Question

prove that sin6alpha+cos6alpha=1-3sin2alpha+3sin4alpha

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Answer 1

we have to prove that,

sin^6α + cos^6α = 1 - 3sin²α + 3sin⁴α

LHS = sin^6α + cos^6α

= (sin²α)³ + (cos²α)³

using algebraic identity,

• a³ + b³ = (a + b)(a² + b² - ab)

= (sin²α + cos²α)(sin⁴α + cos⁴α - sin²α.cos²α)

• we know, sin²α + cos²α = 1

= 1 × (sin⁴α + cos⁴α - sin²α.cos²α)

= sin⁴α + cos⁴α - sin²α.cos²α

= sin⁴α + (1 - sin²α)² - sin²α(1 - sin²α)

= sin⁴α + 1 + sin⁴α - 2sin²α - sin²α + sin⁴α

= 1 + 3sin⁴α - 3sin²α

= 1 - 3sin²α + 3sin⁴α = RHS

[hence proved ]

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