Suppose that you measure three independent variables as $x = 6.5 \pm 0.8; y = 3.1 \pm 0.3; \theta = 40^\circ \pm 3^\circ $ and use these vales to compute
$$q = \frac{x^2 + y\sin\theta + 2}{x + y\cos4\theta}$$
What should be your answer for $q$ and its uncertainty?
I'm not sure if I should plug in the data without the the $\pm$ into the variables and am also not sure if I should find the uncertainty such that it is $ x_{\max} - x_{\min} = ((6.5+.8) - (40^\circ - 3^\circ)) / 2$
to get $q$ just plug in the values without the $\pm$ $$q = \frac{x^2 + ysin\theta + 2}{x + ycos4\theta}$$
for the uncertainty $\Delta q$ , first convert $\theta$ and $\Delta \theta$ from degrees to radians
and then use the formula ...
$$ \Delta q = \sqrt{ \left( \frac{\partial q}{\partial x} \Delta x \right)^2 +\left( \frac{\partial q}{\partial y} \Delta y \right)^2 +\left( \frac{\partial q}{\partial \theta} \Delta \theta \right)^2 } $$
where ...
$$\frac{\partial q}{\partial x} = \frac{2x(x+y\cos 4\theta)-(x^2 + ysin\theta + 2)}{(x + ycos4\theta)^2}$$
$$\frac{\partial q}{\partial y} = \frac{\sin\theta (x+y\cos 4\theta)-\cos 4\theta(x^2 + ysin\theta + 2)}{(x + ycos4\theta)^2}$$
$$\frac{\partial q}{\partial \theta} = \frac{y\cos\theta (x+y\cos 4\theta)+4y\sin 4\theta(x^2 + ysin\theta + 2)}{(x + ycos4\theta)^2}$$
are all evaluated by plugging in the values without the $\pm$