Prove $\sin(A-B)/\sin(A+B)=(a^2-b^2)/c^2$

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Solution 1

HINT:

Use Prove $ \sin(A+B)\sin(A-B)=\sin^2A-\sin^2B $

and $\sin(A+B)=\sin(\pi-C)=?$

and Sine Law

Solution 2

$$\frac{\sin(A-B)}{\sin(A+B)}=\frac{\sin A\cos B-\sin B\cos A}{\sin A\cos B+\sin B\cos A} = \frac{2ac\cos B-2bc\cos A}{2ac\cos B+2bc\cos A}=\frac{(a^2-b^2+c^2)-(-a^2+b^2+c^2)}{(a^2-b^2+c^2)+(-a^2+b^2+c^2)}=\frac{a^2-b^2}{c^2}.$$

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Rwal27
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Updated on August 15, 2022

Comments

  • Rwal27
    Rwal27 about 1 year

    prove

    For any triangle $\triangle ABC$, prove that

    $$\frac{\sin(A-B)}{\sin(A+B)}=\frac{a^2-b^2}{c^2}$$

    • Rwal27
      Rwal27 over 7 years
      I haven't tried anything. I'm confused where to start