The needle on a broken car speedometer is free to swing, and bounces perfectly off the pins at either end, so that if you give it a flick it is equally likely to come to rest at any angle between 0 tox.

1. What is the probability density? Hint: $\rho \left(\theta \right)d\theta$ is the probability that the needle will come to rest between$\theta$ and$\theta +d\theta$ .
2. Compute$⟨\theta ⟩$ ,$⟨{\theta }^2⟩$ , and$\sigma$ , for this distribution.
3. Compute$⟨sin\theta ⟩$ ,$⟨cos\theta ⟩$ , and$⟨co{s}^2\theta ⟩$

$$\begin{gathered} 1=\underset{-\infty }{\overset{\infty }{\int }}\rho \left(\theta \right)d\theta \backslash \mathrm{hfill} \\ 1=\underset{0}{\overset{\pi }{\int }}\rho \left(\theta \right)d\theta \backslash \mathrm{hfill} \end{gathered}$$

It is clear from this integration that the integrand must be constant,

So,

$\rho \left(\theta \right)=\frac{1}{\pi },0⩽\theta ⩽\pi$

$\rho \left(\theta \right)=0$,otherwise

The localid="1658292402403" $\rho \left(\theta \right)$ for the graph is zero except in the interval localid="1658292406704" $0⩽\theta ⩽\pi$

Calculating the expectation values.

$$\begin{gathered} ⟨\theta ⟩=\underset{-\infty }{\overset{\infty }{\int }}\theta \rho \left(\theta \right)d\theta \\ ⟨\theta ⟩=\underset{0}{\overset{\pi }{\int }}\theta \frac{1}{\pi }d\theta \\ ⟨\theta ⟩=\frac{1}{\pi }{\left.\frac{{\theta }^2}{2}\right|}_0^{\pi } \end{gathered}$$

$⟨\theta ⟩=\frac{\pi }{2}$

Expectation value is given by,

$$\begin{gathered} ⟨{\theta }^2⟩=\underset{-\infty }{\overset{\infty }{\int }}{\theta }^2\rho \left(\theta \right)d\theta \\ ⟨{\theta }^2⟩=\underset{0}{\overset{\pi }{\int }}{\theta }^2\frac{1}{\pi }d\theta \\ ⟨{\theta }^2⟩=\frac{1}{\pi }{\left.\frac{{\theta }^3}{3}\right|}_0^{\pi } \\ ⟨{\theta }^2⟩=\frac{{\pi }^2}{3} \end{gathered}$$

Standard deviation is given by,

$$\begin{gathered} \sigma =\sqrt{⟨{\theta }^2⟩-{⟨\theta ⟩}^2} \\ \sigma =\sqrt{\frac{{\pi }^2}{3}-\frac{{\pi }^2}{4}} \\ \sigma =\frac{\pi }{2\sqrt{3}} \end{gathered}$$

For $\sin \theta$ ,

$$\begin{gathered} ⟨\sin \theta ⟩=\underset{-\infty }{\overset{\infty }{\int }}\sin \theta .\rho \left(\theta \right)d\theta \\ ⟨\sin \theta ⟩=\frac{1}{\pi }\underset{0}{\overset{\pi }{\int }}\sin \theta d\theta \\ ⟨\sin \theta ⟩=-{\left.\frac{1}{\pi }\cos \theta \right|}_0^{\pi } \\ ⟨\sin \theta ⟩=\frac{2}{\pi } \end{gathered}$$

For $\cos \theta$,

$⟨\cos \theta ⟩=\underset{-\infty }{\overset{\infty }{\int }}\cos \theta \rho \left(\theta \right)d\theta$

$$\begin{gathered} ⟨\cos \theta ⟩=\frac{1}{\pi }\underset{0}{\overset{\pi }{\int }}\cos \theta d\theta \\ ⟨\cos \theta ⟩={\left.\frac{1}{\pi }\sin \theta \right|}_0^{\pi } \\ ⟨\cos \theta ⟩=0 \end{gathered}$$

And for ${\cos }^2\theta$

$$\begin{gathered} ⟨\cos \theta ⟩=\frac{1}{\pi }\underset{0}{\overset{\pi }{\int }}\cos \theta d\theta \\ ⟨\cos \theta ⟩={\left.\frac{1}{\pi }\sin \theta \right|}_0^{\pi } \\ ⟨\cos \theta ⟩=0 \end{gathered}$$

Since, ${\cos }^2\theta =\frac{1}{2}\cos \left(2\theta \right)+\frac{1}{2}$

Therefore,

$⟨{\cos }^2\theta ⟩=\frac{1}{\pi }\underset{0}{\overset{\pi }{\int }}\left(\frac{1}{2}\cos \left(2\theta \right)+\frac{1}{2}\right)d\theta$

For the first integral, taking

$2\theta =u$

Then,$d\theta =\frac{1}{2}du$$d\theta =\frac{1}{2}du$

And the bound of integration becomes$\theta \to \pi ,u\to 2\pi$and as$\theta \to 0,u\to 0$

$$\begin{gathered} ⟨{\cos }^2\theta ⟩=\frac{1}{2\pi }\underset{0}{\overset{2\pi }{\int }}\cos udu+{\left.\frac{1}{2\pi }\right|}_0^{2\pi } \\ ⟨{\cos }^2\theta ⟩={\left.\frac{1}{2\pi }\sin u\right|}_0^{2\pi }+\frac{1}{2} \\ ⟨{\cos }^2\theta ⟩=0+\frac{1}{2} \\ ⟨{\cos }^2\theta ⟩=\frac{1}{2} \end{gathered}$$

Hence the solution is :

$$\begin{gathered} \rho \left(\theta \right)=\left\{\begin{array}{ll}\frac{1}{\pi },& 0\le \theta \le \pi \\ 0& otherwise\end{array}\right. \\ ⟨\theta ⟩=\frac{\pi }{2},⟨{\theta }^2⟩=\frac{{\pi }^2}{3},\sigma =\frac{\pi }{2\sqrt{3}} \\ ⟨\sin \theta ⟩=\frac{2}{\pi },⟨\cos \theta ⟩=0,⟨{\cos }^2\theta ⟩=\frac{1}{2}. \end{gathered}$$