Question
In $\triangle \mathrm{ABC}$ prove that $\cos \left(\frac{A-B}{2}\right)=\left(\frac{a+b}{c}\right) \sin \frac{C}{2}$
$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k$ $\therefore a=k \sin \mathrm{A}, b=k \sin \mathrm{B}, c=k \sin C$
$\begin{aligned} \text { RHS } & =\left(\frac{a+b}{c}\right) \sin \frac{\mathrm{C}}{2} \\ & =\left(\frac{k \sin \mathrm{A}+k \sin \mathrm{B}}{k \sin \mathrm{C}}\right) \sin \frac{\mathrm{C}}{2} \\ & =\left(\frac{\sin \mathrm{A}+\sin \mathrm{B}}{\sin \mathrm{C}}\right) \sin \frac{\mathrm{C}}{2}\end{aligned}$
$\begin{aligned} &=k \times 2 \sin \left(\frac{\mathrm{A}+\mathrm{B}}{2}\right) \cdot \cos \left(\frac{\mathrm{A}-\mathrm{B}}{2}\right) \times \\ & 2 \cos \left(\frac{\mathrm{A}+\mathrm{B}}{2}\right) \cdot \sin \left(\frac{\mathrm{A}-\mathrm{B}}{2}\right)\end{aligned}$
$\begin{aligned} &=k \times 2 \sin \left(\frac{\mathrm{A}+\mathrm{B}}{2}\right) \cdot \cos \left(\frac{\mathrm{A}+\mathrm{B}}{2}\right) \times \\ & 2 \sin \left(\frac{\mathrm{A}-\mathrm{B}}{2}\right) \cdot \cos \left(\frac{\mathrm{A}-\mathrm{B}}{2}\right)\end{aligned}$
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$\bar{b}-(\bar{a} \cdot \bar{b}) \bar{c}$