If sin A + Cos A = a and sin³A + cos³A = b ,then find the value of 3a – 2b

Answers (1)

3a - 2b = 3 sin A + 3 Cos A  - 2 Sin³ A - 2 Cos³ A
            =  sin A + Cos A + 2 Sin A  (1 - Sin² A) + 2  Cos A (1 - Cos ² A)
            = sin A + Cos A + 2 sin A cos² A + 2 cos A Sin² A
            = (sin A + Cos A)  + 2 sin A cos A ( sin  A + Cos A)
            =  (sin A + Cos A) (1 + sin 2 A)

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