Answer:
[tex]E_{r}=0.005%[/tex] %
Step-by-step explanation:
The Taylor series of a function around a value ("a" is this case) is:
[tex]f(x)=f(a)+\frac{f'(a)(x-a)}{1!}+\frac{f''(a)(x-a)^{2}}{2!}+...[/tex]
If f(x)=sin(x), the Taylor series will be:
[tex]sin(x)=sin(0)+\frac{cos(0)(x-0)}{1!}-\frac{sin(0)(x-0)^{2}}{2!}-\frac{cos(0)(x-0)^{3}}{3!}+\frac{sin(0)(x-0)^{4}}{4!}+\frac{cos(0)(x-0)^{5}}{5!}[/tex]
We truncate to 6 term as the question says.
Now, sin(0)=0 and cos(0)=1.
[tex]sin(x)=x-\frac{x^{3}}{6}+\frac{x^{5}}{120}[/tex]
Let's find sin(π/4) using the Taylor series and with the calculator, and then calculate the relative error.
Taylor series
[tex]sin(\frac{\pi}{4})=\frac{\pi}{4}-\frac{(\frac{\pi}{4})^{3}}{6}+\frac{(\frac{\pi}{4})^{5}}{120}=0.707143058[/tex]
Calculator
[tex]sin(\frac{\pi}{4})=0.7071067812[/tex]
The relative error will be:
[tex]E_{r}=\left |1-\frac{V_{aprox}}{V_{real}} \right |*100=0.005\%[/tex]
I hope it helps you!
