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.  

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}
