#### 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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Solution If A, B and C are interior angles of a triangle ABC, then show that sin((B+C)/2)=cos(A/2) Concept: Trigonometric Ratios of Some Specific Angles.

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