# Solve : If the bisector of angle $A$ of $\triangle ABC$ makes angle $\phi$ with $BC$, what is $\sin \phi$?

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.

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 $$\sin \theta = \cos ( B-C)/2$$