Calculate uncertainty of sine function result I have an angle given in degrees:
$$\theta_{\min} = 63^{\circ} \pm 0.5^{\circ}$$
I need to calculate it's sine and still know the uncertainty of the value:
$$n = 2\sin(\theta_{\min}) = 1.7820130483767356 \pm ???$$
How do I calculate the value represented by ????
Edit: I cheated and had a look in my friends work. This is how he did it:
$$u_C=\sqrt{\left(\dfrac{\partial n}{\partial \theta_\min}u_C(\theta_\min)\right)^2}=\sqrt{\left(2\cos63^\circ\cdot\dfrac{0.5^\circ}{\sqrt{12}}\right)^2}=\sqrt{(0.908\cdot0.144)^2}=0.131$$
But I don't seem to understand that, though I encountered similar thing before.
 A: Let's write your stuff in a cleaner way:
$$n_\text{avg} = 2\sin(63°) = 1.7820130483767356$$
$$n = n_\text{avg} \pm^{u}_l \ .$$
Then
$$u = 2\sin(63.5°) - 2\sin(63°)$$
$$l = 2\sin(63°) - 2\sin(62.5°)$$
The way your friend does it is via first order Taylor approximation:
$$\Delta n \approx \left.\frac{dn}{d\theta}\right|_{\theta=\theta_\text{min}} \cdot \Delta\theta$$
Your buddy uses the absolute value in a sloppy notation. Evaluate the derivative, use $|\Delta\theta| = 0.5°$ and take absolute values to your convenience. I have no idea where the $\sqrt{12}$ that your buddy uses is from, so you might not wanna trust his result.
A: Use a trigonometry identity
$$\sin(a\pm b)=\sin a\cos b\pm\sin b \cos a$$
So, you will get
$$2\sin\theta=2(\sin63^{\circ}\cos0.5^{\circ}\pm\sin0.5^{\circ}\cos63^{\circ})\approx1.78195\pm0.00792353$$
A: Why not to use Taylor; around $x=a$, $$\sin(x+a)=\sin (a)+(x-a) \cos (a)+O\left((x-a)^2\right)$$ So, $$\sin(x+a)-\sin (a)=(x-a) \cos (a)+O\left((x-a)^
2\right)$$ using $\cos(a)=\sqrt{1-\sin^2(a)}$
