Question

Prove the following identities
\[\sin^6 x + \cos^6 x = 1 - 3 \sin^2 x \cos^2 x\]

Answer 1

\[\text{LHS} = \sin^6 x + \cos^6 x\]
\[ = \left( \sin^2 x \right)^3 + \left( \cos^2 x \right)^3 \]
\[ = \left( \sin^2 x + \cos^2 x \right)\left[ \left( \sin^2 x \right)^2 + \left( \cos^2 x \right)^2 - \sin^2 x \cos^2 x \right] \left[ \because a^3 + b^3 = \left( a + b \right)\left( a^2 + b^2 - ab \right) \right]\]
\[ = 1 \times \left[ \left( \sin^2 x + \cos^2 x \right)^2 - 2 \sin^2 x \cos^2 x - \sin^2 x \cos^2 x \right] \left[ \because \sin^2 x + \cos^2 x = 1\text{ and }a^2 + b^2 = \left( a + b \right)^2 - 2ab \right]\]
\[ = 1^2 - 3 \sin^2 x \cos^2 x\]
\[ = 1 - 3 \sin^2 x \cos^2 x\]
 = RHS
Hence proved.