Prove that cos^3A.cos3A+sin^3A.sin3A=cos^32A

Steps to prove cos^3A.cos3A+sin^3A.sin3A=cos^32A

L.H.S

=cos^3A.cos3A+sin^3A.sin3A

=cos^3A.(4cos^3A-3cosA)+sin^3A.(3sinA-4sin^3A)

=4cos^6A-3cos^4A+3sin^4A-4sin^6A

=4(cos^6A-sin^6A)-3(cos^4A-sin^4A)

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

=4(cos^2A-sin^2A){(cos^2A)^2+cos^2A.sin^2A+(sin^2A)^2}-3(cos^2A-sin^2A)(cos^2A+sin^2A)

=4(cos^2A-sin^2A)[4{(cos^2A+sin^2A)^2-2cos^2A.sin^2A+cos^2A.sin^2A}-3.1]

=cos2A(4-4sin^2A.cos^2A-3)

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

=cos2A(1-sin^2A)

=cos2A.cos^2(2A)=cos^32A

Hence cos^3A.cos3A+sin^3A.sin3A=cos^32A is proved.

Detail Information:-

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