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Byju's Answer
Standard XII
Mathematics
Trigonometric Ratios of Compound Angles
Solve sin 7...
Question
Solve
sin
7
A
+
sin
A
cos
5
A
−
cos
3
A
=
?
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Solution
given,
sin
7
A
+
sin
A
cos
5
A
−
cos
3
A
⇒
2
sin
(
7
A
+
A
2
)
cos
(
7
A
−
A
2
)
−
2
sin
(
5
A
+
3
A
2
)
sin
(
5
A
−
3
A
2
)
=
2
sin
4
A
cos
6
A
2
−
2
sin
4
A
sin
A
=
cos
3
A
−
sin
A
.
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0
Similar questions
Q.
Solve the problem:-
sin
A
−
sin
3
A
+
sin
5
A
−
sin
7
A
cos
A
−
cos
3
A
−
cos
5
A
+
cos
7
A
=
2
cot
A
Q.
c
o
s
A
+
c
o
s
3
A
+
c
o
s
5
A
+
c
o
s
7
A
s
i
n
A
+
s
i
n
3
A
+
s
i
n
5
A
+
s
i
n
7
A
=
Q.
Prove that
sinA
+
s
i
n
3
A
+
s
i
n
5
A
+
s
i
n
7
A
c
o
s
A
+
cos
3
A
+
cos
5
A
+
cos
7
A
=
tan
4
A
Q.
sin
A
+
sin
3
A
+
sin
5
A
+
sin
7
A
cos
A
+
cos
3
A
+
cos
5
A
+
cos
7
A
is equal to -
Q.
Prove that:
(i)
sin
A
+
sin
3
A
+
sin
5
A
cos
A
+
cos
3
A
+
cos
5
A
=
tan
3
A
(ii)
cos
3
A
+
2
cos
5
A
+
cos
7
A
cos
A
+
2
cos
3
A
+
cos
5
A
=
cos
5
A
cos
3
A
(iii)
cos
4
A
+
cos
3
A
+
cos
2
A
sin
4
A
+
sin
3
A
+
sin
2
A
=
cot
3
A
(iv)
sin
3
A
+
sin
5
A
+
sin
7
A
+
sin
9
A
cos
3
A
+
cos
5
A
+
cos
7
A
+
cos
9
A
=
tan
6
A
(v)
sin
5
A
-
sin
7
A
+
sin
8
A
-
sin
4
A
cos
4
A
+
cos
7
A
-
cos
5
A
-
cos
8
A
=
cot
6
A
(vi)
sin
5
A
cos
2
A
-
sin
6
A
cos
A
sin
A
sin
2
A
-
cos
2
A
cos
3
A
=
tan
A
(vii)
sin
11
A
sin
A
+
sin
7
A
sin
3
A
cos
11
A
sin
A
+
cos
7
A
sin
3
A
=
tan
8
A
(viii)
sin
3
A
cos
4
A
-
sin
A
cos
2
A
sin
4
A
sin
A
+
cos
6
A
cos
A
=
tan
2
A
(ix)
sin
A
sin
2
A
+
sin
3
A
sin
6
A
sin
A
cos
2
A
+
sin
3
A
cos
6
A
=
tan
5
A
(x)
sin
A
+
2
sin
3
A
+
sin
5
A
sin
3
A
+
2
sin
5
A
+
sin
7
A
=
sin
3
A
sin
5
A
(xi)
sin
θ
+
ϕ
-
2
sin
θ
+
sin
θ
-
ϕ
cos
θ
+
ϕ
-
2
cos
θ
+
cos
θ
-
ϕ
=
tan
θ
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