use the following information to complete parts a. and b. below. f(x)=sin x, a = π/3 a. find the first three…

Explanation:

Step1: Recall Taylor - series formula

The Taylor series of a function $f(x)$ centered at $a$ is given by $f(x)=\sum_{n = 0}^{\infty}\frac{f^{(n)}(a)}{n!}(x - a)^n=f(a)+f^{\prime}(a)(x - a)+\frac{f^{\prime\prime}(a)}{2!}(x - a)^2+\frac{f^{(3)}(a)}{3!}(x - a)^3+\cdots$.

Step2: Find the function value at $a=\frac{\pi}{3}$

First, find $f(a)$ where $f(x)=\sin x$ and $a = \frac{\pi}{3}$. So $f(\frac{\pi}{3})=\sin(\frac{\pi}{3})=\frac{\sqrt{3}}{2}$.

Step3: Find the first - derivative and its value at $a$

The derivative of $f(x)=\sin x$ is $f^{\prime}(x)=\cos x$. Then $f^{\prime}(\frac{\pi}{3})=\cos(\frac{\pi}{3})=\frac{1}{2}$.

Step4: Find the second - derivative and its value at $a$

The second - derivative of $f(x)$ is $f^{\prime\prime}(x)=-\sin x$. So $f^{\prime\prime}(\frac{\pi}{3})=-\sin(\frac{\pi}{3})=-\frac{\sqrt{3}}{2}$.

Step5: Write the first three non - zero terms

The first non - zero term of the Taylor series is $f(a)=\frac{\sqrt{3}}{2}$.

Answer:

$\frac{\sqrt{3}}{2}$