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From the fact that if $z=xy$ then the errors add in quadrature: $$\left(\frac{\sigma_{z}}{\bar{z}}\right)^2=\left(\frac{\sigma_{x}}{\bar{x}}\right)^2+\left(\frac{\sigma_{y}}{\bar{y}}\right)^2$$ Now if $z=x^2$$\sigma_{x^2}=\sqrt{2\frac{\bar{x}^2}{\sigma_{x}^2}}$$ 0 It is not exactly true that the error in measuring an angle "must be in radians" even if the degrees? The error of any measurement must be given in the same units as the measurement itself. HOWEVER, you are not asking about the error in measuring an angle, you are asking about the error in the value of a function of the angle. To do that, you are using ... 0 They don't have to be. It is just if you don't you will have a bunch of$\frac{\pi}{180}$factors when you differentiate$\sin \frac{\pi x}{180}$which is the expression for$\sin$in degrees. You could use whatever coordinate you want, but it would be silly to not use the one that makes the computation simpler. 1 Because, if you don't use radians the derivative of$\sin\theta$is not$\cos\theta$, and so your formula$dy=\cos\theta\,d\theta$doesn't hold (it needs a coefficient). 0 Try methods for first-order homogeneous difference equations? Define$n_k = N( t+k\,t_D )$,$a_k = A( t + k\,t_D )\$. $$n_{k+1} - a_k n_k = 0.$$ There is a closed-form solution, but I don't want to deprive you of all the fun! :-)