How do you propagate uncertainty for the equation $\sqrt{\left( \frac{2 \sin{\theta}}{\sqrt{3}}+\frac{1}{2} \right)^{2}+\frac{3}{4}}$

Question

I am calculating $n(\theta) = \sqrt{\left( \frac{2 \sin{\theta}}{\sqrt{3}}+\frac{1}{2} \right)^{2}+\frac{3}{4}}$. But I have a problem as $n = 59.96 \pm 0.01$.

I have tried using a first order Taylor approximation (based on this post): $$\Delta n \approx \left.\frac{dn}{d\theta}\right|_{\theta=59.96} \Delta\theta \approx (-0.24125)(0.01) \approx -0.002$$ Is this a valid method to obtain new uncertainty values? If someone knows a better way please let me know.

Answer 1

Using Taylor is for sure the good idea but I have the feeling that you are working with degrees instead of radians.

So by Taylor,around $\theta=a$, the error is $$\Delta=\frac{3 \cos (a)+4 \sqrt{3} \sin (a) \cos (a)}{3 \sqrt{4 \sin ^2(a)+2 \sqrt{3} \sin (a)+3}}\Delta \theta$$ and we have $$a=59.96 \times \frac \pi{180}\qquad \text{and} \qquad \Delta \theta=0.01\times \frac \pi{180}$$ This makes $$\Delta=0.5005708179\times0.01\times \frac \pi{180}\sim 0.000087$$

Let us check computing in degrees

• for $\theta=59.96 {}^{\circ}$, the expression is $1.731701542$
• for $\theta=59.95 {}^{\circ}$, the expression is $1.731614164$
• for $\theta=59.97 {}^{\circ}$, the expression is $1.731788896$

which confirm the given result.

Please, make me a favor : in mathematics, forget degrees for ever and always use radians.