Q2 of 118 Page 5

Prove the following identities

sin6x + cos6x = 1 – 3 sin2x cos2x

LHS = sin6x + cos6x


= (sin2x)3 + (cos2x)3


We know that a3 + b3 = (a + b) (a2 + b2 – ab)


= (sin2x + cos2x) [(sin2x)2 + (cos2x)2 – sin2x cos2x]


We know that sin2x + cos2x = 1 and a2 + b2 = (a + b)2 – 2ab


= 1 × [(sin2x + cos2x)2 – 2sin2x cos2x – sin2x cos2x


= 12 - 3sin2x cos2x


= 1 - 3sin2x cos2x = RHS


Hence proved.


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