Solving the system $\cos\frac{a+b}2\cos\frac{a}2=\sin\frac{a}2\sin\frac{a+b}2$, $\cos\frac{b}2\cos\frac{a+b}2=\sin\frac{b}2\sin\frac{a+b}2$

Question

I have the following equations for $0<\alpha,\beta \ <\pi$ $$ \cos\frac{\alpha +\beta}{2}\;\cos\frac{\alpha}{2}=\sin\frac{\alpha}{2}\;\sin\frac{\alpha+\beta}{2}$$ $$\cos\frac{\beta}{2}\;\cos\frac{\alpha+\beta}{2}= \sin\frac{\beta}{2}\;\sin\frac{\alpha+\beta}{2}$$

I subtitude the second into the first and got $$\sin\frac{\beta}{2}\cos\frac{\alpha}{2}= \sin\frac{\alpha}{2}\cos\frac{\beta}{2}$$

I have no idea how to procced from this point, would for some help,

thank you kindly

Answer 1

Hint:

From where you have left off,

$$\sin\left(\dfrac\alpha2-\dfrac\beta2\right)=0$$

$$\implies\dfrac{\alpha-\beta}2=n\pi$$ where $n$ is any integer

$\implies\alpha-\beta=2n\pi$

But $-\pi<\alpha-\beta<\pi\implies n=0$

So, we can replace $\alpha$ with $\beta$ in any one of the given two equations to find $$\cos\left(\beta+\dfrac\beta2\right)=0$$

Answer 2

Hint:

From the first equation

$$\cos\left(\dfrac\alpha2+\dfrac{\alpha+\beta}2\right)=0$$

$\implies\dfrac{2\alpha+\beta}2=(2n+1)\dfrac\pi2$ where $n$ is any integer

$2\alpha+\beta=(2n+1)\pi$

As $0<2\alpha+\beta<3\pi\implies n=0$

Can you play with the second equation to find a similar relation ?