Question

If \[\sin x + \cos x = m\], then prove that \[\sin^6 x + \cos^6 x = \frac{4 - 3 \left( m^2 - 1 \right)^2}{4}\], where \[m^2 \leq 2\]

Answer 1

\[\sin x + \cos x = m\] (Given)

\[\text{To prove:} \sin^6 x + \cos^6 x = \frac{4 - 3 ( m^2 - 1 )^2}{4},\text{ where }m^2 \leq 2\]

Proof: 

LHS: 

\[ \sin^6 x + \cos^6 x \]

\[ = \left( \sin^2 x \right)^3 + \left( \cos^2 x \right)^3 \]

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

\[ = 1 - 3 \sin^2 x \cos^2 x\]

RHS:

\[ \frac{4 - 3 ( m^2 - 1 )^2}{4} \]

\[ = \frac{4 - 3 \left[ \left( \sin x + \cos x \right)^2 - 1 \right]^2}{4}\]

\[ = \frac{4 - 3 \left[ \sin^2 x + \cos^2 x + 2\sin x \cos x - 1 \right]^2}{4}\]

\[ = \frac{4 - 3 \left[ \sin^2 x - \left( 1 - \cos^2 x \right) + 2 \sin x \cos x \right]^2}{4}\]

\[ = \frac{4 - 3 \times 4 \sin^2 x \cos^2 x}{4}\]

\[ = 1 - 3 \sin^2 x \cos^2 x\]

LHS = RHS

Hence proved