In a triangle ABC,(b+c) sin frac{A}{2} = | vedclass

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(B) Using the Sine rule,we have $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k$. Thus,$b = k \sin B$ and $c = k \sin C$. Substituting the

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$a \cos \left(\frac{B-C}{2}\right)$

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(B) Using the Sine rule,we have $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k$. Thus,$b = k \sin B$ and $c = k \sin C$. Substituting these into the expression: $(b+c) \sin \frac{A}{2} = k(\sin B + \sin C) \sin \frac{A}{2}$. Using the sum-to-product formula $\sin B + \sin C = 2 \sin \frac{B+C}{2} \cos \frac{B-C}{2}$: $= k \left[ 2 \sin \frac{B+C}{2} \cos \frac{B-C}{2} \right] \sin \frac{A}{2}$. Since $A+B+C = \pi$,we have $\frac{B+C}{2} = \frac{\pi}{2} - \frac{A}{2}$,so $\sin \frac{B+C}{2} = \cos \frac{A}{2}$. $= k \left[ 2 \cos \frac{A}{2} \cos \frac{B-C}{2} \right] \sin \frac{A}{2}$. $= k \left[ 2 \sin \frac{A}{2} \cos \frac{A}{2} \right] \cos \frac{B-C}{2}$. $= k \sin A \cos \frac{B-C}{2}$. Since $a = k \sin A$,the expression becomes $a \cos \frac{B-C}{2}$.

In a triangle $ABC$,$(b+c) \sin \frac{A}{2} =$

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