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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: \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 \(\frac{90}{16}\)