Question

Cos^6a+sin^6a=1-3sin^2acos^2a

Answer 1

Answer:

It isn't specified here, but I suppose we've got to prove this as an identity !

So,

LHS = sin^6 A + cos^6 A

= (sin^2 A)^3 + (cos^2 A)^3

= (sin^2 A+cos^2 A) ( sin^4 A + cos^4 A - sin^2 A cos^2 A)

= (1)(((sin^2 A)^2 + (cos^2 A)^2 - sin^2 A cos^2 A))

= ((sin^2 A + cos^2 A)^2 - 2 sin^2 A cos^2 A - sin^2 A cos^2 A)

= ((1)^2 - 3 sin^2 A cos^2 A)

= 1 - 3 sin^2 A cos^2 A

= RHS

and, hence

LHS = RHS

Hence proved.

because

a^3 + b^3 = (a+b)(a^2+b^2-ab)

Since,

a^2 + b^2 = (a+b)^2 - 2ab

Answer 2

Hope it will help you..........