Question

If A ,B and C are interior angles of atriangle ABC then show that sin (B+C​)/2= cosA /2
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Answer 1

Sin[(B+C)/2]

Since A+B+C=180 for interior angles of triangle ABC.

then B+C=180-A.

NOW Sin [(180-A)/2]

=Sin[90-(A/2)]

Sin(90-A)=COSA

since

Answer 2

Solution:-

Given △ABC

We know that sum of three angles of a triangle is 180

Hence ∠A+∠B+∠C=180o

or A+B+C=180o

B+C=180o−A

Multiply both sides by 21

21(B+C)=21(180o−A)

21(B+C)=90o−2A(1)

Now 

21(B+C)

Final Answer:-

cosA/2