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Sum

A, B and C are interior angles of a triangle ABC. Show that

sin `(("B"+"C")/2) = cos "A"/2`

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#### 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)`

Is there an error in this question or solution?

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