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Using the sine rule: $$ \frac{a}{\sin(A)} = \frac{b}{\sin(B)} = \frac{c}{\sin(C)}$$

prove, for triangle ABC:

$$\sin\left(\frac{B-C}{2}\right) = \frac{b-c}{a} \cos\left(\frac A2\right)$$

Using the sine rule it's easy to translate the RHS into:

$$\sin\left(\frac{B-C}{2}\right) = \frac{\sin(B)-\sin(C)}{\sin(A)} \cos\left(\frac A2\right)$$

Yes, I can expand out the LHS, and use the difference of 2 sines in the RHS, but neither makes an obvious equality, especially with terms in a and A in the RHS.

A nudge in the right direction would really be appreciated. Thanks.


marked as duplicate by Blue trigonometry Dec 9 '18 at 19:47

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.


Since $A+B+C=\pi$, the difference formula for $\sin$ gives \begin{align}\frac{\sin B-\sin C}{\sin A} \cos\frac A2&=\frac{2\cos\frac{B+C}2\sin\frac{B-C}2}{\sin(\pi-B-C)}\cos\frac{\pi-B-C}2\\&=\frac{2\sqrt{\frac{1+\cos(B+C)}2}\sin\frac{B-C}2}{\sin(B+C)}\sqrt{\frac{1+\cos(\pi-B-C)}2}\\&=\frac{2\sqrt{\frac{1+\cos(B+C)}2}\sin\frac{B-C}2}{\sin(B+C)}\sqrt{\frac{1-\cos(B+C)}2}\\&=\frac{\sin\frac{B-C}2}{\sin(B+C)}\sqrt{1-\cos^2(B+C)}\\&=\frac{\sin\frac{B-C}2}{\sin(B+C)}\sin(B+C)=\sin\frac{B-C}2\end{align} as required.


Just assume that $a/sin(A)=b/sin(B)=c/sin(C)=1/k$.

So $sin(A)=ka, sin(B)=kb,sin(C)=kc$.

Now substitute this and you will get the desired result.

  • $\begingroup$ Are you sure this leads anywhere different from what OP's tried? $\endgroup$ – TheSimpliFire Dec 9 '18 at 19:20
  • $\begingroup$ Well he has given here his method and his last step. Now if we do what I have stated we will get the RHS. $\endgroup$ – Avanish Singh Dec 9 '18 at 19:22
  • $\begingroup$ I might have overlooked this, but how would you handle $\sin(B-C/2)$ and $\cos(A/2)$ using those substitutions? $\endgroup$ – TheSimpliFire Dec 9 '18 at 19:25

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