I am trying to do a measurement uncertainty calculation. I have a gaussian distributed phase angle (theta) with a mean of 0 and standard deviation of 16.6666 micro radians. The variance is the square of the standard. The formula for the measurment uses cos(theta) in the calculation. I need to know the mean, the variance and the distribution function that result from taking the cosine of theta in order to do my calculations correctly.

I wrote out the first few terms in the power series for $ \cos \theta $ and then the first few terms of the series for $ \cos^2 \theta .$ I used your hypothesis of normal distribution, the mean of $ \theta $ is $ \mu = 0$ while the variance is some $ \sigma^2 .$ Then I looked up the expected values of $ \theta^2, \; \theta^4, \; \theta^6, \; \theta^8 $ at http://en.wikipedia.org/wiki/Gaussian_distribution#Moments and used that to find good approximations for your new mean $\mu_1$ and variance $\sigma_1^2$ in $$ \mu_1 = E[ \cos \theta ] = 1  \frac{\sigma^2}{2} + \frac{\sigma^4}{8}  \frac{\sigma^6}{48} + \cdots $$ and $$ \mu_1^2 + \sigma_1^2 = E[ \cos^2 \theta ] = 1  \sigma^2 + \sigma^4  \frac{2 \sigma^6}{3} + \cdots $$ So when you subtract you get $ \sigma_1^2 \approx \frac{\sigma^4}{2} $ I will think about it some more, there is a large theory for calculating moments. But I do not see much to be done in the way of an explicit pdf or cdf. 


Given a normal distribution with mean $\mu$ and variance $\sigma^2$, $X = \mathcal{N}(\mu,\sigma^2)$, if you pass it through trigonometric functions, you can approximate the result with the new normal distributions below 1) normal distribution passed through Cosine function: $X_{\cos} = \mathcal{N}(\cos(\mu),\sigma^2\sin^2(\mu))$ so the new average is $\cos(\mu)$ and the new standard deviation is $\sigma\sin(\mu)$. 2) normal distribution passed through a Sine function: $X_{\sin} = \mathcal{N}(\sin(\mu),\sigma^2\cos^2(\mu))$ so the new average is $\sin(\mu)$ and the new standard deviation is $\sigma\cos(\mu)$. The Matlab script that I used to find these relations is below.



Hi, I know this was asked a long time ago but I have just discovered it because I require a similar solution. It is possible to generate an expression, albeit as an infinite summation. For practical purposes, the first few terms of the summation should suffice. Let $X$ denote a random variable with pdf $f_X(x)$. Let $Y=g(X)$ be a function of $X$. We can specify the cdf of $Y$, denoted $F_Y(y)$ as follows: $F_Y(y)=\mathbb{P}(g(X)\leq y)=\int\limits_{\Omega}f_X(x)\text{d}x$, where the domain of integration $\Omega$ is defined as $\Omega=\left\lbrace x:g(x)\leq y \right\rbrace$ In our case, $g(x)=\cos x$, so we need an expression for the domain of $x\in\mathbb{R}$ such that $\cos x\leq y$. This is given by $2k\pi+\arccos(y) \leq x < 2(k+1)\pi\arccos(y)\, k\in\mathbb{Z}$ So integrating over this domain, we obtain $F_Y(y)=\sum\limits_{k=\infty}^{\infty} \int\limits_{2k\pi+\arccos(y)}^{2(k+1)\pi\arccos(y)} f_X(x)\text{d}x$ Now in our case $X\sim\mathcal{N}(0,\sigma)$, so $f_X(x)=\dfrac{1}{\sigma\sqrt{2\pi}}\exp\left(\dfrac{x^2}{2\sigma^2}\right)$ and the integral of this pdf between limits is given by the cdf of the normal distribution, which we denote $\Phi$: $\int\limits_{a}^{b}f_X(x)\text{d}x = \Phi(b/\sigma)\Phi(a/\sigma)$ The cdf of $Y$ is therefore $F_Y(y)=\sum\limits_{k=\infty}^{\infty} \Phi\left(\dfrac{2(k+1)\pi\arccos(y)}{\sigma}\right)  \Phi\left(\dfrac{2k\pi\arccos(y)}{\sigma}\right)$ To compute the pdf, take the derivative with respect to $y$: $f_Y(y)=\dfrac{dF_Y(y)}{dy} = \sum\limits_{k=\infty}^{\infty} \dfrac{1}{\sqrt{1y^2}}\left( f_{X}(2(k+1)\pi\arccos(y) ) + f_{X}(2k\pi+\arccos(y)) \right)$ There are probably better ways to do this. It's possible the final summation can be rewritten or simplified. But this seems to match with a numerical check. 


A quick way to find the mean of $\cos(\theta)$, where $\theta\sim \mathcal{N}(0, \sigma^2)$, is through calculating the mean of a complex variable $e^{j\theta}=\cos(\theta)+j\sin(\theta)$. We have $E [e^{j\theta}]=e^{0+(j\sigma)^2/2}=e^{\sigma^2/2}$ which implies that the mean of the imaginary part $E [\sin(\theta)]$ equals zero and the mean of the real part $E[\cos(\theta)]$ equals $e^{\sigma^2/2}$. The answer $\mu_1$ derived by Will Jagy is in fact the Taylor series expansion of $e^{\sigma^2/2}$. The variance of $\cos(\theta)$ can be obtained as: $E[\cos^2(\theta)]E[\cos(\theta)]^2= E[\frac{1}{2}+\frac{\cos(2\theta)}{2}] E[\cos(\theta)]^2= \frac{1}{2}[1e^{\sigma^2}]^2$ 

