Question #21c71

1 Answer
Feb 1, 2018

Please refer to a Proof in Explanation.

Explanation:

I hope, the Problem is to prove :

#sin^6A+cos^6A=1-3sin^2Acos^2A#.

We know that, #sin^2A+cos^2A=1#.

Cubing, # (sin^2A+cos^2A)^3=(1)^3....................(square)#.

Using, #(x+y)^3=x^3+y^3+3xy(x+y)#, we get from #(square)#,

#(sin^2A)^3+(cos^2A)^3+3sin^2Acos^2A(sin^2A+cos^2A)=1#,

#:. sin^6A+cos^6A+3sin^2Acos^2A(1)=1, or, #,

#sin^6A+cos^6A=1-3sin^2Acos^2A#.