If A,B and C are the interior angles of a triangle ABC then show that sin[B+C/2]=cos[A/2]
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ANSWER
Given △ABC
We know that sum of three angles of a triangle is 180
Hence ∠A+∠B+∠C=180
o
or A+B+C=180
o
B+C=180
o
−A
Multiply both sides by
2
1
2
1
(B+C)=
2
1
(180
o
−A)
2
1
(B+C)=90
o
−
2
A
...(1)
Now
2
1
(B+C)
Taking sine of this angle
sin(
2
B+C
)[
2
B+C
=90
o
−
2
A
]
sin(90
o
−
2
A
)
cos
2
A
[sin(90
o
−θ)=cosθ]
Hence sin(
2
B+C
)=cos
2
A
proved
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