Question

for ∆ABC,Sin[(A+B)÷2]=?​

Answer 1

Answer:

cos(C/2)

Step-by-step explanation:

Given ,

ABC is a triangle

To Find :-

Value of :-

$sin\left(\dfrac{A+B}{2}\right)$

How To Do :-

As they said that ABC forms a triangle we need to equate that sum of all angles in a triangle = 180° and after we need to simplify it and we need to make A and B as subject and after applying 'sin' on both sides we will get the value of sin(A+B/2).

Formula Required :-

1)  Sum of all the angles in a triangle = 180°

2)  sin(90° - α) = cosα

Solution :-

As ABC forms a triangle :-

∠A + ∠B + ∠C = 180°

[ ∴ Sum of all the angles in a triangle = 180° ]

Dividing by '2' with the whole equation :-

$\dfrac{\angle A+\angle B+\angle C}{2}=\dfrac{180^{\circ}}{2}$

$\dfrac{\angle A+\angle B+\angle C}{2}=90^{\circ}$

$\dfrac{\angle A}{2}+\dfrac{\angle B}{2}+\dfrac{\angle C}{2}=90^{\circ}$

$\dfrac{\angle A}{2}+\dfrac{\angle B}{2}=90^{\circ}-\dfrac{\angle C}{2}$

$\dfrac{\angle A+ \angle B}{2}=90^{\circ}-\dfrac{\angle C}{2}$

Applying 'sin' on both sides :-

$sin\left(\dfrac{\angle A+ \angle B}{2}\right)=sin\left(90^{\circ}-\dfrac{\angle C}{2}\right)$

$sin\left(\dfrac{A+B}{2}\right)=cos\left(\dfrac{C}{2}\right)$

[ ∴ sin(90° - α) = cosα ].

∴ sin(A+B/2) = cos(C/2)