Q. 225.0( 5 Votes )

# Find the value of

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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