Prove that cos^6A-sin^6A=cos2A{1-(1/4)sin^2A}

 Steps to prove cos^6A-sin^6A=cos2A{1-(1/4)sin^2A}

Take L.H.S

=cos^6A-sin^6A

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

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

=(cos^2A-sin^2A){(cos^2A)^2+(sin^2A)^2+sin^2A.cos^2A}

=cos2A{(cos^2A+sin^2A)^2-2sin^2A.cos^2A+sin^2A.cos^2A}

=cos2A{1-sin^2A.cos^2A}

[Multiply 4 and divide 1/4 at sin^2A.cos^2A]

=cos2A{1-(1/4)4sin^2A.cos^2A}

=cos2A{1-(1/4)sin^2A}

Hence cos^6A-sin^6A=cos2A{1-(1/4)sin^2A} is proved.

Detail Information:-

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