4 added Matlab script that helped me discover this relationship

The Matlab script that I used to find these relations is below.

%% Cody Martin% 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 degreesfor theta = -pi/2:pi/36:pi/2    theta1sig = pi/36;                          % standard deviation of orinigal normal distribution    vtheta = theta + theta1sig*randn(9999,1);   % create 9999 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 9999 points    plot(sort(vctheta),[1:length(vctheta)]/length(vctheta),'k','LineWidth',2); % plot cdf of cos(9999 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 9999','Cos(Norm CDF 9999)','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 9999','Sin(Norm CDF 9999)','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)];figure(2); clf;subplot(211); hold on;title('Standard Deviation of Cosine of a Normal Distribution as a Function of the Original Average');legend('From 9999 Points','Fit: std = |\sigmasin(\mu)|');ylabel('std(cos(\theta_{vector})) [rad]');xlabel('\theta [rad]');subplot(212); hold on;title('Standard Deviation of Sine of a Normal Distribution as a Function of the Original Average');legend('From 9999 Points','Fit: std = |\sigmacos(\mu)|');ylabel('std(sin(\theta_{vector})) [rad]');xlabel('\theta [rad]');

3 clarified equations in words

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 $|\sigma\cos(\mu)|$.

2 deleted 4 characters in body

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:

$X_{\cos} = \mathcal{N}(\cos(\mu),\sigma^2\sin^2(\mu))$

2) normal distribution passed through a Sine function:

$X_{\sin} = \mathcal{N}(\sin(\mu),\sigma^2\cos^2(\mu))$

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