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prove that:- sin^6(A) + cos^6(A) = 1 - 3sin^2(A) cos^2(A)
sin6x+cos6x=(sin2x)3+(cos2x)+3sin2xcos2x(sin2x+cos2x)-3sin2xcos2x(sin2x+cos2x) =(sin2x+cos2x)3-3sin2xcos2x =1-3sin2xcos2x
(A+B)^3=A^3+B^3+3AB(A+B) =>A^3+B^3=(A+B)^3-3AB(A+B) PUT A=SIN^2(A) AND B=COS^2(A) SO A^3+B^3 WILL BE SIN^6(A)+COS^6(A) NOW SOLVE R.H.S. TO GET THE ANSWER
(A+B)^3=A^3+B^3+3AB(A+B)
=>A^3+B^3=(A+B)^3-3AB(A+B)
PUT A=SIN^2(A) AND B=COS^2(A)
SO A^3+B^3 WILL BE SIN^6(A)+COS^6(A)
NOW SOLVE R.H.S. TO GET THE ANSWER
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