It's easy to show that sin ⁡ A − sin ⁡ B = 2 cos...

Dale Tate

Dale Tate

Answered

2022-06-26

It's easy to show that
sin A sin B = 2 cos ( A + B 2 ) sin ( A B 2 ) ,
sin ( 2 k + 1 ) x sin ( 2 k 1 ) x = 2 cos 2 k x sin x .

Answer & Explanation

rioolpijpgp

rioolpijpgp

Expert

2022-06-27Added 19 answers

The double angle formula gives sin 2 ( k x ) = 1 cos ( 2 k x ) 2
So we have
k = 1 n sin 2 ( k x ) = k = 1 n 1 cos ( 2 k x ) 2 = n 2 1 2 k = 1 n cos ( 2 k x )
= n 2 1 4 sin x k = 1 n 2 cos ( 2 k x ) sin x
= n 2 1 4 sin x k = 1 n [ sin ( 2 k + 1 ) x sin ( 2 k 1 ) x ]
= n 2 1 4 sin x [ sin ( 2 n + 1 ) x sin x ]
= n 2 1 4 sin x [ 2 cos ( n + 1 ) x sin n x ]
= n 2 cos ( n + 1 ) x sin n x 2 sin x
Provided sin x 0, i.e. x is not an integer multiple of π. Though in these cases the equality is still true if taken as a limit.

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