In a triangle ABC, write cos (b+c/2) in terms of angle A

1.	In a triangle ABC, write cos (b+c/2) in terms of angle A
Answer» Sum of three angles of triangle is\xa0{tex}180^{\\circ}{/tex}A + B + C = 180°or, B + C = 180° - ADivide both sides by 2{tex}\\frac{B+C}{2}=\\frac{180^{\\circ}}{2}-\\frac{A}{2}{/tex}{tex}\\frac{B+C}{2}=90-\\frac{A}{2}{/tex}{tex}\\therefore \\quad \\cos \\left( \\frac { B + C } { 2 } \\right) {/tex}{tex}= \\cos \\left( 90 - \\frac { A } { 2 } \\right){/tex}{tex}= \\sin \\frac { A } { 2 }{/tex}

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