Question

Express cos 6 theta in terms of cos theta

Answer 1

$$\begin{gathered} \cos 6\theta \\ =\cos \left(2\left(3\theta \right)\right) \\ USe\cos 2x=2{\cos }^{2}x-1 \\ =2{\cos }^{2}3\theta -1 \\ =2{\left(\cos 3\theta \right)}^2-1 \\ USe\cos 3x=4{\cos }^{3}x-3\cos x \\ =2{\left(4{\cos }^3\theta -3\cos \theta \right)}^2-1 \\ =2\left({\left(4{\cos }^3\theta \right)}^2+{\left(3\cos \theta \right)}^2-2\times 4{\cos }^3\theta \times 3\cos \theta \right)-1 \\ =2\left(16{\cos }^6\theta +9c{\mathrm{os}}^2\theta -24{\cos }^4\theta \right)-1 \\ =32{\cos }^6\theta +18c{\mathrm{os}}^2\theta -48{\cos }^4\theta -1 \end{gathered}$$