My try: Let the bisector of angle $A$ touch the side $BC$ at $D$. $ADC$ & $ADB$ form a triangle. Apply sine rule.
$$\begin{align}
\dfrac{\sin\phi}b &= \dfrac{\sin\frac A2}{DC} \\
\dfrac{\sin (180 – \phi)}c &= \dfrac{\sin\frac A2}{BD} \\
\dfrac{\sin\phi}c &= \dfrac{\sin\frac A2}{BD} \\
BD + DC &= \dfrac{\sin\frac A2 \cdot[b + c]}{\sin\phi} \\
a &= \dfrac{\sin\frac A2 \cdot[b + c]}{\sin\phi} \\
\sin\phi &= \dfrac{\sin\frac A2 \cdot[b + c]}a
\end{align}$$
But the answer is $\cos (B – C)/2$, I can't simplify my answer further.
Best Answer
Note that $$\theta = C+ A/2$$
Thus $$\sin \theta = \sin C \cos A/2 +\cos C \sin A/2$$
Also we have
$$ \pi - \theta = B + A/2 $$
Thus $$ \sin \theta = \sin (\pi - \theta) = \sin B \cos A/2 +\cos B \sin A/2$$
Adding the two results $\sin \theta $ we get
$$ 2 \sin \theta = (\sin C + \sin B)\cos A/2 +(\cos C + \cos B)\sin A/2 $$
$$ = 2\sin (B+C)/2 \ cos (B-C)/2 \cos A/2$$
$$ + 2\cos (B+C)/2 \ cos (B-C)/2 \sin A/2$$
$$= 2 \cos (B-C)/2 \sin (A+B+C)/2 $$
$$= 2 \cos (B-C)/2$$
Thus $$ \sin \theta = \cos ( B-C)/2 $$