Question

in a triangle ABC right angle ,write cos(B+C)/2 in term of angle A

Answer 1

A, B, C are the angles of a triangle

A + B +C = 180°

=>B +C = 180 - A

=>B+C/2 = 90 - A/2

so COS(B+C/2) = COS(90-A/2)=sin a/2

Answer 2

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In ∆ ABC

angle A + angle B +angle C =180° (angle sum property of a triangle)

angle B+C =180°-Angle A

Dividing the equation by 2 .

$\implies \sf \frac{b + c}{2} = \frac{180}{2} - \frac{a}{2}$

$\implies \: \sf \: \frac{b + c}{2} = 90^{ \circ} - \frac{a}{2}$

Multiplying by Cos

$\implies \sf \: cos(\frac{b + c}{2} ) = cos(90 {}^{ \circ} - \frac{a}{2} )$

$\textbf{We know the identity}$

Cos (90-Ø) = Sin Ø

$\implies \sf \: cos( \frac{b + c}{2} ) = sin \frac{a}{2}$