Answer :

Let, y =


We know that-


sin(3π/2 – α) = -cos α


sin(3π + α) = -sin α


sin(π/2 + α) = cos α


sin(5π – α) = sin α


y =


y = 3 [cos4α + sin4α] – 2[sin6α + cos6α]


y = 3[(sin2α + cos2α)2 – 2sin2α cos2α] – 2[(sin2α)3 + (cos2α)3]


sin2α + cos2α = 1


y = 3[1 – 2sin2α cos2α] – 2[(sin2α + cos2α)( cos4α + sin4α- sin2α cos2α)] { a3+b3 = (a+b)(a2 – ab + b2)}


y = 3[1 – 2sin2α cos2α] – 2[cos4α + sin4α- sin2α cos2α]


y = 3[1 – 2sin2α cos2α] – 2[(sin2α + cos2α)2 – 2sin2α cos2α - sin2α cos2α]


y = 3[1 – 2sin2α cos2α] – 2[1 – 3sin2α cos2α]


y = 3 – 6sin2α cos2α – 2 + 6 sin2α cos2α


y = 1


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