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I need help with this sum:

If $$\dfrac {\sin {A}}{\sin {B}} = \dfrac {\sin {(A-B)}}{\sin {(B-C)}} $$ prove that $a^2$, $b^2$ and $c^2$ are in A. P.

In my attempts, I tried to apply componendo and dividendo on both sides and then applied transformation rules for converting sums of sines to their products, and tried to get to something, but it was a failure.

Wrichik Basu
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    HINT: $$A=\pi-(B+C)$$ and use https://math.stackexchange.com/questions/175143/prove-sinab-sina-b-sin2a-sin2b and https://brilliant.org/wiki/sine-rule/ – lab bhattacharjee May 14 '17 at 16:57
  • I believe you want $\frac{\sin A}{\sin C}$ on the left-hand side. (That is $\sin C$ instead of $\sin B$ in the denominator.) – Blue May 15 '17 at 00:25

2 Answers2

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HINT: using the Theorem of sines we get $$\frac{a}{b}=\frac{\sin(A)\cos(B)-\cos(A)\sin(B)}{\sin(B)\cos(C)-\cos(B)\sin(C)}$$ now set $$\sin(A)=\frac{a}{b}\sin(B)$$ and $$\sin(C)=\frac{c}{b}\sin(B)$$ and cancelling $$\sin(B)\ne 0$$ we get $$\frac{a}{b}=\frac{a\cos(B)-b\cos(A)}{b\cos(C)-c\cos(B)}$$ plugin the Theorem of cosines and simplifying we have $$\frac{a}{b}=\frac{\frac{a^2-b^2}{c}}{\frac{b^2-c^2}{a}}$$ can you proceed?

Dr. Sonnhard Graubner
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As suggested in the comment ,

as $A=\pi-(B+C),\sin A=\sin(B+C)$

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

$$P=\sin A\sin(B-C)=\sin(B+C)\sin(B-C)=\sin^2B-\sin^2C$$

Using Law of Sines,$$P=\dfrac{b^2-c^2}{(2R)^2}$$

So $\dfrac {\sin {A}}{\sin C} = \dfrac {\sin {(A-B)}}{\sin {(B-C)}}\implies\dfrac{b^2-c^2}{(2R)^2}=\dfrac{a^2-b^2}{(2R)^2} $

lab bhattacharjee
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