Resultant probability distribution when taking the cosine of gaussian distributed variable 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.  
 A: 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 $y$:
$f_Y(y)=\dfrac{dF_Y(y)}{dy} = \sum\limits_{k=-\infty}^{\infty} \dfrac{1}{\sqrt{1-y^2}}\left( f_{X}(2(k+1)\pi-\arccos(y) ) + f_{X}(2k\pi+\arccos(y)) \right)$
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: Stochastic Calculus Approach:
If $W_t$ is a standard Wiener process, we know that the increment $W_t=W_0$ is normally distributed (with mean 0 and variance $t$).
Let 
$
\begin{equation}
f(t,x):=e^{t/2}cos(x)
\end{equation}
$
Then by Ito's lemma we have that $X_t:=f(t,W_t)$ satisfies:
\begin{equation}
e^{t/2}cos(W_t) = 1+ \int_0^t \frac{-e^{t/2}}{2}sin(W_t)dW_t.  
\end{equation}


*

*Mean Taking expectation on both sides of the above equation yields:
\begin{align}
\mathbb{E}[e^{t/2}cos(W_t)] & = \mathbb{E}[1+ \int_0^t \frac{-e^{t/2}}{2}sin(W_t)dW_t]
\end{align}
since $\int_0^t sin(W_t)dW_t$ is a stochastic integral, it must be a martingale; wherefrom it follows that the above expectation reduces to:
\begin{align}
\mathbb{E}[e^{t/2}cos(W_t)] & = \mathbb{E}[1]+ \mathbb{E}[\int_0^t \frac{-e^{t/2}}{2}sin(W_t)dW_t] = 1 \\
\therefore \mathbb{E}[cos(W_t)] & = e^{-t/2}.  
\end{align}
Since $\theta:=W_t-W_0$ is a $\mathscr{N}_1(0,t)$-distributed random variable.  Then we may conclude that:
\begin{align}
\mathbb{E}[cos(\theta)] & = e^{-t/2}.  
\end{align}

*Variance:
To calculate the variance we note that by the Ito isometry we have that:
\begin{align}
\mathbb{E}[\left(e^{t/2}cos(W_t)\right)^2] & = \int_0^t \mathbb{E}[\left(e^{t/2}cos(W_t)\right)^2] dt \\
\leq \int_0^t \mathbb{E}[\left(-e^{t/2}\right)^2] dt & = e^{t/2}-1.  \\
\therefore \left(e^{t/2}\right)^2\mathbb{E}[\left(cos(\theta)\right)^2] & \leq e^{t/2}-1 \\
\therefore \mathbb{E}[\left(cos(\theta)\right)^2] &\leq e^{-t/2}(1-e^{-t/2}) \\
\end{align}
Therefore we have that:
\begin{align}
Var(e^{t/2}cos(\theta))& = Var(e^{t/2}cos(W_t-W_0)) = \mathbb{E}[\left(e^{t/2}cos(W_t)\right)^2] - \mathbb{E}[cos(\theta)]^2 \\
& =\mathbb{E}[\left(e^{t/2}cos(W_t)\right)^2]- \left(e^{-t/2}\right)^2 \\
& \leq e^{-t/2}(1-e^{-t/2}) - \left(e^{-t/2}\right)^2.  
\end{align}
Since the lower-bound of $e^{t/2}cos(W_t)$ is $0$ then we have the following estimate on the variance:
\begin{align}
\therefore 0 \leq Var(cos(\theta)) & \leq e^{-t/2}\left(1-2e^{-t/2}\right).
\end{align}

A: Original Approach
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.
%% Cody Martin
% 9/2/2010
% m-file used to discover the mean and variance of a normal distribution
% passed through cosine and sine functions...results:
%   - N(mu,sig^2) -> cos(N(mu,sig^2)) = N(cos(mu),sig^2*sin^2(mu))
%   - N(mu,sig^2) -> sin(N(mu,sig^2)) = N(sin(mu),sig^2*cos^2(mu))

%% distribution of cosine and sine of a normal distribution?
cresults = zeros(0,5);
sresults = zeros(0,5); 
% loop from an average angle -90 degrees to +90 degrees
for theta = -pi/2:pi/180:pi/2
    theta1sig = pi/36;                          % standard deviation of orinigal normal distribution
    vtheta = theta + theta1sig*randn(99999,1);  % create 99999 points using this avg and std
    vctheta = cos(vtheta);                      % take the cosine of those points
    vstheta = sin(vtheta);                      % take the sine of those points
    theta_ = min(vtheta):0.01:max(vtheta);      % for plotting ideal distributions
    ctheta_ = min(vctheta):0.01:max(vctheta);   % for plotting
    stheta_ = min(vstheta):0.01:max(vstheta);   % for plotting

    figure(1); clf;
    subplot(211); hold on;
    plot(theta_,cdf('normal',theta_,theta,theta1sig),':');  % plot cdf of normal distribution with avg and std
    plot(sort(vtheta),[1:length(vtheta)]/length(vtheta));   % plot cdf of 99999 points
    plot(sort(vctheta),[1:length(vctheta)]/length(vctheta),'k','LineWidth',2); % plot cdf of cos(99999 points)
    plot(ctheta_,cdf('normal',ctheta_,cos(theta),...        % plot cdf of norm dist with new avg and std after being passed through cos()
         sqrt(theta1sig^2*sin(theta)^2)),'r:');
    plot(cos(theta)*[1 1],[0 1],'k:');                      % vertical line @ cos(theta) - shows new average matches cos(old avg)
    title('Cosine of a Normal Distribution (for Different Initial Averages)');
    legend('Norm CDF Theory','Norm CDF 99999','Cos(Norm CDF 99999)','Cos(Norm CDF) Theory');
    axis([-pi/2 pi/2 0 1])

    subplot(212); hold on;
    plot(theta_,cdf('normal',theta_,theta,theta1sig),':');
    plot(sort(vtheta),[1:length(vtheta)]/length(vtheta));
    plot(sort(vstheta),[1:length(vstheta)]/length(vstheta),'k','LineWidth',2);
    plot(stheta_,cdf('normal',stheta_,sin(theta),...
         sqrt(theta1sig^2*cos(theta)^2)),'r:');
    plot(sin(theta)*[1 1],[0 1],'k:');
    title('Sine of a Normal Distribution (for Different Initial Averages)');
    legend('Norm CDF Theory','Norm CDF 99999','Sin(Norm CDF 99999)','Sin(Norm CDF) Theory');
    axis([-pi/2 pi/2 0 1])

%   fprintf('theta: %3.0f\tstd: %5.3f\tsin(theta): %5.3f\tavg: %5.3f\tstd: %5.3f\n',theta*180/pi,theta1sig,sin(theta),mean(vstheta),std(vstheta));
    cresults = [cresults; theta theta1sig cos(theta) mean(vctheta) std(vctheta)];
    sresults = [sresults; theta theta1sig sin(theta) mean(vstheta) std(vstheta)];
end

figure(2); clf;
subplot(211); hold on;
plot(cresults(:,1),cresults(:,end));
plot(cresults(:,1),abs(theta1sig*sresults(:,3)),'r:');
title('Standard Deviation of Cosine of a Normal Distribution as a Function of the Original Average');
legend('From 99999 Points','Fit: std = |\sigmasin(\mu)|');
ylabel('std(cos(\theta_{vector})) [rad]');
xlabel('\theta [rad]');

subplot(212); hold on;
plot(sresults(:,1),sresults(:,end));
plot(sresults(:,1),abs(theta1sig*cresults(:,3)),'r:');
title('Standard Deviation of Sine of a Normal Distribution as a Function of the Original Average');
legend('From 99999 Points','Fit: std = |\sigmacos(\mu)|');
ylabel('std(sin(\theta_{vector})) [rad]');
xlabel('\theta [rad]');

figure(3); clf;
subplot(211); hold on;
plot(cresults(:,1),abs(theta1sig*sresults(:,3))-cresults(:,end));
title('Error Between \sigma^2sin^2(\mu) and std of 99999 Draws of cos(\theta)')
ylabel('Residual [rad]');
xlabel('\theta [rad]');


subplot(212); hold on;
plot(sresults(:,1),abs(theta1sig*cresults(:,3))-sresults(:,end));
title('Error Between \sigma^2cos^2(\mu) and std of 99999 Draws of cos(\theta)')
ylabel('Residual [rad]');
xlabel('\theta [rad]');

Update
As others have pointed out, this fails where $\cos(\mu)$ and $\sin(\mu)$ are near 0. Residuals between my proposed solution and the empirical results from 99999 draws are shown below.

A: 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:
$E[\cos^2(\theta)]-E[\cos(\theta)]^2= E[\frac{1}{2}+\frac{\cos(2\theta)}{2}]- E[\cos(\theta)]^2= \frac{1}{2}[1-e^{-\sigma^2}]^2$
A: 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. 
A: Given that $x\sim \mathcal{N}(\mu,\sigma^2)$, I used Mathematica to explicitly compute the integral corresponding to the expectation of $\sin(ax)$ and $\cos(ax)$. This would generalize all of the previous responses already given. 
\begin{align*} &\mathbb{E}\left \{ \sin(ax) \right \}=\sin( a\mu)  \exp \left ( \dfrac{1}{2} a^2 \sigma^2 \right ). \\
& \mathbb{E}\left \{ \cos(ax) \right \}=\cos( a\mu)  \exp \left ( \dfrac{1}{2} a^2 \sigma^2 \right ).
\end{align*}
Putting $\mu=0$ and $a=1$ will recover the desired result. 
A: For $y=\cos(x)$, the CDF of $y$
\begin{array}{l}
 F_Y \left( y \right) = \left\{ {\begin{array}{*{20}l}
   {0,y <  - 1}  \\
   {P\left( {2k\pi  + \arccos y \le x \le 2\left( {k + 1} \right)\pi  - \arccos y} \right),k \in \Bbb Z, - 1 \le y \le 1}  \\
   {1,y > 1}  \\
\end{array}} \right. \\ 
 P\left( {2k\pi  + \arccos y \le x \le 2\left( {k + 1} \right)\pi  - \arccos y} \right) \\ 
  = \sum\limits_{k =  - \infty }^{ + \infty } {\int_{2k\pi  + \arccos y}^{2\left( {k + 1} \right)\pi  - \arccos y} {f_X (x)} } dx \\ 
 f_Y (x) = \sum\limits_{k =  - \infty }^{ + \infty } {\left[ { - \left( { - \frac{1}{{\sqrt {1 - y^2 } }}} \right)f_X \left( {2\left( {k + 1} \right)\pi  - \arccos y} \right) - \left( { - \frac{1}{{\sqrt {1 - y^2 } }}} \right)f_X \left( {2k\pi  + \arccos y} \right)} \right]}  \\ 
  = \frac{1}{{\sqrt {1 - y^2 } }}\sum\limits_{k =  - \infty }^{ + \infty } {\left[ {f_X \left( {2\left( {k + 1} \right)\pi  - \arccos y} \right) + f_X \left( {2k\pi  + \arccos y} \right)} \right]}  \\ 
 \end{array}
