# Use the sine rule to prove trig identity [duplicate]

This question already has an answer here:

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 StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Dec 9 '18 at 19:47

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 $$sin(A)=ka, sin(B)=kb,sin(C)=kc$$.
• I might have overlooked this, but how would you handle $\sin(B-C/2)$ and $\cos(A/2)$ using those substitutions? – TheSimpliFire Dec 9 '18 at 19:25