If a+b+c =pi/2 show that sina+sinb+sinc =1 +4sin((a+b)/2)sin((b+c)/2)sin((a+c)/2)
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Answer:
sin
2
A
+sin
2
B
+sin
2
C
=2sin
4
A+B
cos
4
A−B
+sin
2
c
=2sin
4
A+B
cos
4
A−B
+1−2sin
2
4
A+B
[∵A+B+C=Π]
[
2
C
=
2
Π
−(
2
A
+
2
B
)]
∴ 1+2sin(
4
Π
−
4
C
)[cos(
4
A−B
)−cos(
2
Π
−(
4
A+B
))]
⇒1+2sin(
4
Π
−
4
C
)2sin(
4
Π
−
4
A
)sin(
4
Π
−
4
B
)
⇒1+4sin
4
Π−C
sin
4
Π−A
sin
4
Π−B
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