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 .

Question

It&#039;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  .

rioolpijpgp · Accepted Answer

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.

It's easy to show that sin ⁡ A − sin ⁡<!-- ⁡

Answered question

Answer & Explanation

New Questions in Calculus 2