Fastest way to find $\cos{6\theta}$ or higher in terms of powers of $\cos$ 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.
 A: Note that $\cos(6\theta)=\cos(2(3\theta))$, so if you know the polynomials for $2\theta$ and for $3\theta$ then you can combine them:
$$\cos(6\theta)=2\cos^{2}(3 \theta)-1=2(4\cos^{3}(\theta)-3\cos(\theta))^{2}-1.$$
Expanding this is easy. Obviously, the same idea works for higher multiples of $\theta$, provided that the multiple is a composite number. To be honest, though, in practice you should just be able to look such things up, or get a computer to calculate them for you. Doing it by hand really is tedious.
A: 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

*

*Powers of $c,s$ change by $2$ every term. Power of $c$ goes down by $2$, while that of $s$ goes up by $2$.

*The coefficients are first, third, fifth etcetera of the corresponding Pascal row (and second, fourth, sixth etcetera for $\sin n\theta$)

*Signs alternate between positive and negative.

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.
A: $$\cos{(6x)}=2\cos^2{(3x)}-1=$$
$$=2(4\cos^3x-3\cos x)^2-1=$$
$$=32\cos^6 x-48\cos^4 x+18\cos^2 x-1$$
$$\cos3x=4\cos^3 x- 3\cos x$$
A: Write the coefficients of the polynomials $1$ and $x$. Then shift the last polynomial to one power higher, double the coefficients and subtract the before-last polynomial. Every other coefficient is zero, and you can omit them.
$$\begin{align}
1\\
1\\
-1&&2\\
-3&&4\\
1&&-8&&8\\
5&&-20&&16\\
-1&&18&&-48&&32\\\cdots
\end{align}$$
To obtain the $n+1$ degree-$2n$ coefficients from those for degrees $2n-1$ and $2n-2$, you perform $n$ doublings and $n$ subtractions. This is an $O(n^2)$ process, with simple operations.
A: We can use a bootstrapping approach to obtain $\cos(nx)$.
We start with the sum-product relation:
$\cos(mx)\cos(nx)=(1/2)[\cos((m+n)x)+\cos((m-n)x)]$
Thus
$\cos((m+n)x)=2\cos(mx)\cos(nx)-\cos((m-n)x)$
So, starting with $\cos(0x)=1$ and $\cos(1x)=\cos(x)$:
$\cos(2x)=2\cos(x)\cos(x)-\cos(0x)=2\cos^2(x)-1$
$\cos(3x)=2\cos(2x)\cos(x)-\cos(x)=4\cos^3(x)-3\cos(x)$ (after first substituting for $\cos(2x)$)
$\cos(4x)=2\cos(2x)\cos(2x)-\cos(0x)=8\cos^4(x)-8\cos^2(x)+1$
$\cos(7x)=2\cos(4x)\cos(3x)-\cos(x)=64\cos^7(x)-112\cos^5(x)+56\cos^3(x)-7\cos(x)$
Thus $\cos(7x)$ is obtained without stepping through all the lower multiples; we skipped $\cos(5x)$ and $\cos(6x)$.
A: Having seen answers using the expressions in polynomials of $ \ \cos \theta \ $ for $ \ \cos (2 \theta) \ $ and $ \ \cos (3 \theta) \ \ , $  I was a bit surprised not to see
$$ \cos (6 \theta) \ \ = \ \ \cos(3·[2 \theta]) \ \ = \ \ 4·(\cos [2 \theta])^3 \ - \ 3·\cos([2 \theta])  $$
$$  = \ \ 4·( \ 2·\cos^2  \theta \ - \  1 \ )^3 \ - \ 3· ( \ 2·\cos^2  \theta \ - \  1 \ ) $$
$$  = \ \ 4·8·\cos^6  \theta \ - \ 4·3·4·\cos^4 \theta \ + \ 4·3·2·\cos^2 \theta \ -  \  4   \ - \ 6·\cos^2  \theta \ + \  3  $$
$$  = \ \ 32·\cos^6  \theta \ - \ 48·\cos^4 \theta \ + \ 18·\cos^2 \theta \ -  \  1   \ \ ,  $$
since "binomial-cubes" are not difficult to construct.
This sort of "piling-up" of identities is fairly quick for $ \ \cos(m \theta) \ \ $ when the multiplier is a product of low powers of $ \ 2 \ $ and $ \ 3 \ \ , $ but not much help if the multiplier is prime.  The "angle-addition" or "sum-to-product" formulas extend your reach a bit, as long as the multiplier doesn't get too big.  All of these are accessible if your trig course gave you decent coverage of the useful identities.  (I mention this as I've had students in second-semester calculus who had never seen the "sum-to-product" and "product-to-sum" relations.)  The more sophisticated methods mentioned aren't generally encountered earlier than in "upper-division" undergraduate courses.
Since $ \ \cos (m \theta) \ $ is an even function, one can always get a polynomial in $ \ \cos \theta \ \   $  (even/odd powers only if $ \ m \ $ is even/odd), What doesn't work quite as nicely is $ \ \sin (m \theta) \ $ with $ \ m \ $ even, for which the best you'll get is $ \ \cos \theta \ $ times a
polynomial in odd powers of $ \ \sin \theta \ \   $  (for $ \ m \ $ odd, we do just have an odd-powers-of-sine polynomial).
