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Byju's Answer
Standard XII
Mathematics
Properties Derived from Trigonometric Identities
If Tn=sin n+c...
Question
If
T
n
=
sin
n
x
+
cos
n
x
, prove that
(i)
T
3
-
T
5
T
1
=
T
5
-
T
7
T
3
(ii)
2
T
6
-
3
T
4
+
1
=
0
(iii)
6
T
10
-
15
T
8
+
10
T
6
-
1
=
0
Open in App
Solution
(i) LHS:
T
3
-
T
5
T
1
=
sin
3
x
+
cos
3
x
-
sin
5
x
+
cos
5
x
sin
x
+
cos
x
=
sin
3
x
-
sin
5
x
+
cos
3
x
-
cos
5
x
sin
x
+
cos
x
=
sin
3
x
1
-
sin
2
x
+
cos
3
x
1
-
cos
2
x
sin
x
+
cos
x
=
sin
3
x
.
cos
2
x
+
c
os
3
x
.
sin
2
x
sin
x
+
cos
x
=
sin
2
x
.
cos
2
x
sin
x
+
c
os
x
sin
x
+
cos
x
=
sin
2
x
.
cos
2
x
RHS:
T
5
-
T
7
T
3
=
sin
5
x
+
cos
5
x
-
sin
7
x
+
cos
7
x
sin
3
x
+
cos
3
x
=
sin
5
x
-
si
n
7
x
+
cos
5
x
-
cos
7
x
sin
3
x
+
cos
3
x
=
sin
5
x
1
-
sin
2
x
+
cos
5
x
1
-
cos
2
x
sin
3
x
+
cos
3
x
=
sin
5
x
cos
2
x
+
cos
5
x
sin
2
x
sin
3
x
+
cos
3
x
=
sin
2
x
.
cos
2
x
LHS = RHS
Hence proved.
(ii) LHS:
2
T
6
-
3
T
4
+
1
2
sin
6
x
+
cos
6
x
-
3
sin
4
x
+
cos
4
x
+
1
2
sin
2
x
+
cos
2
x
sin
4
x
+
cos
4
x
-
sin
2
x
cos
2
x
-
3
sin
4
x
+
cos
4
x
+
1
2
.
1
.
sin
4
x
+
cos
4
x
-
sin
2
x
cos
2
x
-
3
sin
4
x
+
cos
4
x
+
1
2
sin
4
x
+
2
cos
4
x
-
2
sin
2
x
cos
2
x
-
3
sin
4
x
-
3
cos
4
x
+
1
-
sin
4
x
+
cos
4
x
-
sin
2
x
cos
2
x
+
1
-
(
sin
2
x
+
cos
2
x
)
2
+
1
-
1
+
1
0
Hence proved.
(iii) LHS:
6
T
10
-
15
T
8
+
10
T
6
-
1
6
sin
10
x
+
cos
10
x
-
15
sin
8
x
+
cos
8
x
+
10
sin
6
x
+
cos
6
x
-
1
Suggest Corrections
2
Similar questions
Q.
If
T
n
=
sin
n
θ
+
cos
n
θ
, prove that
(i)
T
3
-
T
5
T
1
=
T
5
-
T
7
T
3
(ii)
2
T
6
-
3
T
4
+
1
=
0
(iii)
6
T
10
-
15
T
8
+
10
T
6
-
1
=
0
Q.
If
T
n
=
s
i
n
n
θ
+
c
o
s
n
θ
, prove that
(
i
)
T
3
−
T
5
T
1
=
T
5
−
T
7
T
3
(
i
i
)
2
T
6
−
3
T
4
+
1
=
0
(
i
i
i
)
6
T
10
−
15
T
8
+
10
T
6
−
1
=
0
Q.
If
T
n
=
sin
n
θ
+
cos
n
θ
, prove that
(i)
2
T
6
−
3
T
4
+
1
=
0
(ii)
6
T
10
−
15
T
8
+
10
T
6
−
1
=
0
Q.
If
T
n
=
sin
n
x
+
cos
n
x
, then
find the value of
6
T
10
−
15
T
8
+
10
T
6
.
Q.
If
T
n
=
sin
n
x
+
cos
n
x
, then f
ind the value of
2
T
6
−
3
T
4
+
10.
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