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Question
If A, B and C are interior angles of a triangle ABC, then show that `\sin( \frac{B+C}{2} )=\cos \frac{A}{2}`
Solution
∵ A + B + C = 180° (a.s.p. of ∆)
B + C = 180° – A
`( \frac{B+C}{2})=90^\circ -\frac{A}{2}`
`\sin ( \frac{B+C}{2})=\sin ( 90^\circ -\frac{A}{2})`
`\sin ( \frac{B+C}{2} )=\cos \frac{A}{2} `
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