Question

Hey!!

In a triangle ABC, write cos(B+C/2) in terms of angle A.

Answer 1

$\textsf{Answer :-}$

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$\textsf{}$

$\textsf{Given :-}$

$\textsf{ABC is a triangle.}$

$\textsf{To write :-}$

$\mathsf{cos \frac{ B + C }{2} \: in \: terms \: of \: angle \: A.}$

$\textsf{Salutation :-}$

$\textsf{As a triangle , angle A + B + C = 180°}$

$\textsf{Then , B + C = 180° - A}$

$\textsf{Now , }$

$\mathsf{cos( \frac{ B + C }{2})}$

$\mathsf{=cos (\frac{ 180° - A }{2})}$

$\mathsf{=cos (\frac{ 180°}{2} - \frac{ A }{2})}$

$\mathsf{=cos( 90° - \frac{A}{2})}$

= $\mathsf{ = sin \frac{A}{2}}$            [ ★ Required answer ]

[ • As we know , cos ( 90° - θ ) = sinθ ]

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