Question

If A, B and C are interior angles of a triangle ABC, then show that sin (B+C/2) = cos A/2. ​

Answer 1

Solution:

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

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

A + B + C = 180° ….(1)

Now we can write the above equation as;

⇒ B + C = 180° – A

Dividing by 2 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

Hence proved.

Answer 2

Step-by-step explanation:

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