Question

If A, B, C are the interior angles of a ∆ABC, show that :
(i)$sin\frac{B+C}{2}=cos\frac{A}{2}$
(ii)$cos\frac{B+C}{2}=sin\frac{A}{2}$

Answer 1

SOLUTION :  

(i) Given : If A, B, C are the interior angles of a ∆ ABC.

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

In ∆ ABC,

A + B + C=180°

[Sum of the angles of a ∆ is 180°]

B + C = 180° - A

(B+ C)/2 = 180° /2 − A/2

[On Dividing both Sides by 2]

(B+ C)/2 = 90° /2 −A/2

On taking sin both sides,

sin (B+C)/2 = sin (90° - A)/2

sin(B+C)/2 = cosA/2

[sin (90 - θ) = cos θ]

Hence proved

(ii)  

Given : If A, B, C are the interior angles of a ∆ABC.

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

In ∆ ABC,

A + B + C = 180°

[Sum of the angles of a ∆ is 180°]

B + C = 180° - A

(B+ C)/2 = 180° /2 − A/2

[On Dividing both Sides by 2]

(B+ C)/2 = 90° /2 −A/2

On taking cos both sides,

cos (B+C)/2 = cos (90° - A)/2

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

[cos (90 - θ) = sin θ]

Hence proved

HOPE THIS ANSWER WILL HELP YOU...

Answer 2

refer to the attachment please.