Question

Sinpower6thita + cospower6thita=1-3sin square thita.cos sqare thita​

Answer 1

Answer:

Step-by-step explanation:

Sin⁶θ + Cos⁶θ = 1 - 3Sin²θCos²θ

L.H.S:

Sin⁶θ + Cos⁶θ

=> (Sin²θ)³ + (Cos²θ)³

we know that a³ + b³ = (a + b) (a² - ab + b²)

=> (Sin²θ + Cos²θ)(Sin⁴θ - Sin²θCos²θ + Cos⁴θ)

=> Sin⁴θ + Cos⁴θ - Sin²θCos²θ  (∵ Sin²θ + Cos²θ = 1)

=> (Sin²θ)² + (Cos²θ)² - Sin²θCos²θ

We know that a² + b² = (a + b)² - 2ab

=> (Sin²θ + Cos²θ)² - 2Sin²θCos²θ - Sin²θCos²θ

=> 1 - 3 Sin²θCos²θ           (∵ Sin²θ + Cos²θ = 1)

= R.H.S.

Hence proved