Question

Prove that:
$\cos 4x=1-8{\cos }^{2}x+8{\cos }^{4}x$

Answer 1

$$\begin{gathered} \mathrm{LHS}=\cos 4x \\ =\cos \left(2\times 2x\right) \\ =2{\cos }^2\times 2x-1\left[\because \cos 2\mathrm{\theta }=2{\cos }^2\mathrm{\theta }-1\right] \\ =2{\left(2{\cos }^{2}x-1\right)}^2-1\left[\because {\cos }^{2}2\mathrm{\theta }={\left(2{\cos }^2\mathrm{\theta }-1\right)}^2\right] \end{gathered}$$

$$\begin{gathered} =2\left(4{\cos }^{4}x-4{\cos }^{2}x+1\right)-1 \\ =8{\cos }^{4}x-8{\cos }^{2}x+1 \\ =1-8{\cos }^{2}x+8{\cos }^{4}x=\mathrm{RHS} \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$