It's easy to show that sin &#x2061;<!-- ⁡ --> A &#x2212;<!-- − --> sin &#x2061;<!-- ⁡

Dale Tate

Dale Tate

Answered question

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

Beginner2022-06-27Added 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.

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?