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

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Hello from MrBillDoesMath!

Answer:

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

User RScottCarson
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