The formula for $\cos nx$ without Demoivre's theorem? De Moirve's theorem easily derives $\cos\left(nx\right)$ in terms of decreasing powers of $\cos\left(x\right)$ and increasing powers of $\sin\left(x\right)$. 
But I'd like to use trignomoetry to derive this simple recursion.
I have tried but the recursion soon gets too complex.
Is there a trignometric way ?.
 A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,}
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Use the Recursive Identity:

\begin{align}
\cos\pars{nx} & =
-\cos\pars{\bracks{n - 2}x} + 2\cos\pars{\bracks{n - 1}x}\cos\pars{x}
\end{align}

For instance,

\begin{align}
\cos\pars{4x} &= -\cos\pars{2x} + 2\cos\pars{3x}\cos\pars{x}
\\[5mm] & =
-\cos\pars{2x} +
2\bracks{-\cos\pars{x} + 2\cos\pars{2x}\cos\pars{x}}\cos\pars{x}
\\[5mm] & =
\bracks{-1 + 4\cos^{2}\pars{x}}\cos\pars{2x} - 2\cos^{2}\pars{x}
\\[5mm] &=
\bracks{-1 + 4\cos^{2}\pars{x}}\bracks{-1 + 2\cos^{2}\pars{x}} - 2\cos^{2}\pars{x}
\\[5mm] & =\bbx{\ds{%
8\cos^{4}\pars{x}} - 8\cos^{2}\pars{x} + 1}
\end{align}
A: We may prove that $\cos(nx)$ is always a polynomial in $\cos(x)$ in a quite simple way.

Claim: for every $n\in\mathbb{N}$, $\cos(nx)=T_n(\cos x)$ where $T_n(z)$ is a polynomial with degree $n$.

Proof: The claim is trivial for $n=0$ ($T_0(z)=1$) and for $n=1$ ($T_1(z)=z$). 
By the cosine addition formulas,
$$\cos(nx)+\cos((n+2)x) = 2\cos(x)\cos((n+1)x) \tag{1}$$
or
$$ \cos((n+2)x) = 2\cos(x)\cos((n+1)x)-\cos(nx).\tag{2} $$
The last identity provides a way for proving the claim by induction, since it gives:
$$ T_{n+2}(z) = 2z\cdot T_{n+1}(z)-T_{n}(z).\tag{3}$$
If $T_n(z)$ and $T_{n+1}(z)$ are polynomials with degree $n$ and $n+1$, by $(3)$ we have that $T_{n+2}(z)$ is a polynomial with degree $1+\partial T_{n+1}=n+2$.

These polynomials are commonly known as Chebyshev polynomials of the first kind.
The explicit formula
$$ T_n(z) = \sum_{r=0}^{\left\lfloor\frac{n}{2}\right\rfloor}\frac{(-1)^r}{n-r}\binom{n-r}{r}(2z)^{n-2r}\tag{4}$$
is equivalent to
$$\cos(nx) = \sum_{r=0}^{\left\lfloor\frac{n}{2}\right\rfloor}\frac{(-1)^r}{n-r}\binom{n-r}{r}(2\cos x)^{n-2r}\tag{5} $$
and $(4)$, too, can be proved by induction on $n$ through $(3)$.

If we consider that, by construction, the roots of $T_n(z)$ are given by $\cos\left(\frac{\pi(2k-1)}{2n}\right)$ for $k=1,2,\ldots,n$, and that Vieta's formulas gives a neat set of relations between the elementary symmetric function of the roots and the coefficients of a polynomial, $(4)$ gives us a way for evaluating many interesting trigonometric sums, like, for instance:
$$ \sum_{k=1}^{n}\cos^2\left(\frac{\pi(2k-1)}{2n}\right)=\frac{n}{2}.$$
A: Check out Chebyshev polynomials
A: Is this OK? Trigonometry Cosine Multiple Angle Formulae from Ken Ward's Mathematics Pages.
