Question

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

 

Answer 1

\[LHS = \cos^6 A - \sin^6 A\]

\[ = \left( \cos^2 A \right)^3 - \left( \sin^2 A \right)^3 \]

\[ = \left( \cos^2 A - \sin^2 A \right)\left( \cos^4 A + \sin^2 A . \cos^2 A + \sin^4 A \right)\]

\[= \cos2A\left( \cos^4 A + 2 \sin^2 A \cos^2 A + \sin^4 A - \sin^2 A \cos^2 A \right)\]

\[ = \cos2A\left\{ \left( \sin^2 A + \cos^2 A \right)^2 - \frac{1}{4} \times 4 \sin^2 A \cos^2 A \right\}\]

\[= \cos2A\left\{ \left( \sin^2 A + \cos^2 A \right)^2 - \frac{1}{4} \left( 2\text{ sin } A\text{ cos } A \right)^2 \right\}\]

\[ = \cos2A\left\{ 1 - \frac{1}{4} \left( \sin2A \right)^2 \right\}\]

\[ = \cos2A\left\{ 1 - \frac{1}{4} \sin^2 2A \right\} = RHS\]

\[\text{ Hence proved } .\]