Question

Cos(a+c)/2=sinb/2 where a,b,c are interior angles of triangle abc

Answer 1

Given that a,b and c are the interior angles of triangle abc.


sum of all angles of a triangle = 180°

a + b + c = 180°

a + c = 180° - b


Divide both sides by 2


( a + c ) / 2 = ( 180 - b ) / 2

( a + c ) / 2 = 90 - b / 2



Hence,

= > cos( a + c ) / 2

= > cos( 90 - b / 2 )


From trigonometric identities, cos( 90 - ∅ ) = sin∅


= > sin( b / 2 )



$\:$