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Question
Prove that: \[\cos^6 A - \sin^6 A = \cos 2A\left( 1 - \frac{1}{4} \sin^2 2A \right)\]
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Solution
\[LHS = \cos^6 A - \sin^6 A\]
\[ = \left( \cos^2 A \right)^3 - \left( \sin^2 A \right)^3 \]
\[ = \left( \cos^2 A - \sin^2 A \right)\left( \cos^4 A + \sin^2 A . \cos^2 A + \sin^4 A \right)\]
\[= \cos2A\left( \cos^4 A + 2 \sin^2 A \cos^2 A + \sin^4 A - \sin^2 A \cos^2 A \right)\]
\[ = \cos2A\left\{ \left( \sin^2 A + \cos^2 A \right)^2 - \frac{1}{4} \times 4 \sin^2 A \cos^2 A \right\}\]
\[= \cos2A\left\{ \left( \sin^2 A + \cos^2 A \right)^2 - \frac{1}{4} \left( 2\text{ sin } A\text{ cos } A \right)^2 \right\}\]
\[ = \cos2A\left\{ 1 - \frac{1}{4} \left( \sin2A \right)^2 \right\}\]
\[ = \cos2A\left\{ 1 - \frac{1}{4} \sin^2 2A \right\} = RHS\]
\[\text{ Hence proved } .\]
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