In a ∆ABC, write cos[(B+C)÷2] in terms of angle A

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Solution

A+B+C= 180 (ASP of triangle)

~ B+C = 180 - A.......

~ cos (B+C)/2 = cos (180 - A)/2

~cos (180/2-A/2)

~cos (90 - A/2)

~sin (A/2)

therefore, cos(B+C/2)= sin(A/2)

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