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Byju's Answer
Standard XII
Mathematics
Higher Order Derivatives
sin 3θ + 2sin...
Question
sin
3
θ
+
2
sin
5
θ
+
sin
7
θ
sin
θ
+
2
sin
3
θ
+
sin
5
θ
=
cos
2
θ
+
sin
2
θ
.
cot
3
θ
.
Open in App
Solution
s
i
n
3
θ
+
2
s
i
n
θ
+
s
i
n
7
θ
s
i
n
θ
+
2
s
i
n
3
θ
+
s
i
n
5
θ
∴
s
i
n
3
θ
+
s
i
n
7
θ
+
2
s
i
n
5
θ
s
i
n
θ
+
s
i
n
5
θ
+
2
s
i
n
3
θ
=
2
s
i
n
5
θ
c
o
s
2
θ
+
2
s
i
n
5
θ
2
s
i
n
3
θ
c
o
s
2
θ
+
2
s
i
n
3
θ
.
.
.
.
.
(
s
i
n
A
+
s
i
n
B
=
2
s
i
n
(
A
+
B
)
2
c
o
s
(
A
−
B
)
2
)
=
2
s
i
n
5
θ
(
c
o
s
2
θ
+
1
)
2
s
i
n
3
θ
(
c
o
s
2
θ
+
1
)
=
s
i
n
5
θ
s
i
n
3
θ
→
L
H
S
Now, RHS ,
c
o
s
2
θ
+
s
i
n
2
θ
c
o
t
3
θ
=
c
o
s
2
θ
+
s
i
n
2
θ
×
c
o
s
3
θ
s
i
n
3
θ
s
i
n
3
θ
c
o
s
2
θ
+
s
i
n
2
θ
c
o
s
3
θ
s
i
n
3
θ
=
s
i
n
5
θ
s
i
n
3
θ
.
.
.
.
.
.
(
s
i
n
(
A
+
B
)
=
s
i
n
A
c
o
s
B
+
c
o
s
A
s
i
n
B
)
=
L
H
S
Hence LHS=RHS
Hence proved.
Suggest Corrections
0
Similar questions
Q.
Solve the following equations:
(i)
cos
θ
+
cos
2
θ
+
cos
3
θ
=
0
(ii)
cos
θ
+
cos
3
θ
-
cos
2
θ
=
0
(iii)
sin
θ
+
sin
5
θ
=
sin
3
θ
(iv)
cos
θ
cos
2
θ
cos
3
θ
=
1
4
(v)
cos
θ
+
sin
θ
=
cos
2
θ
+
sin
2
θ
(vi)
sin
θ
+
sin
2
θ
+
sin
3
=
0
(vii)
sin
θ
+
sin
2
θ
+
sin
3
θ
+
sin
4
θ
=
0
(viii)
sin
3
θ
-
sin
θ
=
4
cos
2
θ
-
2
(ix)
sin
2
θ
-
sin
4
θ
+
sin
6
θ
=
0
Q.
Solve:
sin
5
θ
+
sin
θ
=
sin
3
θ
Q.
If
sin
θ
+
sin
2
θ
+
sin
3
θ
=
sin
α
and
cos
θ
+
cos
2
θ
+
cos
3
θ
=
cos
α
, then
θ
is equal to -
Q.
Solve
sin
θ
+
sin
5
θ
=
sin
3
θ
;
0
≤
θ
≤
π
.
Q.
Find the general solution of the equation
s
i
n
7
θ
=
s
i
n
3
θ
+
s
i
n
θ
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