# Prove that cos &#x2061;<!-- ⁡ --> ( 2 n x ) = <msubsup> &#x2211;

Prove that $\mathrm{cos}\left(2nx\right)={\sum }_{k=0}^{n}\left(-1{\right)}^{k}\left(\genfrac{}{}{0}{}{2n}{2k}\right){\mathrm{cos}}^{2\left(n-k\right)}\left(x\right)\cdot {\mathrm{sin}}^{2k}\left(x\right):=p\left(n\right)$
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Calvin Oneill
Not by induction:
$\begin{array}{rl}\mathrm{cos}2nx& =\mathrm{\Re }\left[\mathrm{cos}2nx+i\mathrm{sin}2nx\right]\\ & =\mathrm{\Re }\left[\left(\mathrm{cos}x+i\mathrm{sin}x{\right)}^{2n}\right]\\ & =\mathrm{\Re }\left[\sum _{j=0}^{2n}\left(\genfrac{}{}{0}{}{2n}{j}\right){\mathrm{cos}}^{2n-j}x\cdot {i}^{j}{\mathrm{sin}}^{j}x\right]\\ & =\sum _{k=0}^{n}\left(\genfrac{}{}{0}{}{2n}{2k}\right){\mathrm{cos}}^{2n-2k}x\cdot {i}^{2k}{\mathrm{sin}}^{2k}x& & \left(j=2k\right)\\ & =\sum _{k=0}^{n}\left(-1{\right)}^{k}\left(\genfrac{}{}{0}{}{2n}{2k}\right){\mathrm{cos}}^{2\left(n-k\right)}x\cdot {\mathrm{sin}}^{2k}x\end{array}$