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Trigonometry find sin(alpha)+cos(beta)

DrunkenOldFool

New member
Feb 6, 2012
20
If $\alpha>0$, $\beta< \pi$ and $\cos(\alpha)+\cos(\beta)-\cos(\alpha+\beta)=3/2$, then what is the value of $\sin(\alpha)+\cos(\beta)$?
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
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.
 

DrunkenOldFool

New member
Feb 6, 2012
20
Thank You Sudharaka! Can you suggest any other simpler method?(f)
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
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.
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,706
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 $\sin(\alpha)+\cos(\beta)=\frac{1+\sqrt{3}}{2}.$
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
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.