Question

if A,B,C are interiors angles of a triangle ABC then show that
sin (B+C/2)=cosA/2........

Answer 1

Given that A, B, C are interior angles of the triangle

A+ B+ C= 180° (Angle sum property)

B+ C= 180° - A


Multiply both sides by 1/2, we get

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

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

(B+C) / 2 = 90 - A/2


Taking LHS

Sin ( B+ C) /2 

= Sin ( 90- A/2)

= Cos (A/ 2)

=RHS

Hence proved. 

Answer 2

Heya☺.

Angle 1+ 2 +3 = 180° [ Angle sum property of triangle]


Therefore
A+B+C = 180°
B+C = 180- A

Dividing each side by 2

B+C/2 = 180°-A/2

B+C/2 = 2(90- A/2) ÷2

Multiplying both the sides by sinA

sinA( B+C÷2) = SinA (90- A/2)
sinA( B+C÷2) = cosA/2

{ sin(90-a) = cosa }

Hence proved
Hope it helps u!!

✌✌All the best for ur exam
..... Shanaya