Question

in a triangle abc prove that sin
[ a+bby 2 = cos c by 2​

Answer 1

Answer:

we know

in triangle abc

Step-by-step explanation:

a + b + c = 180

so a + b = 180 - c

now, consider

$sin( \frac{a \: + b}{2} ) = sin \frac{180 - c}{2} = sin( \frac{180}{2} - \frac{c}{2} ) \\ = sin(90 - \frac{c}{2} ) = cos \frac{c}{2}$

hence proved