Triangle $ABC$ have two sides of length $8$ and $17$. If $\sin 2A=\sin 2B$, find every possible value for the third side. Ok, I know that the third side has to be smaller than the sum of the other two sides, and larger than the difference of the two sides.
But the problem places a limit on the value for the third side.
$2\sin A\cos A-2\sin B\cos B$, I don't see how I could use this.
 A: With $\sin 2A=\sin 2B$ you have one of:


*

*$A=B$, which is an isosceles triangle, 

*$2A=180-2B$, meaning $A+B=90$. It follows that $C=90$ so you have a right triangle.


How can you assign the lengths in those 2 cases?
A: Let's draw a graph of the sine function:

Now, the coordinates are in radians, so we might convert the units using $180°=\pi$.  We know that every angle $x$ in a triangle satisfies $0=0°<x<180°=\pi$. If we draw an horizontal line in that interval, we can see that it intersects the sine function at most two times. We are trying to solve $\sin x = \sin y$ for (say) $x$.$x=y$ is an immediate solution. But is there another one?
By using $\sin(a-b)=\sin a\cos b - \cos a\sin b$, we can verify that $\sin x=\sin y=\sin(180°-y)$. So $x=180-y$ is another solution.
But, are those solutions actually different? Yes, except if $x=y=180-y$, or $x=y=90$. Since in the graph we can verify that in fact that is the only solution there (in the peak), we have found all solutions.  So if $\sin x=\sin y$ and $0<x,y<180°$, then $x=y$ or $x=180-y$.
