Prove the following : ^6A + ^6A = 1 – 3 ^2A + 3sin^4A - Vidyadip

Question

Prove the following : $\sin^6A + \cos^6A = 1 – 3 \sin^2A + 3sin^4A$

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Answer

L.H.S. $= \sin^6A + \cos^6A$
$= (\sin^2 A)^3 + (\cos^2 A)^3$
$= (\sin^2 A + \cos^2 A)^3$
$– 3sin^2A \cos^2A(\sin^2 A + \cos^2 A)$
$…[ a^3 + b^3 = (a + b)^3 – 3ab(a + b)]$
$= 1^3 – 3sin^2A \cos^2A (1)$
$= 1 – 3sin^2A \cos^2A$
$= 1 – 3 \sin^2A (1 – \sin^2A)$
$= 1 – 3 \sin^2A + 3sin^4A$
= R.H.S.

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