If A, B and C are interior angles of triangle ABC, then show that sin (B + C)/2 = cos A/2
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Hi ,
It is given that ,
A , B and C are interior angles of triangle
ABC ,
A + B + C = 180°
B + C = 180° - A
( B + C )/2 = 180°/2 - A/2
( B + C ) / 2 = 90° - A/2
Now ,
sin( B + C )/2 = Sin ( 90° - A/2 )
sin ( B + C )/2 = cos A/2
Hence proved.
: )
It is given that ,
A , B and C are interior angles of triangle
ABC ,
A + B + C = 180°
B + C = 180° - A
( B + C )/2 = 180°/2 - A/2
( B + C ) / 2 = 90° - A/2
Now ,
sin( B + C )/2 = Sin ( 90° - A/2 )
sin ( B + C )/2 = cos A/2
Hence proved.
: )
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