Showing that Newton's method converges I'm studying newton's method for the equation $\cos(x) = x$ over $[0, \pi/2]$ and I'm asked to support the argument that given the values $x_0 = 0.78539816$, $x_1 = 0.73953613$, $x_2 = 0.73908518$ further iterations of the method will be convergent to the functions root.
I think I'm suppose to prove the 4 steps that are sufficient for convergence:


*

*Show that $f(0) \cdot f(\pi/2) < 0$. $\implies$ $f(0) \cdot f(\pi/2) = -\pi/2$.

*Show that $f'(x) \neq 0$ on $[0, \pi/2]$. $\implies$ $f'(x) = -\sin(x) - 1$, is it enough to say that $-\sin(x)$ only takes the value $1$ at $3\pi/2 + k \cdot 2\pi$ and that's never in the domain of $f(x)$?

*Show that $f''(x)$ does not change sign in $[0, \pi/2]$. $\implies$ $f''(x) = -\cos(x)$, range of $-\cos(x)$ on $[0, \pi/2]$ is only $[0, -1]$.

*A simple calculation like part 1.


I'm really struggling with parts 2 and 3. I don't know how to show this differently and I'm pretty sure it's wrong or at least insufficient.
 A: First, it helps to explicitly define $f(x) = \cos x - x$.
Second, you can just show some algebra for step 2, like so:
$f'(x) = -\sin x - 1 = 0 \\
\sin x = -1 \\
x = \frac{3\pi}{2}+2k\pi \mathrm{\ for\ } k \in \mathbb{Z}$
As you've already noted, the solutions don't fall in the domain you are looking at so $f'(x) \neq 0$ on $[0,\frac{\pi}{2}]$.
Third step seems fine to me but if this is for a class, then you should check with the teacher and see if you have the details they are looking for in an answer.
A: Not sure why you need 2. and 3 in your question.
What you need to show that the map 
$$x_{n+1} = g(x_n)$$ is a contraction.
We have
$$g\left(x\right)={{x\,\sin x+\cos x}\over{\sin x+1}}$$
$$\frac{g(x)}{dx}=-{{\cos x\,\left(\cos x-x\right)}\over{
 \left(\sin x+1\right)^2}}$$
For $x$ close to the solution $|g'(x)| < 1$. So if you start close to the answer, you will converge. For the initial value of $x =0.78539816$ $g'(x) \approx 0.018996673485909$. SO you know that $|g'(x)| < 0.02$. The rest follows.
