If a b c are interior angle of triangle abc then show that sin ( b +c / 2 ) = cos a /2

Hello from MrBillDoesMath!

See Discussion below

Discussion:

As a,b,c are interior angles of a triangle

a + b + c = 180 =>

b + c = 180 -a =>

(b +c)/2 = (1/2) (180 -a) =>

(b+c)/2  = (1/2) 180 * (1/2) (-a) =>

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

In general, for any angle @

   sin (90 - @ ) = sin(90) cos(@) - cos(90)sin(@)

                        = 1 * cos(@) -  0                                 as cos(90) = 0, sin(90) = 1

                        = cos(@)                  (exhibit B)

So

sin( (b+c)/2)  = sin(90 - a/2)        from Exhibit A

                    = cos (a/2)          

where, in the final step,  we let @ = a/2 in Exhibit B

Thank you,

MrB

i am unable to solve this problem

step-by-step explanation:

-7/8

step-by-step explanation:

instead of subtracting you add the -6/8 and -1/8