Question

If A, B and C are interior angles of a triangle ABC, then show that

sin (B+C/2) = cos A/2​

Answer 1

Question:-

• If A,B and C are interior angles of a triangle ABC, then show that
• $\tt sin\dfrac{B + C}{2} = cos\dfrac{A}{2}$

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Solution:-

•In ∆ABC

Sum of angles of a triangle = $\tt 180^{0}$ [Angle Sum Property = 180°]

$\mapsto \tt A + B + C = 180^{0} \\ \\ \mapsto \tt B + C = 180^{0} - A$

Multiply both sides by $\dfrac{1}{2}$

$\mapsto \tt \dfrac{ B + C}{2} = \dfrac{180^{0} - A}{2} \\ \\ \mapsto \tt \dfrac{B + C}{2} = \dfrac{180^{0}}{2} - \dfrac{A}{2} \\ \\ \mapsto \tt \dfrac{B + C}{2} = 90^{0} - \dfrac{A}{2}$

Now by taking LHS

$\mapsto \tt sin \dfrac{B + C}{2} \\ \\ \mapsto \tt sin 90^{0}- \dfrac{A}{2} \\ \\ \mapsto \tt cos \dfrac{A}{2}$

=RHS

Hence proved.

Note:-

Refer above picture for diagram

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${\pink{\star {\boxed {\boxed {\tt {More\ to \ Know}}}}}} {\pink {\star}}$

• by comparing the equations of trigonometry we will finally with the answer
• By checking the correct identities which match the problem will reach the solution.
• And hence we are done✓