Question

if A, B, C are the interior angles of a triangle of a ABC, show that sin(B+C/2) cosA/2+cos(B+C/2)sin A/2=1.​

Answer 1

Answer:

cos A/2

Step-by-step explanation:

We know that, for a given triangle, the sum of all the interior angles of a triangle is equal to 180°

A + B + C = 180° …………………..(1)

To find the value of (B + C)/2, simplify the equation (1)

⇒ B + C = 180° – A

⇒ (B + C)/2 = (180° – A)/2

⇒ (B + C)/2 = 90° – A/2

Now, multiply both sides by sin functions, we get

⇒ sin (B+C)/2 = sin (90°-A/2)

sin (90°-A/2) = cos A/2

The above equation is equal to

sin (90° – A/2) = sin A/2

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

MARK IT AS BRAINIEST

Answer 2

Given :-

A, B and C are the interior angles of a triangle ABC.

To find :-

sin (B+C)/2 cos A/2 + cos (B+C)/2 sin A/2 = 1

Solution :-

Given that

The interior angles of a triangle ABC are A,B and C

We know that

"The sum of the three interior angles in a triangle is equal to 180° ".

=> A+B+C = 180° ---------------(1)

On dividing by 2 both sides then

=> (A+B+C)/2 = 180°/2

=> (A/2)+(B+C)/2 = 90°

=> (B+C)/2 = 90° - (A/2)----------(2)

On taking 'sine' trigonometric ratio both sides then

=> sin (B+C)/2 = sin [90°-(A/2)]

We know that

sin (90°-A) = cos A

Therefore,sin (B+C)/2 = cos A/2 ---------(3)

On taking ' cosine ' trigonometric ratio both sides of (2) then

=> cos (B+C)/2 = cos [90° - (A/2)]

We know about

cos (90°-A) = sin A

Therefore, cos (B+C)/2 = sin A/2 --------(4)

Now,

On taking LHS :

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

=> cos A/2 cos A/2 + sin A/2 sin A/2

( From (1)&(2) )

=> cos² A/2 + sin² A/2

=> sin² A/2 + cos² A/2

=> 1

Since, sin² A + cos² A = 1

=> RHS

=> LHS = RHS

Hence, Proved.

Answer :-

sin (B+C)/2 cos A/2+cos (B+C)/2 sin A/2 = 1

Used formulae:-

• sin² A + cos² A = 1

• sin (90°-A) = cos A

• cos (90°-A) = sin A

Used Property:-

• "The sum of the three interior angles in a triangle is equal to 180° ".