Question

Prove that:
${\cos }^{3}2x+3\cos 2x=4\left({\cos }^{6}x-{\sin }^{6}x\right)$

Answer 1

$$\begin{gathered} \mathrm{RHS}=4\left({\cos }^{6}x-{\sin }^{6}x\right) \\ =4\left[{\left({\cos }^{2}x\right)}^3-{\left({\sin }^{2}x\right)}^3\right] \end{gathered}$$

Using the identity $a^3-b^3=\left(a-b\right)\left(a^2+ab+b^2\right)$, we get

$$\begin{gathered} =4\left({\cos }^{2}x-{\sin }^{2}x\right)\left({\cos }^{4}x+{\sin }^{4}x+{\sin }^{2}x{\cos }^{2}x\right) \\ =4\left({\cos }^{2}x-{\sin }^{2}x\right)\left({\cos }^{4}x+{\sin }^{4}x+{\sin }^{2}x{\cos }^{2}x\right) \end{gathered}$$

$$\begin{gathered} =4\cos 2x\left[{\left({\cos }^{2}x-{\sin }^{2}x\right)}^2+2{\sin }^{2}x{\cos }^{2}x+{\sin }^{2}x{\cos }^{2}x\right] \\ =4\cos 2x\left[{\cos }^{2}2x+3{\sin }^{2}x{\cos }^{2}x\right] \\ =4\cos 2x\left[{\cos }^{2}2x+3\left(\frac{1-{\cos }^{2}x}{2}\right)\left(\frac{1+{\cos }^{2}x}{2}\right)\right] \end{gathered}$$
$$\begin{gathered} =4\cos 2x\left[{\cos }^{2}2x+\frac{3}{4}\left(1-{\cos }^{2}2x\right)\right] \\ =\cos 2x\left[4{\cos }^{2}2x+3\left(1-{\cos }^{2}2x\right)\right] \\ =\cos 2x\left[4{\cos }^{2}2x+3-3{\cos }^{2}2x\right] \\ =\cos 2x\left[{\cos }^{2}2x+3\right] \\ ={\cos }^{3}2x+3\cos 2x=\mathrm{LHS} \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$