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Byju's Answer
Standard XII
Mathematics
Principal Solution of Trigonometric Equation
Evaluate: si...
Question
Evaluate:
s
i
n
2
π
10
+
s
i
n
2
2
π
5
+
s
i
n
2
3
π
5
+
s
i
n
2
9
π
10
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Solution
Given,
sin
2
π
10
+
sin
2
2
π
5
+
sin
2
3
π
5
+
sin
2
9
π
10
=
sin
2
π
10
+
sin
2
2
π
5
+
sin
2
(
π
−
2
π
5
)
+
sin
2
(
π
−
π
10
)
=
sin
2
π
10
+
sin
2
2
π
5
+
sin
2
2
π
5
+
sin
2
π
10
since
sin
(
π
−
θ
)
=
sin
θ
=
2
[
sin
2
π
10
+
sin
2
2
π
5
]
=
2
[
cos
2
(
π
2
−
π
10
)
+
sin
2
2
π
5
]
=
2
[
cos
2
5
π
−
π
10
+
sin
2
2
π
5
]
=
2
[
cos
2
4
π
10
+
sin
2
2
π
5
]
=
2
[
cos
2
2
π
5
+
sin
2
2
π
5
]
=
2
×
1
since
sin
2
θ
+
cos
2
θ
=
1
=
2
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Q.
If
sin
2
π
10
+
sin
2
2
π
5
+
sin
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5
+
sin
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.Find
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Q.
Find the value of
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Q.
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Q.
Write equations whose roots are equal to numbers
s
i
n
2
π
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1
,
s
i
n
2
2
π
2
n
+
1
,
s
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Q.
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i
n
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(
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)
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i
n
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π
4
)
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