Question

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

Answer 1

A+B+C=180

B+C=180-A

B+C/2=90-A/2

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

Sin(B+C/2)=CosA/2

Answer 2

${\huge{\bold{\boxed{\tt{\color{magenta}{answer}}}}}}$

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....:)