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Mathematics
Mathematics
Problem
Prove that for any positive integer
K
K
K
,
sin
2
K
x
sin
x
=
2
[
cos
x
+
cos
3
x
+
⋯
+
cos
(
2
K
−
1
)
x
]
\frac{\sin 2Kx}{\sin x} = 2[\cos x + \cos 3x + \dots + \cos(2K-1)x]
sin
x
sin
2
K
x
=
2
[
cos
x
+
cos
3
x
+
⋯
+
cos
(
2
K
−
1
)
x
]
Hence, prove that
∫
0
π
/
2
sin
2
K
x
⋅
cot
x
d
x
=
π
2
\int_0^{\pi/2} \sin 2Kx \cdot \cot x dx = \frac{\pi}{2}
∫
0
π
/2
sin
2
K
x
⋅
cot
x
d
x
=
2
π
.
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