Prove
that 2(sin6o + cass6o)
3 (sin4o + cas4o )+1 =
0​

1 thought on “Prove<br />that 2(sin6o + cass6o)<br />3 (sin4o + cas4o )+1 = <br />0​”

1. 2(sin6θ + cos6θ) – 3 (sin4θ + cos4θ) + 1

   =2[(sin2θ)3 + (cos2θ)3] -3[(sin2θ)2 + (cos2θ)2]+1

   =2[(sin2θ + cos2θ)3 -3sin2θcos2θ (sin2θ + cos2θ)] -3[(sin2θ + cos2θ)2

   -2sin2θcos2θ]+1

   The algebraic identity

   a3 + b3 = (a+b)3 – 3ab(a+b) and

   a2 + b2 = (a+b)2 – 2ab

   are used in the above step where

   a = sin2θ and b = cos2θ.

   writing sin2θ + cos2θ = 1, we have

   = 2[1-3 sin2θ cos2θ] -3[-2 sin2θcos2θ] + 1

   = 2-6 sin2θcos2θ -3 + 6 sin2θcos2θ + 1

   = -3+3=0

   Reply