Please help me to understand how to solve the P(theta)min

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In summary: So the minimum of ##P## is ##1^{-1} = 1##.In summary, we discussed the process of finding the minimum value of P(theta) and discovered two possible methods to solve it - by differentiation or by recognizing a double-angle formula and using trigonometric identities. Ultimately, the minimum value of P(theta) is 1.
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tracker890 Source h
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Dear Everyone:
Q:Please help me to understand how to solve the P(theta)min is
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?

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Thank you for your time and consideration.
 
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You're trying to prove that ##\sqrt{3} w/2## is the smallest value P can take? What have you done so far to try to solve this? We can help identify what the next step is easier if we see what your work is so far.
 
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Dear Office_Shredder
Thank you for reminding, and this problem has been solved.
(ref.https://mathhelpforum.com/threads/please-help-me-to-understand-how-to-solve-the-p-theta-min.286784/#post-958901)

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Problem:
Determine the Pmin for the sys equilibrium.sol/
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tracker890 Source h said:
Dear Everyone:
Q:Please help me to understand how to solve the P(theta)min is View attachment 268338 ?

Thank you for your time and consideration.

I know this has been solved now, but another method to do this (other than differentiation) is to notice that the denominator can be expressed as a double-angle formula. The coefficients [itex] \sqrt 3 /2 [/itex] and [itex] 1/2 [/itex] seem to stand out to me. For example:

[tex] \frac{1}{2} cos\theta + \frac{\sqrt 3}{2} sin \theta = sin(\theta + \frac{\pi}{6}) [/tex]
We can now see that we want the minimum of [itex] P [/itex] which occurs when the denominator is a maximum. Max(sin) = 1 and thus we get the answer as required (and we can see that [itex] \theta = \pi / 3 [/itex] for the sin() to equal 1)
 
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An even shorter option: If two sine waves are shifted by pi/2 (just like sine and cosine are) then they can be combined to a single sine function by adding the individual amplitudes in quadrature. As formula: ##a\sin(\theta)+b\cos(\theta) = \sqrt{a^2+b^2} \sin(\theta+x)## with some phase x we don't need to care about. This function has a maximum of ##\sqrt{a^2+b^2}##. That means we just need to calculate ##\sqrt{\left(\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2} = \sqrt{\frac 1 4 + \frac 3 4}=1##.
 
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1. What does P(theta)min represent?

P(theta)min represents the minimum possible value of the probability distribution function, which is the likelihood of a particular event or outcome occurring.

2. How do you solve for P(theta)min?

To solve for P(theta)min, you need to use mathematical techniques such as calculus or optimization methods. This involves finding the derivative of the probability distribution function and setting it equal to zero to find the minimum value.

3. Why is it important to find P(theta)min?

Finding P(theta)min allows us to understand the likelihood of a specific outcome occurring and can help us make predictions or decisions based on this information. It is also a key component in statistical analysis and modeling.

4. Can P(theta)min be negative?

No, P(theta)min cannot be negative. Probability values must be between 0 and 1, so the minimum value of the probability distribution function cannot be negative.

5. Are there any real-world applications of P(theta)min?

Yes, P(theta)min has many real-world applications in fields such as statistics, physics, and engineering. It is used to analyze data, make predictions, and optimize processes or systems.

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