If  sin2⁡A=x, then sin⁡Asin⁡2Asin⁡3Asin⁡4A is a polynomial in x, the sum of whose coefficients is

We have sin⁡Asin⁡2Asin⁡3Asin⁡4A=sin⁡A(2sin⁡Acos⁡A)3sin⁡A−4sin3⁡A×2sin⁡2Acos⁡2A=2sin2⁡Acos⁡A×sin⁡A3−4sin2⁡A×2×2sin⁡Acos⁡A1−2sin2⁡A=8sin4⁡Acos2⁡A3−4sin2⁡A1−2sin2⁡A=8x2(1−x)(3−4x)(1−2x)=24x2−104x3+144x4−64x5The required sum =24−104+144−64=0.