<s>[INST] Finding the length of one side of a triangle In $\triangle\,$ABC if BC is unity and $\sin\frac{A}{2} = x_1, \sin\frac{B}{2} = x_2, \cos\frac{A}{2} = x_3, \cos\frac{B}{2} = x_4$ with $$ \left(\frac{x_1}{x_2}\right)^ {2007} - \left(\frac{x_3}{x_4}\right)^{2006} = 0 $$ then the length of AC is equal to $\ldots$.
I tried using sine rule and arrive at an expression for b but have not been able to do so.
Any help would be appreciated. Thanks in advance. 
 [/INST] In a given $\triangle$ both $\displaystyle \frac{A}{2},\frac{B}{2}\in\bigg(0,\frac{\pi}{2}\bigg).$
For $x\in(0,\dfrac{\pi}{2})$, $\sin x$ is a strictly increasing function whereas $\cos x$ is a strictly decreasing function.
Now if $\displaystyle \frac{A}{2}>\frac{B}{2}$ then $\displaystyle \sin\frac{A}{2}>\sin\frac{B}{2}$ and $\displaystyle \cos\frac{A}{2}<\cos\frac{B}{2}$. So $x_{1}>x_{2}$ and $x_{3}<x_{4}$.
So $x^{2007}_{1}x^{2006}_{4}=x^{2007}_{2}x^{2006}_{3}$ not possible.
Similarly if  $\displaystyle \frac{A}{2}<\frac{B}{2}$ then $\displaystyle \sin\frac{A}{2}<\sin\frac{B}{2}$ and $\displaystyle \cos\frac{A}{2}>\cos\frac{B}{2}$. So $x_{1}<x_{2}$ and $x_{3}>x_{4}$.
So $x^{2007}_{1}x^{2006}_{4}=x^{2007}_{2}x^{2006}_{3}$ not possible.
Therefore the only possibility is $\displaystyle \frac{A}{2}=\frac{B}{2}.$
So $x_{1}=x_{2}$ and $x_{3}=x_{4}.$
In that case $\triangle ABC$ is isosceles with $\angle ABC=\angle CAB.$ So $AC=BC=1$ unit.
</s>