Question

IF a , b ,c are interior angles of a ∆abc, then show that cos(B+C)/2=SINA/2​

Answer 1

$\huge\frak\pink answer$

A,B,C are the interior angles of triangle ABC

the sum of the interior angles of a triangle is 180°

therefore-

A+B+C=180°

B+C=180°-A

(dividing by 2 both sides)

B+C/2=180°/2-A/2

B+C/2=90°-A/2

(taking cos both sides)

cos(B+C/2)=cos(90°-A/2)

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

{cos(90°-A)=sin A}

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

[HENCE PROVED]

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