Find the value of 3(sinA-cosA)^4 + 6(sinA+cosA)^2 + 4(sin^6A+cos^6A)

Asked by Ayesha Maria Zahir | 21st Jul, 2013, 12:02: PM

Expert Answer:

3(sinA-cosA)^4 + 6(sinA+cosA)^2 + 4(sin^6A+cos^6A) 
= 3(1- sin2A)^2 + 6(1+sin2A) +4 [(sin^2x+cos^2x)^3 -3sin^2xcos^2x(sin^2x+cos^2x)]
= 3(1 + sin^2(2A) - 2sin2A) + 6 + 6sin2A +4 [1 - 3sin^2(2A)/4]
= 13 + 3sin^2(2A) - 3sin^2(2A)
= 13

Answered by  | 21st Jul, 2013, 10:25: PM

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