if a,b,c are interior angle of a triangle abc then show that sin(b+c)/2=cosa/2
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a+b+c=180°
a+b+c/2=180°/2
a/2+(b+c)/2=90°
(b+c)/2=90°-a/2
taking sin bot sides
sin(b+c)/2=sin(90°-a/2)
sin(b+c)/2=cos a/2
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a+b+c/2=180°/2
a/2+(b+c)/2=90°
(b+c)/2=90°-a/2
taking sin bot sides
sin(b+c)/2=sin(90°-a/2)
sin(b+c)/2=cos a/2
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