Question

Prove that: ${\cos }^{6}A-{\sin }^{6}A=\cos 2A\left(1-\frac{1}{4}{\sin }^{2}2A\right)$

Answer 1

$$\begin{gathered} \mathrm{LHS}={\cos }^{6}A-{\sin }^{6}A \\ ={\left({\cos }^{2}A\right)}^3-{\left({\sin }^{2}A\right)}^3 \\ =\left({\cos }^{2}A-{\sin }^{2}A\right)\left({\cos }^{4}A+{\sin }^{2}A.{\cos }^{2}A+{\sin }^{4}A\right) \end{gathered}$$

$$\begin{gathered} =\cos 2A\left({\cos }^{4}A+2{\sin }^{2}A{\cos }^{2}A+{\sin }^{4}A-{\sin }^{2}A{\cos }^{2}A\right) \\ =\cos 2A\left\{{\left({\sin }^{2}A+{\cos }^{2}A\right)}^2-\frac{1}{4}\times 4{\sin }^{2}A{\cos }^{2}A\right\} \end{gathered}$$

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