Question 1077680
That's really beyond the scope of this website. 
You need to work on integrating to get your coefficients.
{{{a[0]=(1/(2pi))*int(f(t),dt,0,2pi)=(1/(2pi))*int(sin^3(t),dt,0,2pi)=0}}}
{{{a[1]=(1/(2pi))*int(f(t)*cos(t),dt,0,2pi)=(1/(2pi))*int(cos(t)*sin^3(t),dt,0,2pi)=0}}}
Actually all of the a coefficients are zero.
{{{b[1]=(1/(2pi))*int(f(t)*sin(t),dt,0,2pi)=(1/(2pi))*int(sin^4(t),dt,0,2pi)=(1/(2pi))((3pi)/4)=3/8}}}
{{{b[2]=(1/(2pi))*int(f(t)*sin^2(t),dt,0,2pi)=(1/(2pi))*int(sin^5(t),dt,0,2pi)=(1/(2pi))((3pi)/4)=0}}}
{{{b[3]=(1/(2pi))*int(f(t)*sin^3(t),dt,0,2pi)=(1/(2pi))*int(sin^6(t),dt,0,2pi)=(1/(2pi))((5pi)/8)=5/16}}}
{{{b[4]=(1/(2pi))*int(f(t)*sin^4(t),dt,0,2pi)=(1/(2pi))*int(sin^7(t),dt,0,2pi)=(1/(2pi))((3pi)/4)=0}}}
{{{b[5]=(1/(2pi))*int(f(t),dt,0,2pi)=(1/(2pi))*int(sin^8(t),dt,0,2pi)=(1/(2pi))((35pi)/64)=35/128}}}
That's a start on the first one. 
You can find the formula also by plugging in,
{{{b[n]= (1/(2pi))*int(f(t)*sin^(n)(t),dt,0,2pi)=(1/(2pi))*int(sin^(n+3)(t),dt,0,2pi))}}}
and solving for when {{{n}}} is even.