Question about uncertainties after applying cos

[ZedCar]

Homework Statement

I have conducted a simple experiment about polarized light.

The angles were 0, 10, 20, 30 etc. The device used to measure the angles has units of 1 degree. Therefore the uncertainty is +-0.5 for each angle measurement.

I was then required to produce a column beside this which gave [cos(angle)]^2

Does this therefore mean the uncertainty in this second column is +-cos(0.5)^2
That would be approximately +- 0.77 which is a very large uncertainty considering the [cos(angle)]^2 values are working out in the 0 to 1 range.

Thank you

Homework Equations

The Attempt at a Solution

 

[CompuChip]

The uncertainty in f(x), given the uncertainty \Delta x in x, is
\Delta f = \left|\frac{df(x)}{dx}\right| \cdot |\Delta x|

and, in general, if f depends on x₁, ..., x_n with corresponding uncertainties \Delta x_i,
\Delta f = \sqrt{ \left( \frac{\partial f}{\partial x_1} \right)^2 (\Delta x_1)^2 + \cdots + \left( \frac{\partial f}{\partial x_n} \right)^2 (\Delta x_n)^2}

 

[gneill]

You can get an approximation for the uncertainty in the values of cos²(x) when x has uncertainty ±Δx by plugging the extreme values of x, namely (x + Δx) and (x - Δx) into the formula to see how much the result changes. For example, suppose that the angle was 37° and the uncertainty in the angle ±0.2°. Then:

cos²(37 + 0.2) = 0.6345
cos²(37 - 0.2) = 0.6412

The difference between these extreme values is 0.007 . So the estimated uncertainty is ±0.007.

A better value can be obtained by taking the partial derivative of the given function w.r.t. the variable and multiplying by the uncertainty in that variable. If f is some function of x, i.e. y = f(x), then the uncertainty Δf(x) given uncertainty Δx in the variable x is:
\Delta f = \left| \frac{ \partial f}{\partial x}\right| \Delta x
The absolute value is taken to make the result a magnitude (positive value).

If the function has more than one variable each with its own uncertainty, f = (A,B,C...) for A±ΔA, B±ΔB, C±ΔC,... and so on, then the total uncertainty in the result of the function is given by summing the individual uncertainties in quadrature (square root of sum of squares, like vector components):

{\Delta f}^2 = \left| \frac{ \partial f}{\partial A}\right|^2 {\Delta A}^2 + \left| \frac{ \partial f}{\partial B}\right|^2 {\Delta B}^2 + \left| \frac{ \partial f}{\partial C}\right|^2 {\Delta C}^2 ...

 

[ZedCar]

That's great. Thanks very much for clarifying that for me. I can see what I was doing wrong now. Thanks again!