Question

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

Answer 1

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

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