Question
Prove that: $2(\sin^6\theta + \cos^6\theta ) - 3 ( \sin^4\theta + \cos^4\theta ) + 1 = 0.$
✓
Answer
$LHS = 2(\sin^6\theta + \cos^6\theta ) - 3 ( \sin^4\theta + \cos^4\theta ) + 1$
$= 2( \sin^2\theta + \cos^2\theta ) ( \sin^4\theta + \cos^4\theta - \sin^2\theta .\cos^2\theta ) - 3( ( \sin^2\theta + \cos^2\theta )^2- 2sin^2\theta . \cos^2\theta + 1$
$= 2 x 1 ( ( \sin^2\theta + \cos^2\theta )^2^- 2 \sin^2\theta .\cos^2\theta - \sin^2\theta .\cos^2\theta ) - 3( (1)^2 - 2sin^2\theta . \cos^2\theta ) + 1$
$= 2 ( (1)^2 - 3 \sin^2\theta .\cos^2\theta ) - 3 ( 1 - 2 \sin^2\theta . \cos^2\theta ) + 1$
$= 2 - 6 \sin^2\theta . \cos^2\theta - 3 + 6 \sin^2\theta . \cos^2\theta + 1$
$= - 1 + 1 = 0$
$= RHS$
Hence proved.
$= 2( \sin^2\theta + \cos^2\theta ) ( \sin^4\theta + \cos^4\theta - \sin^2\theta .\cos^2\theta ) - 3( ( \sin^2\theta + \cos^2\theta )^2- 2sin^2\theta . \cos^2\theta + 1$
$= 2 x 1 ( ( \sin^2\theta + \cos^2\theta )^2^- 2 \sin^2\theta .\cos^2\theta - \sin^2\theta .\cos^2\theta ) - 3( (1)^2 - 2sin^2\theta . \cos^2\theta ) + 1$
$= 2 ( (1)^2 - 3 \sin^2\theta .\cos^2\theta ) - 3 ( 1 - 2 \sin^2\theta . \cos^2\theta ) + 1$
$= 2 - 6 \sin^2\theta . \cos^2\theta - 3 + 6 \sin^2\theta . \cos^2\theta + 1$
$= - 1 + 1 = 0$
$= RHS$
Hence proved.
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