if sin^3x.cos3x + cos^3x.sin3x=3/8,\nthen the value of sin4x is

sin^3[cos3x]+cos^3x[sin3x] =sin^3x[4cos^3x-3cosx]+cos^3x[3sinx-4sin^3x] now we take sinxcosx common sinxcosx[sin^2x{4cos^2x-3}+cos^2x{3-4sin^2x}] sinxcosx[4sin^2xcos^2x-3sin^2x+3cos^2x-4sin^2xcos^2x] sinxcosx[-3sin^2x+3cos^2x] sinxcosx[3*(cos2x)] 3/2*(sin2x)[cos2x] 3/4sin4x sin^3x.cos3x + cos^3x.sin3x=3/4sin4x=3/8 now sin4x=3/8divided by 3/4 which is =1/2 hence 1/2 is the ans

Approved