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For a measurement $\theta$ with uncertainty $\Delta\theta$, the propagated error of the sine function is given by

$$\Delta(\sin\theta) = (\cos\theta)\cdot\Delta\theta$$

However I note that if $\theta$ is given in terms of radians and degrees, $\Delta\theta$ will be scaled accordingly.

i.e. $$\Delta(\sin\theta_{radian}) = (\cos\theta_{radians})\cdot\Delta\theta_{radians}$$

$$\Delta(\sin\theta_{degrees}) = (\cos\theta_{degrees})\cdot\Delta\theta_{degrees}$$

Which gives different values of error for $\Delta(\sin\theta)$

How should the error of propagation be evaluated? Is the convention for it to be evaluated in radians?

Tian
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  • Radians should be used. Note: this had nothing to do with error function – Yuriy S Feb 17 '18 at 07:28
  • @YuriyS, can I ask if this is by convention? – Tian Feb 17 '18 at 08:31
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    no, radians are the natural measure of the angle. Only for the radians the approximation $\sin x \approx x$ holds for $x<<1$. Degrees are just arbitrary, they do not have any mathematical significance – Yuriy S Feb 17 '18 at 08:33

1 Answers1

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What you are doing is approximating $f(x+h)-f(x) \approx f'(x)h$.

If $f(x) = \sin x$ then $f(x+h)-f(x) \approx (\cos x) h$.

If $f(x) = \sin ({\pi \over 180} x)$, then $f(x+h)-f(x) \approx {\pi \over 180}(\cos ({\pi \over 180} x)) h$.

The third formula in the question is incorrect.

copper.hat
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