Use the sine rule to prove trig identity 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.
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.
Since $A+B+C=\pi$, the difference formula for $\sin$ gives \begin{align}\frac{\sin B-\sin C}{\sin A} \cos\frac A2&amp;=\frac{2\cos\frac{B+C}2\sin\frac{B-C}2}{\sin(\pi-B-C)}\cos\frac{\pi-B-C}2\\&amp;=\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}\\&amp;=\frac{2\sqrt{\frac{1+\cos(B+C)}2}\sin\frac{B-C}2}{\sin(B+C)}\sqrt{\frac{1-\cos(B+C)}2}\\&amp;=\frac{\sin\frac{B-C}2}{\sin(B+C)}\sqrt{1-\cos^2(B+C)}\\&amp;=\frac{\sin\frac{B-C}2}{\sin(B+C)}\sin(B+C)=\sin\frac{B-C}2\end{align} as required.