sin3 (x) = 3sinx - 4sin^3 x
so
sin^3 x = (3sinx - sin3x)/4
General maclaurin series is :
A Maclaurin series is a Taylor series expansion of a function about 0,
f(x)==f(0)+f'(0)x+(f''(0))/(2!)x^2+(f^((3))(0))/(3!)x^3+...+(f^((n))(0))/(n!)x^n+....
f(x) = (3sinx - sin3x)/4
f'(x) = 3cosx/4 - 3cos3x/4
f''(x) = -3sinx/4 +9sin3x/4
f'''(x) = -3cosx/4 + 27 cos3x/4
f''''(x) = 3sinx/4 -81sin3x/4
f'''''(x) = 3cosx/4 -243cos3x/4
f''''''(x) = -3sinx/4 +729sin3x/4
f'''''''(x) = -3cosx/4 +2187cos3x/4
f^n(x) = 0 if n is even
= (-1)^(n+1)/2[3^n -3]/4
f(0) = 0
f'(0) = 0
f''(0) = 0
f'''(0) = 6
f''''(0) = 0
f'''''(0) = -60
f'''''(0) = 0
f'''''(0) = 546
sin^3x=6/3!x^3 - 60/5!x^5...+ 546/7! x^7 + (f^((n))(0))/(n!)x^n+....
sin^3x=x^3 - x^5/2...+ 546/5040 x^7 + (-1)^(n+1)/2[3^n -3]/4 x^n/n!
For all n which is odd that is 1,3,5,7 ...
for even n the terms are 0