What is the Average of Cosine of Theta over a Specific Range?

[Pushoam]

Homework Statement

Homework Equations

The Attempt at a Solution

The average of \$\cos \theta$ for $\theta$ going from 0 to $\pi$ is - $\pi/2$.
Is this correcct?

 

[phyzguy]

The average of \$\cos \theta$ for $\theta$ going from 0 to $\pi$ is - $\pi/2$.
Is this correcct?

Certainly not! Cos(theta) is always between -1 and 1. How could its average over any interval ever be -1.57?

 

[cnh1995]

The average of \cosθcos⁡θ\cos \theta for θθ\theta going from 0 to ππ\pi is - π/2π/2\pi/2.

Plot the graph and see the area under the cos function between 0 to pi. What does that tell you about its average?

 

[haruspex]

Plot the graph and see the area under the cos function between 0 to pi. What does that tell you about its average?

The average required in this question is a weighted average, so its value is not evident from such a graph. Or were you just saying this is a way to see what the limits must be on any such average?

 

[Pushoam]

Certainly not! Cos(theta) is always between -1 and 1. How could its average over any interval ever be -1.57?

Sorry, it is $\frac{-2}{\pi}$. Right?

 

[haruspex]

Sorry, it is $\frac{-2}{\pi}$. Right?

If we ignore the actual question and ask for the average value of cos(θ) over the interval 0 to π then we would assume a uniform distribution of θ over that interval. And, no, the answer is not -2/π. How do you get that?
But this is irrelevant here. The given information alters the probability distribution of θ, leading to a different average.

 

[Pushoam]

$<\cos\theta>= \frac{\int_0^{\pi/2} \cos\theta d\, \theta} {\int_0^{\pi/2} d\, \theta} = \frac2{-\pi}$

 

[haruspex]

$<\cos\theta>= \frac{\int_0^{\pi/2} \cos\theta d\, \theta} {\int_0^{\pi/2} d\, \theta} = \frac2{-\pi}$

Three problems with that:

1. You wrote in post #1 that you wanted the average over 0 to π, not 0 to π/2.
2. The integral above does not produce a negative answer.
3. This is irrelevant to the question at hand. As I wrote, that formula gives the average for a uniform distribution of θ over the range. In this question you have other information, and that results in a non-uniform distribution. You need a weighted average.