Question

use one iteration of newtons method with an initial guess of $x_1=\frac{pi}{2}$ to approximate the solution to $cos(x)=x$. the approximation, $x_2$ equals
0
$\frac{3pi}{13}$
$\frac{pi}{4}$
1
it is not possible to compute $x_2$.

Explanation:

Step1: Rewrite the equation

Rewrite $\cos(x)=x$ as $f(x)=\cos(x)-x = 0$.

Step2: Find the derivative

The derivative of $f(x)$ is $f'(x)=-\sin(x)-1$.

Step3: Apply Newton's Method formula

Newton's Method formula is $x_{n + 1}=x_{n}-\frac{f(x_{n})}{f'(x_{n})}$. Here $n = 1$, $x_1=\frac{\pi}{2}$.
First, find $f(x_1)$:
$f(\frac{\pi}{2})=\cos(\frac{\pi}{2})-\frac{\pi}{2}=0 - \frac{\pi}{2}=-\frac{\pi}{2}$.
Then, find $f'(x_1)$:
$f'(\frac{\pi}{2})=-\sin(\frac{\pi}{2})-1=-1 - 1=-2$.
Now, calculate $x_2$:
$x_2=x_1-\frac{f(x_1)}{f'(x_1)}=\frac{\pi}{2}-\frac{-\frac{\pi}{2}}{-2}=\frac{\pi}{2}-\frac{\pi}{4}=\frac{\pi}{4}$.

Answer:

$\frac{\pi}{4}$