Why Is the Expected Value of cos(t)sin(t) Zero?

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SUMMARY

The expected value of the product of the functions cos(t) and sin(t) over the interval [0, 2π) is definitively zero, as established by the integral E[cos(t)sin(t)] = (1/2π)∫₀²π sin(t)cos(t) dt = 0. This conclusion arises from the properties of the sine and cosine functions, which are periodic and symmetric about the x-axis. The integral evaluates to zero due to the equal areas above and below the x-axis over one complete cycle.

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pamparana
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Hello,

Just came across this that:

E[cos(t)sin(t)] = 0

the expected value of the product of cos(t)sin(t) is 0. However, I am unable to convince myself that is the case. Can anyone help me understand why this is so?

Many thanks,

Luc
 
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I suppose the probability space is [itex][0,2 \pi)[/itex] with normalized Lebesgue measure? In that case, what it means is
[tex] \frac{1}{2\pi}\int_0^{2\pi} \sin(t) \cos(t)\,dt = 0[/tex]
 

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