin(x)^2}$$ $$\sin(2x)^2 = 4 \sin(x)^2 (1 - \sin(x)^2)$$ $$4\sin(x)^4 - 4\sin(x)^2 + \sin(2x)^2 = 0$$
And then we bring out the big gunz... meaning the Quadratic Formula (I'll grab the positive have for now although both choices are valid)
$$\sin(x)^2 = \frac{4 + \sqrt{16 - 16\sin(2x)^2}}{8}$$ $$\sin(x) = \sqrt{\frac{1 \pm \sqrt{1 - \sin(2x)^2}}{2}}$$
Meaning
$$\sin(\frac{x}{2}) = \sqrt{\frac{1 \pm \sqrt{1 - \sin(x)^2}}{2}}$$
Course we can get even fancier by looking at something along the lines of
$$\sin \left(\frac{x}{4} \right) = \sqrt{\frac{1 \pm \sqrt{1 - {\frac{1 \pm \sqrt{1 - \sin(x)^2}}{2}}}}{2}} $$
etc... the possibilities are endless. Once you determine
$$\sin(\frac{x}{prime})$$
the multiples of that prime all become accessible (assuming you know the other prime factors as well)
To subdivide by 3 it seems:
$$\sin\left( \frac{x}{3} \right) = \frac{1}{2} \left( \sqrt[3]{\sqrt{\sin(x)^2 - 1} - \sin(x)} + \frac{1}{\sqrt[3]{\sqrt{\sin(x)^2 - 1} - \sin(x)}} \right) $$
so combining my earlier formula with this one:
$$ \sin \left( \frac{x}{12} \right) = \sqrt{\frac{1 \pm \sqrt{1 - {\frac{1 \pm \sqrt{1 - \left(\frac{1}{2} \left( \sqrt[3]{\sqrt{\sin(x)^2 - 1} - \sin(x)} + \frac{1}{\sqrt[3]{\sqrt{\sin(x)^2 - 1} - \sin(x)}} \right) \right)^2}}{2}}}}{2}} $$
What you need to notice is that, since $U^HU=I$, $$ x^Hx=(Ux)^HUx=(\lambda x)^H\lambda x=\lambda^H\lambda\,x^Hx=|\lambda|^2\,x^Hx. $$ So $|\lambda|^2=1$.
Now, if you have $\lambda^H\beta=1$, you multiply the equality by $\lambda$ to get $\beta=\lambda$.
By the way, $\lambda^H$ is the conjugate of $\lambda$, and it is universally denoted by $\overline\lambda$.
Best Answer
Since the Pascal's triangle $6$th row is $1,6,15,20,15,6,1$
$$\color{blue}{\cos 6\theta} + i\color{red}{\sin 6\theta} = (\cos \theta + i \sin \theta)^6 \equiv (c+is)^6 =$$ $$c^6 + 6c^5\cdot (is) + 15c^4(is)^2 + 20c^3(is)^3 + 15c^2(is)^4 + 6c(is)^5+(is)^6$$ $$=\color{blue}{(c^6-15c^4s^2+15c^2s^4-s^6)} + i\color{red}{(6c^5 - 20c^3s^3+6cs^5)}$$
and you get formulas for both $\cos n\theta$ and $\sin n\theta$ simultaneously (practical for small values of $n$ such as $<15$ or $20$.
You can practice this method and see that you can even go quicker. Realize that for $\cos n\theta$, the pattern is
Using these observations, I can write down in one stroke,
since Pascal's triangle $7$th row is $\color{blue}{1},7,\color{blue}{21},35,\color{blue}{35},21,\color{blue}{7},1$,
$$\cos 7\theta = c^7 - 21c^5s^2 + 35c^3s^4 - 7cs^6$$
You can try your hands at $\sin 7\theta, \cos 8\theta$ et cetera.