Question

Prove the following identities:

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

`(ii) (sin^8 θ – cos^8 θ) = (sin^2 θ – cos^2 θ) (1 – 2sin^2 θ cos^2 θ)`

Answer 1

(i) We have,

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

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

`= 2[(sin^2 θ + cos^2 θ) {(sin^2 θ)^2 + (cos^2 θ)^2 – sin^2 θ cos^2 θ)]}– 3[(sin^2 θ)^2 + (cos^2 θ)^2 + 2 sin^2 θ cos^2 θ –2 sin2 θ cos2 θ] + 1`

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

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

`= 2 (1 – 3 sin^2 θ cos^2 θ) – 3(1 – 2 sin^2 θ cos^2 θ) + 1`

`= 2 – 6 sin^2 θ cos^2 θ –3 + 6 sin^2 θ cos^2 θ + 1`

= 0 = RHS

(ii) We have,

`LHS = (sin^8 θ – cos^8 θ) = (sin^4 θ)^2 – (cos^4 θ)^2`

`= (sin^4 θ – cos^4 θ) (sin^4 θ + cos^4 θ)`

`= (sin^2 θ – cos^2 θ) (sin^2 θ + cos^2 θ) (sin^4 θ + cos^4 θ)`

`= (sin^2 θ – cos^2 θ){(sin^2 θ)^2 + (cos^2 θ)^2 + 2 sin^2 θ cos^2 θ – 2sin^2 θ cos^2 θ`

`= (sin^2 θ – cos^2 θ) {(sin^2 θ + cos^2 θ)^2 – 2sin^2 θ cos^2 θ}`

`= (sin^2 θ – cos^2 θ) (1 – 2sin^2 θ cos^2 θ) = RHS`