sinA+sinB+sinC=4cosA/2cosB/2cosC/2 prove that
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0
Answer:
We have,
sin
A
+
sin
B
−−−−−−−−−−−
+
sin
C
=
2
sin
(
A
+
B
2
)
cos
(
A
−
B
2
)
+
sin
C
...
.
(
⋆
)
.
But,
A
+
B
+
C
=
π
∴
A
+
B
=
π
−
C
∴
(
A
+
B
2
)
=
π
−
C
2
.
∴
(
A
+
B
2
)
=
π
2
−
C
2
.
∴
sin
(
A
+
B
2
)
=
sin
(
π
2
−
C
2
)
=
cos
(
C
2
)
...
.
(
⋆
1
)
.
Also,
sin
C
=
2
sin
(
C
2
)
cos
(
C
2
)
...
...
...
(
⋆
2
)
.
Utilising
(
⋆
1
)
and
(
⋆
2
)
in
(
⋆
)
,
we get,
sin
A
+
sin
B
+
sin
C
,
=
2
cos
(
C
2
)
cos
(
A
−
B
2
)
+
2
sin
(
C
2
)
cos
(
C
2
)
,
=
2
cos
(
C
2
)
{
cos
(
A
−
B
2
)
+
sin
(
C
2
)
}
,
=
2
cos
(
C
2
)
{
cos
(
A
−
B
2
)
+
sin
(
π
−
(
A
+
B
)
2
)
...
[
a
s
,
A
+
B
+
C
=
π
]
,
=
2
cos
(
C
2
)
{
cos
(
A
−
B
2
)
+
sin
(
π
2
−
A
+
B
2
)
}
,
=
2
cos
(
C
2
)
{
cos
(
A
−
B
2
)
+
cos
(
A
+
B
2
)
}
,
=
2
cos
(
C
2
)
⎧
⎨
⎩
2
cos
⎛
⎝
A
−
B
2
+
A
+
B
2
2
⎞
⎠
cos
⎛
⎝
A
−
B
2
−
A
+
B
2
2
⎞
⎠
,
=
2
cos
(
C
2
)
{
2
cos
(
A
2
)
cos
(
−
B
2
)
}
,
=
4
cos
(
A
2
)
cos
(
B
2
)
cos
(
C
2
)
...
.
[
∵
,
cos
(
−
x
)
=
cos
x
]
.
Hence, the Proof.
Enjoy Maths.!
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