Find the value of the expression
$3\left[\sin ^{4}\left(\frac{3 \pi}{2}-\alpha\right)+\sin ^{4}(3 \pi+\alpha)\right]-2\left[\sin ^{6}\left(\frac{\pi}{2}+\alpha\right)+\sin ^{6}(5 \pi-\alpha)\right]$
According to the question,
Let, y = 3[sin4 (3π/2 – α) + sin 4 (3π + α)] – 2[sin6 (π/2 + α) + sin6 (5π – α)]
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
Therefore,
y = 3[(– cos α)4 + (– sin α)4] – 2[cos6 α + sin6 α]
⇒ y = 3 [cos4α + sin4α] – 2[sin6α + cos6α]
⇒ y = 3[(sin2α + cos2α)2 – 2sin2α cos2α] – 2[(sin2α)3 + (cos2α)3]
Since, we know that,
sin2α + cos2α = 1
Also, we know that,
a3+b3 = (a+b)(a2 – ab + b2)
⇒ y = 3[1 – 2sin2α cos2α] – 2[(sin2α + cos2α)( cos4α + sin4α- sin2α cos2α)]
⇒ 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