If the measurement of an object is in the form $\sin(\theta)=.256 \pm.004$, how do we calculate the error of $\theta$?
Using the computational method for $\sin(\theta)=\frac{H}{L}$, for $H=.254$ ,$\delta_H=.001$, $L=.992$ and $\delta_L=.001$, I get
$$\delta_{\sin\left(\theta\right),H}=\frac{H+\delta_H}{L}-\frac{H}{L}=.004....$$
and
$$\delta_{\sin\left(\theta\right),L}=\frac{H}{L+\delta_{L}}-\frac{H}{L}=.255...$$
I want to calculate $\delta_{\theta,H}$ and $\delta_{\theta,L}$ to get $\delta_{\theta}=\sqrt{\left(\delta_{\theta,H}\right)^2+\left(\delta_{\theta,L}\right)^2}=$
I looked online but could not find an answer. My assumption is too take $\pm\sin^{-1}(.004)$ but I have feeling this is incorrect.
Let's assume we are given $\sin \theta=x+\Delta x$, $x$ and $\Delta x$ and we wanna find the maximum error this would provide. Obviously $\theta$ is a function of $x$ like $f(x)$ and according to calculus we know that:$$d\theta=f'(x)dx$$Also for partial deviations (and small enough) $\Delta x$ and $\Delta \theta$ we have the latter formula approximately as following:$$\Delta \theta\approx f'(x)\Delta x$$since $\theta=f(x)=\sin^{-1}x$ we have $f'(x)=\dfrac{1}{\sqrt{1-x^2}}$. By substituting $x=0.256$ and $\Delta x=0.004$ we obtain:$$\Delta\theta\approx 0.00413788773$$ and we know that the real deviation is $$\theta(x+\Delta x)-\theta(x)=\sin^{-1}(0.260)-\sin^{-1}(0.256)=0.00414016924$$