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Question
A, B and C are interior angles of a triangle ABC. Show that
sin `(("B"+"C")/2) = cos "A"/2`
Solution
We know that for a triangle ABC,
∠A + ∠B + ∠C = 180°
∠B + ∠C = 180° − ∠A
`(angle "B" +angle "C")/2 = 90° - (angle"A")/2`
`((sin "B+C")/2) = sin (90°- "A"/2)`
= cos `("A"/2)`
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