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Byju's Answer
Standard XII
Mathematics
Complex Numbers
∑k=13cos2 2k-...
Question
∑
3
k
=
1
cos
2
(
(
2
k
−
1
)
π
12
)
=
A
0
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B
1
2
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C
−
1
2
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D
3
2
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Solution
The correct option is
C
3
2
cos
2
π
12
+
cos
2
3
π
12
+
cos
2
5
π
12
=
cos
2
π
12
+
cos
2
π
4
+
cos
2
5
π
12
=
cos
2
π
12
+
cos
2
(
π
2
−
π
12
)
+
(
1
√
2
)
2
=
cos
2
π
12
+
sin
2
π
12
+
1
2
=
1
+
1
2
=
3
2
Therefore, Answer is
3
2
Suggest Corrections
0
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k
x
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k
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x
2
-
2
k
+
1
x
+
k
+
4
=
0