Prove cos^3xsin3x + sin^3xcos3x = (3/4)*sin3x

Hello there!

I tried my hand at this problem and seem to be getting a rather odd conclusion,

cos^3xsin3x + sin^3xcos3x
= (1/4)*{4cos^3xsin3x + 4sin^3xcos3x}
= (1/4)*{(cos3x + 3cosx)sin3x + (3sinx-sin3x)cos3x}
= (1/4)*{ cos3xsin3x + 3cosxsin3x + 3sinxcos3x - cos3xsin3x}
= (1/4)*{3sin3xcosx + 3cos3xsinx}
= (3/4)*{sin3xcosx + cos3xsinx}
= (3/4)*{sin(3x+x)}
= (3/4)*{sin4x}

So, while the question asks to prove cos^3xsin3x + sin^3xcos3x = (3/4)*sin3x , I have ended up proving [at least I cannot see the fault in this proof] that cos^3xsin3x + sin^3xcos3x = (3/4)*sin4x

Could anyone kindly point out my mistake, or verify that this is indeed a misprint in the problem itself?

Thank you.

Hi
i tried the same problem & am getting the answer as (3/4) x (sin4x)..
so this one the real solution not the other..the only possibility to get (3/4)*sin3x is,in the step where u got (3/4)*{sin(3x+x)},here instead of 3x u must get 2x,so if u get so the previous steps goes meaningless..(3/4) x (sin4x)} is the correct answer..
so it may be due to print mistake..