Advertisement Remove all ads

Prove that sin^6θ + cos^6θ = 1 – 3 sin^2θ. cos^2θ. - Geometry

Prove that sin6θ + cos6θ = 1 – 3 sin2θ. cos2θ.

Advertisement Remove all ads

Solution

LHS = sin6θ + cos6θ

        = (sin2θ)3 + (cos2θ)3

        = (sin2θ + cos2θ) (sin4θ + cos4θ - sin2θ⋅cos2θ)

        = (1)[(sin2θ + cos2θ)2 - 2sin2θ⋅cos2θ - sin2θ⋅cos2θ]

        = (1)[(1)2 - 3sin2θ⋅cos2θ]

        = 1 - 3sin2θ ⋅ cos2θ

        = RHS

  Is there an error in this question or solution?
Advertisement Remove all ads
Advertisement Remove all ads
Advertisement Remove all ads
Share
Notifications

View all notifications


      Forgot password?
View in app×