sin(b+2c) + sin(c+2a) + sin(a+2b)= 4sin(b-c/2)sin(c-a/2) cos(a-b/2) if a+b+c=π
Answers
Answer:
Explanation:
A
+
B
+
C
=
π
⇒
A
+
B
=
π
−
C
.
∴
A
+
2
B
=
(
A
+
B
)
+
B
=
(
π
−
C
)
+
B
=
π
−
(
C
−
B
)
.
∴
sin
(
A
+
2
B
)
=
sin
(
π
−
(
C
−
B
)
)
=
sin
(
C
−
B
)
.
Similarly,
sin
(
B
+
2
C
)
=
sin
(
A
−
C
)
,
and
,
sin
(
C
+
2
A
)
=
sin
(
B
−
A
)
.
∴
sin
(
A
+
2
B
)
+
sin
(
B
+
2
C
)
+
sin
(
C
+
2
A
)
,
=
sin
(
C
−
B
)
+
sin
(
A
−
C
)
+
sin
(
B
−
A
)
.
Let,
C
−
B
=
x
,
A
−
C
=
y
and
B
−
A
=
z
.
Then,
x
+
y
+
z
=
0
.
∴
x
+
y
=
−
z
...
...
(
1
)
and
z
=
−
(
x
+
y
)
...
...
(
2
)
.
The L.H.S.=
sin
x
+
sin
y
+
sin
z
,
=
2
sin
(
x
+
y
2
)
cos
(
x
−
y
2
)
+
sin
z
,
=
2
sin
(
−
z
2
)
cos
(
x
−
y
2
)
+
sin
z
,
...
...
...
...
[
∵
(
1
)
]
,
=
−
2
sin
(
z
2
)
cos
(
x
−
y
2
)
+
2
sin
(
z
2
)
cos
(
z
2
)
,
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
[
∵
sin
2
a
=
2
sin
a
cos
a
]
,
=
2
sin
(
z
2
)
{
cos
(
z
2
)
−
cos
(
x
−
y
2
)
}
,
=
2
sin
(
z
2
)
{
cos
(
−
x
+
y
2
)
−
cos
(
x
−
y
2
)
}
,
...
...
.
[
∵
(
2
)
]
,
=
2
sin
(
z
2
)
{
cos
(
x
+
y
2
)
−
cos
(
x
−
y
2
)
}
,
=
2
sin
(
z
2
)
{
−
2
sin
(
x
2
)
sin
(
y
2
)
}
,
=
−
4
sin
(
x
2
)
sin
(
y
2
)
sin
(
z
2
)
,
=
−
4
sin
(
C
−
B
2
)
sin
(
A
−
C
2
)
sin
(
B
−
A
2
)
,
=
−
4
sin
(
−
B
−
C
2
)
sin
(
−
C
−
A
2
)
sin
(
−
A
−
B
2
)
,
=
+
4
sin
(
B
−
C
2
)
sin
(
C
−
A
2
)
sin
(
A
−
B
2
)
,
=The R.H.S.
Enjoy Maths.!
A+B+C=pi rArr A+B=pi-C.#
#:. A+2B=(A+B)+B=(pi-C)+B=pi-(C-B).#
#:. sin(A+2B)=sin(pi-(C-B))=sin(C-B).#
Similarly,
# sin(B+2C)=sin(A-C), and, sin(C+2A)=sin(B-A).#
#:. sin(A+2B)+sin(B+2C)+sin(C+2A),#
#=sin(C-B)+sin(A-C)+sin(B-A).#
Let, #C-B=x, A-C=y and B-A=z.#
Then, #x+y+z=0.#
#:. x+y=-z......(1) and z=-(x+y)......(2).#
#"The L.H.S.="sinx+siny+sinz,#
=2sin((x+y)/2)cos((x-y)/2)+sinz,
=2sin(-z/2)cos((x-y)/2)+sinz,............[because (1)],
=-2sin(z/2)cos((x-y)/2)+2sin(z/2)cos(z/2),...........................................................................[because sin2a=2sinacosa],
=2sin(z/2){cos(z/2)-cos((x-y)/2)},
=2sin(z/2){cos(-(x+y)/2)-cos((x-y)/2)},.......[because (2)],
=2sin(z/2){cos((x+y)/2)-cos((x-y)/2)},
=2sin(z/2){-2sin(x/2)sin(y/2)},
=-4sin(x/2)sin(y/2)sin(z/2),
=-4sin((C-B)/2)sin((A-C)/2)sin((B-A)/2),
=-4sin(-(B-C)/2)sin(-(C-A)/2)sin(-(A-B)/2),
=+4sin((B-C)/2)sin((C-A)/2)sin((A-B)/2),
=The R.H.S."