If $A, B$ and $C$ are interior angles of a triangle $A B C$ , then show that
$sin (\frac{B + C}{2}) = cos \frac{A}{2} .$

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Solution

$sin (\frac{B + C}{2})$
$∵ A + B + C = 180 °$ (interior angles of a $△ A B C$ )
$B + C = 180 ° - A$
$sin (\frac{180 ° - A}{2})$ [ $B + C = 180 ° - A$ ]
$sin (90 ° - \frac{A}{2})$
$cos (\frac{A}{2})$

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sin (\frac{B + C}{2}) = cos \frac{A}{2}

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For a triangle ABC, show that sin $(\frac{B + C}{2})$ = cos $(\frac{A}{2})$ , where A, B and C are interior angles of $△$ ABC.

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sin \frac{B + C}{2} = cos \frac{A}{2}

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