Question

in triangle ABC prove that sin (A+B/2)=cos c/2​

Answer 1

Step-by-step explanation:

We have:

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

Since for a traingle

A + B + C = 180°

A + B = 180°-C

Substituting for A+B here,

$\\\tt \sin( \dfrac{180°-C}{2} ) \\ \\\tt = \sin( 90° - \dfrac{C}{2} )$

Since,

$\sin( \dfrac{\pi}{2} - x) = \cos(x)$

The final answer will be:

$\\ \\\tt \cos( \dfrac{C}{2} )$

Answer 2

Given :       There is a triangle ABC

now,

           Sum of all angles of triangle = 180°

⇒              ∠A + ∠B + ∠C = 180°

⇒               ∠A + ∠B = 180° - ∠C

         Multipying both sides by 1/2

⇒              (∠A + ∠B) / 2 = (180° - ∠C) / 2

⇒               (∠A + ∠B) / 2 = 90° - ∠C / 2                            ...(i)

Now taking LHS,

                   sin (∠A + ∠B) / 2

               =  sin (90° - ∠C / 2)

               =  cos ∠C / 2 .                              [∵ sin (90 - ∅) = cos ∅]

              = RHS .

hence proved.