Question

prove that sin^6A + cos^6A =1 - 3sin^2.cos^3A​

Answer 1

Answer:

solution

Step-by-step explanation:

sin^6A+cos^6A

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

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

=1(sin^4A-sin^2Acos^2A+cos^4A+2sin^2Acos^2A-2sin^2Acos^2A)

=(sin2A+cos2A)=3sin2Acos2A

=1-3sin^2Acos^2A