Question

answer this question​

Answer 1

Answer:

In a ∆ABC,

Sum of angles in a triangle=180°

$</strong><strong>A</strong><strong> + </strong><strong>B</strong><strong> + </strong><strong>C</strong><strong> = \pi$

$</strong><strong>A</strong><strong> + </strong><strong>B</strong><strong> = \pi - </strong><strong>C</strong><strong>$

In the above equation,

Take LHS,

→ $sin( \frac{A + B}{2} ) = sin( \frac{\pi - C}{2} )$

→ $sin( \frac{\pi - C}{2} ) = sin(90 - \frac{C}{2} ) = cos \frac{C}{2}$

$sin( \frac{</strong><strong>A</strong><strong> + </strong><strong>B</strong><strong>}{2}) = cos \frac{</strong><strong>C</strong><strong>}{2}$

Answer 2

Answer:-

Given:-

• Sin(A+B/2)

To prove:-

• Sin(A+B/2) = cos C/2

Solution:-

We know that for a ΔABC

∠A + ∠B + ∠C = 180°

∠B + ∠A / 2 = 90° - ∠C / 2

Sin(A+B/2) ⇒ sin(90° - c/2)

⇒ cos C/2

Hence proved....