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Thread: Proving type

  1. #1
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    Question 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.
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  2. #2
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    Quote Originally Posted by matsci0000 View Post
    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, \frac a{\sin A} = \frac b{\sin B}. So the left side of the above equation is equal to \frac{\sin^2A-\sin^2B}{\sin^2A+\sin^2B}. Use the addition formulae on the right side of the equation, and it becomes \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 \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 \sin(2A) = \sin(2B) then either 2A+2B = 180^\circ (in which case the triangle is again right-angled), or A=B and the triangle is isosceles.
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  3. #3
    MHF Contributor alexmahone's Avatar
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    Quote Originally Posted by Opalg View Post
    \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.
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  4. #4
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    Quote Originally Posted by alexmahone View Post
    This is incorrect. Either of the first two factors cannot be zero.
    Thanks — stupid mistake. The rest of the argument still works though.
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