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Tardigrade
Question
Mathematics
Let sn = cos ((nπ/10)), n=1,2,3, ldots Then the value of (s1s2 ldots s10/s1+s2+ ldots+s10) is equal to
Q. Let
s
n
=
cos
(
10
nπ
)
,
n
=
1
,
2
,
3
,
…
Then the value of
s
1
+
s
2
+
…
+
s
10
s
1
s
2
…
s
10
is equal to
1861
206
KEAM
KEAM 2014
Trigonometric Functions
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A
2
1
B
2
3
C
2
2
D
0
E
2
1
Solution:
Given,
s
n
=
cos
(
10
nπ
)
Now,
S
1
S
2
S
3
…
S
10
=
cos
(
10
π
)
cos
(
10
2
π
)
…
cos
(
10
5
π
)
…
cos
(
10
10
π
)
=
cos
(
10
π
)
cos
(
5
π
)
…
cos
(
2
π
)
…
cos
(
π
)
=
cos
(
10
π
)
cos
(
5
π
)
…
0
…
cos
π
=
0
∴
S
1
+
s
2
+
…
+
s
10
S
1
S
2
S
3
…
S
10
=
0