1. ## Proving type

If (a^2-b^2)/(a^2+b^2)={sin(A-B)}/{sin(A+B)}then prove that the triangle is either a right angled triangle or an isosceles triangle.

2. Originally Posted by matsci0000
If (a^2-b^2)/(a^2+b^2)={sin(A-B)}/{sin(A+B)}then prove that the triangle is either a right angled triangle or an isosceles triangle.
By the sine rule, $\displaystyle \frac a{\sin A} = \frac b{\sin B}$. So the left side of the above equation is equal to $\displaystyle \frac{\sin^2A-\sin^2B}{\sin^2A+\sin^2B}$. Use the addition formulae on the right side of the equation, and it becomes $\displaystyle \frac{\sin^2A-\sin^2B}{\sin^2A+\sin^2B} = \frac{\sin A\cos B - \cos A\sin B}{\sin A\cos B + \cos A\sin B}.$ Multiply out the fractions, do some cancelling and rearrangement, and you get $\displaystyle \sin A\sin B\bigl(\sin(2A) - \sin(2B)\bigr) = 0$. If either of the first two factors is zero then the triangle is right-angled. If $\displaystyle \sin(2A) = \sin(2B)$ then either $\displaystyle 2A+2B = 180^\circ$ (in which case the triangle is again right-angled), or $\displaystyle A=B$ and the triangle is isosceles.

3. Originally Posted by Opalg
$\displaystyle \sin A\sin B\bigl(\sin(2A) - \sin(2B)\bigr) = 0$. If either of the first two factors is zero then the triangle is right-angled.
This is incorrect. Either of the first two factors cannot be zero.

4. Originally Posted by alexmahone
This is incorrect. Either of the first two factors cannot be zero.
Thanks — stupid mistake. The rest of the argument still works though.