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1998 Belarus Team Selection Test
4
\sum sin (k+1)x / sin kx < 2 cos x / sin^2 x
\sum sin (k+1)x / sin kx < 2 cos x / sin^2 x
Source: 1998 Belarus TST 2.4
December 25, 2020
inequalities
trigonometry
algebra
Problem Statement
Prove the inequality
∑
k
=
1
n
sin
(
k
+
1
)
x
sin
k
x
<
2
cos
x
sin
2
x
\sum_{k=1}^{n}\frac{\sin (k+1)x}{\sin kx}< 2\frac{\cos x}{\sin^2x}
k
=
1
∑
n
sin
k
x
sin
(
k
+
1
)
x
<
2
sin
2
x
cos
x
where
0
<
n
x
<
π
/
2
0 < nx < \pi/2
0
<
n
x
<
π
/2
,
n
∈
N
n \in N
n
∈
N
.
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