Question 192086
{{{sin(4alpha)}}}
={{{2sin(2alpha)cos(2alpha)}}}
={{{2*2sin(alpha)cos(alpha)(cos^2(alpha)-sin^2(alpha))}}}
={{{4sin(alpha)cos(alpha)(cos^2(alpha)-sin^2(alpha))}}}
={{{4sin(alpha)cos^3(alpha)-4sin^3(alpha)cos(alpha)}}}