question 10 of 10 > let f(θ)=sin(10θ) centered at θ = 0. choose the correct taylor polynomial tn for f. 10θ…

Explanation:

Step1: Recall Taylor - series formula

The Taylor series of a function $f(x)$ centered at $a$ is given by $f(x)=\sum_{n = 0}^{\infty}\frac{f^{(n)}(a)}{n!}(x - a)^n$. For $f(\theta)=\sin(10\theta)$ centered at $a = 0$, we know that the Taylor - series of $\sin(x)=\sum_{n=0}^{\infty}\frac{(- 1)^n}{(2n + 1)!}x^{2n+1}=x-\frac{x^{3}}{3!}+\frac{x^{5}}{5!}-\cdots$.

Step2: Substitute $x = 10\theta$

Let $x = 10\theta$. Then $\sin(10\theta)=\sum_{n = 0}^{\infty}\frac{(-1)^n}{(2n+1)!}(10\theta)^{2n + 1}=10\theta-\frac{(10\theta)^{3}}{3!}+\frac{(10\theta)^{5}}{5!}-\cdots$.

Step3: Simplify the terms

We have $\frac{(10\theta)^{3}}{3!}=\frac{1000\theta^{3}}{6}=\frac{500}{3}\theta^{3}$ and $\frac{(10\theta)^{5}}{5!}=\frac{100000\theta^{5}}{120}=\frac{2500}{3}\theta^{5}$. So $\sin(10\theta)=10\theta-\frac{500}{3}\theta^{3}+\frac{2500}{3}\theta^{5}-\cdots$.

Answer:

$10\theta-\frac{500}{3}\theta^{3}+\frac{2500}{3}\theta^{5}-\cdots$