MHB Maximizing the Value of sin(alpha)+cos(beta) in the Interval [0, pi]

  • Thread starter Thread starter DrunkenOldFool
  • Start date Start date
AI Thread Summary
The discussion centers on finding the value of \(\sin(\alpha) + \cos(\beta)\) under the condition that \(\cos(\alpha) + \cos(\beta) - \cos(\alpha + \beta) = \frac{3}{2}\) with \(\alpha > 0\) and \(\beta < \pi\). Initial attempts suggest multiple potential solutions, but further analysis reveals that the only valid solution within the interval \([0, \pi]\) is \(\alpha = \beta = \frac{\pi}{3}\), leading to \(\sin(\alpha) + \cos(\beta) = \frac{1 + \sqrt{3}}{2}\). The method involves evaluating the function \(f(\alpha, \beta)\) and applying critical point analysis to confirm that this pair maximizes the function. Thus, the conclusion is that \(\alpha = \beta = \frac{\pi}{3}\) is the sole solution in the specified range.
DrunkenOldFool
Messages
20
Reaction score
0
If $\alpha>0$, $\beta< \pi$ and $\cos(\alpha)+\cos(\beta)-\cos(\alpha+\beta)=3/2$, then what is the value of $\sin(\alpha)+\cos(\beta)$?
 
Mathematics news on Phys.org
DrunkenOldFool said:
If $\alpha>0$, $\beta< \pi$ and $\cos(\alpha)+\cos(\beta)-\cos(\alpha+\beta)=3/2$, then what is the value of $\sin(\alpha)+\cos(\beta)$?

Hi DrunkenOldFool, :)

I am not sure whether there's a unique solution to your question. For if we take, \(sin(\alpha)+\cos(\beta)=a\) then using \(\cos\alpha+\cos\beta-\cos(\alpha+\beta)=\frac{3}{2}\) we have,

\[\cos\alpha+a-\sin\alpha-\cos\alpha(a-\sin\alpha)+\sin\alpha\sqrt{1-(a-sin\alpha)^2}=\frac{3}{2}\]

Solving this using Wolfram we get,

\[a=\frac{5}{4}\mbox{ and }\alpha=(2n+1)\pi\mbox{ where }n\in\mathbb{Z}\]

\[a=\frac{1+\sqrt{3}}{2}\mbox{ and }\alpha=\frac{1}{3}(6n+1)\pi\mbox{ where }n\in\mathbb{Z}\]

You can verify that both of these are solutions.

Kind Regards,
Sudharaka.
 
Thank You Sudharaka! Can you suggest any other simpler method?(f)
 
DrunkenOldFool said:
Thank You Sudharaka! Can you suggest any other simpler method?(f)

Hi DrunkenOldFool, :)

What I have shown you is that if you are given, \(\cos(\alpha)+\cos(\beta)-\cos(\alpha+\beta)=3/2\) the value of \(\sin(\alpha)+\cos(\beta)\) will take different values depending on the value you choose for \(\alpha\). If you choose \(\alpha=(2n+1)\pi\mbox{ where }n\in\mathbb{Z}\) you have a corresponding \(\beta\) value which you can find from the equation given. For this \(\alpha\) and \(\beta\) values we have \(\sin(\alpha)+\cos(\beta)=\frac{5}{4}\). Similarly for \(\alpha=\frac{1}{3}(6n+1)\pi\mbox{ where }n\in\mathbb{Z}\) and the corresponding \(\beta\) value you have \(\sin(\alpha)+\cos(\beta)=\frac{1+\sqrt{3}}{2}\).

Therefore the value of \(\sin(\alpha)+\cos(\beta)\) is not dependent on the given equation alone. For different values of \(\alpha\) you have different values for \(\sin(\alpha)+\cos(\beta)\). For the moment I cannot think of any simpler method to obtain these solutions. :)

Kind Regards,
Sudharaka.
 
Sudharaka said:
Hi DrunkenOldFool, :)

What I have shown you is that if you are given, \(\cos(\alpha)+\cos(\beta)-\cos(\alpha+\beta)=3/2\) the value of \(\sin(\alpha)+\cos(\beta)\) will take different values depending on the value you choose for \(\alpha\). If you choose \(\alpha=(2n+1)\pi\mbox{ where }n\in\mathbb{Z}\) you have a corresponding \(\beta\) value which you can find from the equation given. For this \(\alpha\) and \(\beta\) values we have \(\sin(\alpha)+\cos(\beta)=\frac{5}{4}\). Similarly for \(\alpha=\frac{1}{3}(6n+1)\pi\mbox{ where }n\in\mathbb{Z}\) and the corresponding \(\beta\) value you have \(\sin(\alpha)+\cos(\beta)=\frac{1+\sqrt{3}}{2}\).
The solution $\alpha=(2n+1)\pi$ does not work, because in that case it would follow that $\sin\alpha=0$. The equation $\sin(\alpha)+\cos(\beta)=\frac{5}{4}$ then implies that $\cos(\beta)=\frac{5}{4}$, which is not possible.

It seems that the only solution with $\alpha$ and $\beta$ lying between 0 and $\pi$ is $\alpha=\beta=\pi/3$, in which case $\sin(\alpha)+\cos(\beta)=\frac{1+\sqrt{3}}{2}.$
 
Opalg said:
The solution $\alpha=(2n+1)\pi$ does not work, because in that case it would follow that $\sin\alpha=0$. The equation $\sin(\alpha)+\cos(\beta)=\frac{5}{4}$ then implies that $\cos(\beta)=\frac{5}{4}$, which is not possible.

It seems that the only solution with $\alpha$ and $\beta$ lying between 0 and $\pi$ is $\alpha=\beta=\pi/3$, in which case $\sin(\alpha)+\cos(\beta)=\frac{1+\sqrt{3}}{2}.$

Hi Opalg, :)

Thank you. I had overlooked that. :)

Here is a method that I thought of to show \(\alpha=\beta=\frac{\pi}{3}\) is the only solution in \(\left[0,\, \pi\right]\).

Let, \(f(\alpha,\,\beta)=\cos\alpha+\cos\beta-\cos(\alpha+\beta)\) where \(0<\alpha,\,\beta<\frac{\pi}{2}\).

\[f_{\alpha}=-sin\alpha+\cos(\alpha+\beta)\mbox{ and }f_{\beta}=-\sin\beta+\cos(\alpha+\beta)\]

When, \(f_{\alpha}=f_{\beta}=0\) we get,

\[\sin\alpha=\sin\beta\]

\[\therefore \alpha=\beta\]

Therefore \((\alpha,\,\alpha)\) is a critical point of \(f\).

Also, (Refer: second partial derivative test)

\[D(\alpha,\,\alpha)=f_{\alpha\alpha}(\alpha,\, \alpha)f_{\beta\beta}(\alpha,\, \alpha)-f^{2}_{\alpha\beta}(\alpha,\,\alpha)\]

\[\therefore D(\alpha,\,\alpha)=\left(\cos 2\alpha-\cos\alpha\right)^{2}>0\]

So \((\alpha,\,\alpha)\) is a relative maximum.

\[f(\alpha,\,\alpha)=\cos\alpha+\cos\alpha-\cos(\alpha+\alpha)=2\cos\alpha-\cos(2\alpha)\]

Using the first derivative test we can show that \(f(\alpha,\,\alpha)\) has a maximum when, \(\cos\alpha=\frac{1}{2}\Rightarrow\alpha=\frac{ \pi}{3}\). Hence \(f(\alpha,\,\beta)\) maximizes at the point \(\left(\frac{\pi}{3},\,\frac{\pi}{3}\right)\). Also, \(f\left(\frac{\pi}{3},\,\frac{\pi}{3}\right)= \frac{3}{2}\). Similarly by defining \(f(\alpha,\,\beta)=\cos\alpha+\cos\beta-\cos(\alpha+\beta)\) where \(\frac{\pi}{2}<\alpha,\,\beta<\pi\) we can show that \(f\) maximizes at \(\alpha=\frac{2\pi}{3}\). However \(f\left(\frac{2\pi}{3},\,\frac{2\pi}{3}\right)=-\frac{1}{2}\).

Therefore \(\alpha=\beta=\frac{\pi}{3}\) is the only solution of the equation in \(\left[0,\,\pi\right] \).

Kind Regards,
Sudharaka.
 
Suppose ,instead of the usual x,y coordinate system with an I basis vector along the x -axis and a corresponding j basis vector along the y-axis we instead have a different pair of basis vectors ,call them e and f along their respective axes. I have seen that this is an important subject in maths My question is what physical applications does such a model apply to? I am asking here because I have devoted quite a lot of time in the past to understanding convectors and the dual...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Back
Top