By chance I found: 2 sin ⁡ ( 2 m + 1 ) x −...

Jameson Lucero

Jameson Lucero

Answered

2022-07-04

By chance I found:
2 sin ( 2 m + 1 ) x sin ( 2 m 1 ) x = sin ( x ) ( 1 + 2 cos ( 2 m x ) + 2 k = 1 m cos ( 2 k x ) )

Answer & Explanation

Immanuel Glenn

Immanuel Glenn

Expert

2022-07-05Added 12 answers

You may write, for any real number x Z π
k = 1 m cos ( 2 k x ) = k = 1 m e 2 i k x = ( e 2 i x e 2 i m x 1 e 2 i x 1 ) = ( e 2 i x e i m x ( e i m x e i m x ) e i x ( e i x e i x ) ) = ( e i ( m + 1 ) x sin ( m x ) sin ( x ) ) = ( ( cos ( ( m + 1 ) x ) + i sin ( ( m + 1 ) x ) ) sin ( m x ) sin ( x ) ) = cos ( ( m + 1 ) x ) sin ( x ) sin ( m x ) .
Thus
sin ( x ) ( 1 + 2 cos ( 2 m x ) + 2 k = 1 m cos ( 2 k x ) ) = sin ( x ) + 2 sin ( x ) cos ( 2 m x ) + 2 cos ( ( m + 1 ) x ) sin ( m x )
and, using 2 sin a cos b = sin ( a + b ) + sin ( a b ) , we get
sin ( x ) ( 1 + 2 cos ( 2 m x ) + 2 k = 1 m cos ( 2 k x ) ) = sin ( x ) + sin ( ( 2 m + 1 ) x ) sin ( ( 2 m 1 ) x ) + sin ( ( 2 m + 1 ) x ) sin ( x ) = 2 sin ( ( 2 m + 1 ) x ) sin ( ( 2 m 1 ) x )
as announced.

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