If the angles A, B, and C of a triangle are in an arithmetic progression and if a, b, and c denote the lengths of the sides opposite to A, B, and C respectively, then the value of the expression acsin2C + casin2A is

A, B, and C are in arithmetic progression ⇒ B = (A+C)/2
In any triangle, A + B + C = π
⇒ A + (A+C)/2 + C = π
⇒ 3A + 3C = 2π
⇒ A + C = 2π/3
⇒ B = π/3
By sine rule, a/sinA = b/sinB = c/sinC = k (constant)
a = ksinA, b = ksinB, c = ksinC
acsin2C + casin2A = k² sinAsinC(2sinCcosC) + k² sinCsinA(2sinAcosA)
= 2k²sinAsinC(sinCcosC + sinAcosA)
= 2k²sinAsinCsin(A+C)
= 2k²sinAsinCsin(2π/3 - B)
= 2k²sinAsinCsin(2π/3 - π/3)
= 2k²sinAsinCsin(π/3)
= 2k²sinAsinC(√3/2)
= √3k²sinAsinC
Since A + C = 2π/3, we have
2RsinA sinC = 2R(sin(2π/3-B)/2) = a/2
So acsin2C + casin2A = √3k²sinAsinC = √3/2 k² (2sinAsinC)
= √3(ac/2R2R) = √3(ac/4R²) = √3/2