Question

SIN^6A + COS^6A= 1-3sin^2​

Answer 1

Answer:

Step-by-step explanation:

Sin⁶A + Cos⁶A

=> (Sin²A)³ + (Cos²A)³

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

=> (Sin²A + Cos²A)(Sin⁴A - Sin²ACos²A + Cos⁴A)

=> Sin⁴A - Sin²ACos²A + Cos⁴A   (∵ Sin²A + Cos²A = 1)

=> (Sin²A)² + (Cos²A)² - Sin²ACos²A

//We know that a² + b² = (a + b)² - 2ab

=> (Sin²A + Cos²A)² - 2Sin²ACos²A - Sin²ACos²A

=> 1 - 3Sin²ACos²A

=> R.H.S

Hence proved.