Question

Prove that: 2(sin⁶ θ + cos⁶ θ) – 3 (sin⁴ θ + cos⁴ θ) + 1 = 0.

Answer 1

2(sin⁶θ + cos⁶θ) – 3(sin⁴θ + cos⁴θ) + 1 = 0

sin⁶θ + cos⁶θ = (sin²θ + cos²θ)³ – 3sin²θ cos²θ(sin²θ + cos²θ)

Since sin⁡²θ + cos⁡²θ = 1,

sin⁶θ + cos⁶θ = 1 − 3sin²θ cos²θ

sin⁴θ + cos⁴θ = (sin²θ + cos²θ)² – 2sin²θ cos²θ

= 1 – 2sin²θ cos²θ

Substitute into the given expression

2(1 – 3sin²θ cos²θ) – 3(1 – 2sin²θ cos²θ) + 1

= 2 – 6sin²θ cos²θ – 3 + 6sin²θ cos²θ + 1

Final simplification

= (2 – 3 + 1) + (–6 + 6) sin²θ cos²θ

= 0

Hence proved:

2(sin⁶θ + cos⁶θ) – 3(sin⁴θ + cos⁴θ) + 1 = 0​

Answer 2

L.H.S.

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

= 2[(sin²θ)³ + (cos²θ)³] – 3(sin⁴θ + cos⁴θ) + 1

= 2[(sin²θ + cos²θ) (sin⁴θ – sin²θ cos²θ + cos⁴θ] – 3(sin⁴θ + cos⁴θ) + 1  ...[∵ a³ + b³ = (a + b) (a² – ab + b²)]

= 2(sin⁴θ – sin²θ cos²θ + cos⁴θ) – 3(sin⁴θ + cos⁴θ) + 1   ...[∵ sin²θ + cos²θ = 1]

= – sin⁴θ – cos⁴θ – 2sin²θ cos²θ + 1

= – (sin⁴θ + cos⁴θ + 2sin²θ cos²θ) + 1

= – (sin²θ + cos²θ)² + 1   ...[∵ (a + b)² = a² + b² + 2ab]

= –1 + 1

= 0 = R.H.S.