# Trigonometryfind sin(alpha)+cos(beta)

#### DrunkenOldFool

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

#### Sudharaka

##### Well-known member
MHB Math Helper
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
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
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.