Prove $\sin(A-B)/\sin(A+B)=(a^2-b^2)/c^2$
9,791
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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Author by
Rwal27
Updated on August 15, 2022Comments
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Rwal27 about 1 year
prove
For any triangle $\triangle ABC$, prove that
$$\frac{\sin(A-B)}{\sin(A+B)}=\frac{a^2-b^2}{c^2}$$
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Rwal27 over 7 yearsI haven't tried anything. I'm confused where to start
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