Trigonometry find sin(alpha)+cos(beta)

[DrunkenOldFool]

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]

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]

Thank You Sudharaka! Can you suggest any other simpler method?(f)

 

[Sudharaka]

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]

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]

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.