Answer:
2.5, -2.61, 0.81, -0.12
Step-by-step explanation:
The taylor series of the function sin(x) around zero is given by
[tex]sin(x)=\sum_{n=0}^{\infty}\dfrac{(-1)^k}{(2k-1)!}x^{2k+1}=x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}-\dfrac{x^7}{7!}[/tex]
Therefore,
[tex]\sin(\frac{5}{2})=\dfrac{5}{2}-\dfrac{[\frac{5}{2}]^3}{3!}+\dfrac{[\frac{5}{2}]^5}{5!}-\dfrac{[\frac{5}{2}]^7}{7!}+...[/tex]
hence the first four nonzero terms of the series are
[tex]\dfrac{5}{2}=2.5\\\\-\dfrac{[\frac{5}{2}]^3}{3!} \approx -2.61\\\\\dfrac{[\frac{5}{2}]^5}{5!} \approx 0.81\\\\-\dfrac{[\frac{5}{2}]^7}{7!} \approx -0.12[/tex]
