Question

prove that
2 [Sin^6 + cos^6] - 3 (sin^4 cos^4) + 1 = 0​

Answer 1

Step-by-step explanation:

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

=2(1-3sin²θcos²θ)-3(1-2sin²θcos²θ)+1

=2-6sin²θcos²θ-3+6sin²θcos²θ+1

=-1+1

=0 (Proved)

I hope it will help u dear

Answer 2

Answer:

Take L.H.S

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

Hence proved.