sinA +cosA = x

Asked by  | 10th Sep, 2012, 10:53: PM

Expert Answer:

We know: a3 + b3 = (a + b)(a2 + b2 - ab)

=> sin6A + cos6A = (sin2A + cos2A)(sin4A + cos4A - sin2Acos2A)

                          = (sin2A + cos2A)2 - 3sin2Acos2A = 1 - 3(sinAcosA)2

As sinA + cosA = x

Squaring both sides

=> sin2A + cos2A + 2sinAcosA = x2

=> sinAcosA = (x2 - 1)/2

Therefore,

sin6A + cos6A = 1 - 3(sinAcosA)2 = 1 - 3(x2 - 1)2/4 = [4 - 3(x2 - 1)2]/4

Hence Proved.

Answered by  | 11th Sep, 2012, 09:53: AM

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