Problem source

The function f is defined by
$$\f(x) = 2\sin x - x\,.$$
Show graphically that the equation $\f(x)=0$ has exactly one root in the interval $[\frac12\pi ,\,{\pi}]\,$. This interval is denoted $I_0$.
In order to determine the root, a sequence of intervals $I_1$, $I_2, \,\ldots$ is generated in the following way. If the interval $I_n=[a_n,b_n]\,$, and $c_n=(a_n+b_n)/2\,$, then
\begin{equation*}
I_{n+1}=
\begin{cases}
[a_n,c_n] & \text{if $\; \f(a_n)\f(c_n)<0 \,$}; \\[5pt]
[c_n,b_n] & \text{if $\; \f(c_n)\f(b_n)<0 \,$}.
\end{cases}
\end{equation*}
By using the approximations $\displaystyle \frac 1{\sqrt{2}} \approx 0.7$ and ${\pi} \approx \sqrt{10} \,$, show that $I_2=[\frac12{\pi},\,\frac58{\pi}]$ and find $I_3\,$.

Solution source

\begin{array}{c|c|c|c|c|c}
n & a_n & b_n & c_n & f(a_n) & f(c_n) & f(b_n) \\ \hline 
0 & \tfrac12 \pi & \pi & \tfrac34\pi & 2\sin(\tfrac12\pi)-\tfrac12\pi = 2-\tfrac12\pi &
2\sin(\tfrac34\pi)-\tfrac34\pi = \frac{2}{\sqrt{2}}-\tfrac34\pi &
2\sin(\pi)-\pi =-\pi \\
0 & \tfrac12 \pi & \pi & \tfrac34\pi & >0 &
 2-\frac{9}{16}10 < 0&
<0 \\ \hline
1 & \frac12 \pi & \frac34\pi & \frac58\pi & 
>0 &2\sin \tfrac58\pi - \tfrac58\pi &
< 0\\
1 & \frac12 \pi & \frac34\pi & \frac58\pi & 
>0 &
\approx 1.4 \cdot \sqrt{1.7} -\frac58\sqrt{10} < 0 & <0 \\ \hline
2 & \frac12 \pi & \frac58\pi & \frac9{16}\pi & >0 & 2\sin \frac{9}{16}\pi-\frac{9}{16}\pi & <0  \\
2 & \frac12 \pi & \frac58\pi & \frac9{16}\pi & >0 & > 0 & <0  \\
\end{array}

Threfore $I_3 = [\frac9{16}\pi,\frac58\pi]$

$\sin \frac{5\pi}{8} = \cos \frac{\pi}{8} = \sqrt{\frac12(\cos \frac{\pi}{4}+1)} = \frac{1}{\sqrt{2}}\sqrt{1 + \frac{1}{\sqrt{2}}} \approx 0.7 \cdot \sqrt{1.7}$
$\sin \frac{9\pi}{16} = \cos \frac{\pi}{16} = \sqrt{\frac12\left ( \cos \frac{\pi}{8}+1 \right)}  $

So we are comparing $2\cos \frac{\pi}{16}$ with $\frac{9}{16}\pi$ or $4 \cos^2 \frac{\pi}{16} = 2\cos \frac{\pi}{8}+2$ with $\frac{90}{16}$