Question

2(sin⁶θ + cos⁶θ) − 3(sin⁴θ + cos⁴θ) is equal to 

A. 0 

B. 1 

C. −1 

D. None of these

Answer 1

To find: 2(sin⁶θ + cos⁶θ) – 3(sin⁴θ + cos⁴θ) 

First, we consider

sin⁶θ + cos⁶θ = (sin²θ)³ + (cos²θ)³ 

Now, as (a + b)³ = a³ + b³ + 3a²b + 3ab²

⇒ a³ + b³ = (a + b)³ – 3a²b – 3ab² 

⇒ sin⁶θ + cos⁶θ 

= (sin²θ)³ + (cos²θ)³ 

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

= 1 – 3 sin⁴θ cos²θ – 3 sin²θ cos⁴θ [∵ sin²θ + cos²θ = 1] 

= 1 – 3 sin²θ cos²θ (sin²θ + cos²θ) 

= 1 – 3 sin²θ cos²θ [∵ sin²θ + cos²θ = 1] ……(i) 

Next, we consider

sin⁴θ + cos⁴θ = (sin²θ)² + (cos²θ)² 

Now, as (a + b)² = a² + b² + 2ab 

⇒ a² + b² = (a + b)² – 2ab 

⇒ sin⁴θ + cos⁴θ 

= (sin²θ)² + (cos²θ)²

= (sin²θ + cos²θ)² – 2 sin² θ cos² θ 

= 1 – 2 sin²θ cos²θ [∵ sin²θ + cos²θ = 1] …(ii) 

Now, using (i) and (ii), we have 

2(sin⁶θ + cos⁶θ) – 3(sin⁴θ + cos⁴θ) 

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

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

= 2 – 3 = – 1