Prove that cos^6A+sin^6A=(1/4)(1+3cos^2(2A))

Steps to prove cos^6A+sin^6A=(1/4)(1+3cos^2(2A))

L.H.S

=cos^6A+sin^6A

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

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

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

={1-3sin^2A.cos^2A}

={1-(3/4)(2sinA.cosA)^2}

={1-(3/4)sin^2(2A)}

={1-(3/4)(1-cos^2(2A))}

=1-(3/4)+(3/4)cos^2(2A)

=(4-3)/4+(3/4)cos^2(2A)

=(1/4)(1+3cos^2(2A))

=R.H.S

Hence cos^6A+sin^6A=(1/4)(1+3cos^2(2A)) is proved. 

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