Start from the MacLaurin expansion of #sint#:
#sint = sum_(n=0)^oo (-1)^n t^(2n+1)/((2n+1)!)#
substitute #t=2x#
#f(x) = sin(2x) = sum_(n=0)^oo (-1)^n (2x)^(2n+1)/((2n+1)!)#
So for #x= 1/2#
#f(1/2) = sum_(n=0)^oo (-1)^n (2*1/2)^(2n+1)/((2n+1)!) = sum_(n=0)^oo (-1)^n /((2n+1)!)#
Using Lagrange's formula for the error we have:
#R_n(1/2) = [d^(n+1)/(dx)^(n+1) sin(2x)]_(x=c)/((n+1)!)(1/2)^(n+1)#
where #0 < c < 1/2#.
Now we have:
#d^(n+1)/(dx)^(n+1) sin(2x) = 2^(n+1)sin(x+((n+1)pi)/2)#
so as #abs sin x <= 1#:
#abs([d^(n+1)/(dx)^(n+1) sin(2x)]_(x=c)) <= 2^(n+1)#
and then:
#abs(R_n(1/2)) <= 2^(n+1)/((n+1)!)(1/2)^(n+1)#
that is:
#abs(R_n(1/2)) <= 1/((n+1)!)#
Thus we have to determine #n# such that:
#1/((n+1)!) <= 0.01 #
which is satisfied for #n >= 4#
In conclusion we have:
#f(1/2) = 1-1/6+1/120-1/5040+R_4 = (5040-840+42-1)/5040+R_4 = 4241/5040+R_4 ~=0.84#
Where #abs(R_4) < 1/120#
