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2006 IberoAmerican Olympiad For University Students
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Trigonometric equation always has root - OIMU 2006 Problem 2
Trigonometric equation always has root - OIMU 2006 Problem 2
Source:
August 30, 2010
function
trigonometry
ratio
arithmetic sequence
algebra proposed
algebra
Problem Statement
Prove that for any positive integer
n
n
n
and any real numbers
a
1
,
a
2
,
⋯
,
a
n
,
b
1
,
b
2
,
⋯
,
b
n
a_1,a_2,\cdots,a_n,b_1,b_2,\cdots,b_n
a
1
,
a
2
,
⋯
,
a
n
,
b
1
,
b
2
,
⋯
,
b
n
we have that the equation
a
1
sin
(
x
)
+
a
2
sin
(
2
x
)
+
⋯
+
a
n
sin
(
n
x
)
=
b
1
cos
(
x
)
+
b
2
cos
(
2
x
)
+
⋯
+
b
n
cos
(
n
x
)
a_1 \sin(x) + a_2 \sin(2x) +\cdots+a_n\sin(nx)=b_1 \cos(x)+b_2\cos(2x)+\cdots +b_n \cos(nx)
a
1
sin
(
x
)
+
a
2
sin
(
2
x
)
+
⋯
+
a
n
sin
(
n
x
)
=
b
1
cos
(
x
)
+
b
2
cos
(
2
x
)
+
⋯
+
b
n
cos
(
n
x
)
has at least one real root.
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