Question

if A, B, C are the interior angles of triangle ABC, then show that cos(B+C/2)= sin A/2​

Answer 1

Here's your 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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Answer 2

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As we know, for any given triangle, the sum of all its interior angles is equals to 180°.

Thus,

A + B + C = 180° ….(1)

Now we can write the above equation as;

⇒ B + C = 180° – A

Dividing by 2 on both the sides;

⇒ (B + C)/2 = (180° – A)/2

⇒ (B + C)/2 = 90° – A/2

Now, put sin function on both sides.

⇒ sin (B + C)/2 = sin (90° – A/2)

Since,

sin (90° – A/2) = cos A/2

Therefore,

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

Hope it's Helpful....:)