Question

Prove that 2 (sin6θ + cos6θ) — 3 (sin4θ + cos4θ) + 1 = 0​

Answer 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

2 nd method

L.H.S. = 2(sin6 θ + cos6 θ) – 3(sin4 θ + cos4 θ) + 1

= 2(sin2 θ + cos2 θ) [sin4 θ + cos4 θ – sin2 θ.cos2 θ] – 3[(sin2 θ + cos2 θ)2 – 2 sin2 θ. cos2 θ] + 1

= 2 × 1[(sin2 θ + cos2 θ)2 – 2 sin2 θ.cos2 θ – sin2 θ.cos2 θ] – 3[(1)2 – 2 sin2 θ cos2 θ] + 1

= 2[(1)2 – 3 sin2 θ cos2 θ] – 3[1 – 2 sin2 θ.cos2 θ] + 1

= 2 – 6 sin2 θ. cos2 θ – 3 + 6 sin2 θ.cos2 θ + 1

= - 1 + 1 = 0 = R.H.S.

Hence proved.