Question

Prove the following identity : 

`2(sin^6θ + cos^6θ) - 3(sin^4θ + cos^4θ) + 1 = 0`

Answer 1

LHS = `2(sin^6θ + cos^6θ) - 3(sin^4θ + cos^4θ) + 1`

= `2(sin^6θ + cos^6θ) - 3(sin^4θ + cos^4θ) + 1`

= `2[(sin^2θ)^3 + (cos^2θ)^3] - 3(sin^4θ + cos^4θ) + 1`

= `2[(sin^2θ + cos^2θ){(sin^2θ)^2 + (cos^2θ)^2 - sin^2θcos^2θ}] - 3(sin^4θ + cos^4θ) + 1`

= `2{(sin^2θ)^2 + (cos^2θ)^2 - sin^2θcos^2θ} - 3(sin^4θ + cos^4θ) + 1`

= `2sin^4θ + 2cos^4θ - 2sin^2θcos^2θ - 3sin^4θ - 3cos^4θ + 1`

= `-sin^4θ - cos^4θ - 2sin^2θcos^2θ + 1`

= `-(sin^4θ + cos^4θ + 2sin^2θcos^2θ) + 1`

= `-(sin^2θ + cos^2θ)^2 + 1 = -1 + 1 = 0`