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If A ,B, and C are interior angles of a triangle ABC then show that sin(B+C/2)=COSA/2

Posted by Dhara Sarda 6 years, 5 months ago

CBSE > Class 08 > Mathematics

1 answers

Sia ? 6 years, 5 months ago

A, B, C, are interior angles of a {tex}\Delta {/tex}
{tex}\because A + B + C = 180 ^ { 0 }{/tex}
{tex}\Rightarrow B + C = 180 ^ { 0 } - A \Rightarrow \frac { B + C } { 2 } = 90 ^ { 0 } - \frac { A } { 2 }{/tex}
{tex}\Rightarrow \sin \frac { \mathrm { B } + \mathrm { C } } { 2 } = \sin \left( 90 ^ { \circ } - \frac { \mathrm { A } } { 2 } \right) \left[ \because \sin \left( 90 ^ { 0 } - \theta \right) = \cos \theta \right]{/tex}
{tex}\Rightarrow \sin \frac { \mathrm { B } + \mathrm { C } } { 2 } = \cos \frac { \mathrm { A } } { 2 } \text { proved }{/tex}
LHS = RHS

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CBSE > Class 08 > Mathematics