Prove resemblance between ∑ n = 1 ∞ sin ⁡ ( 2 n x ) and cot ⁡ x

Question

Prove resemblance between       ∑          n      =      1        ∞    sin  ⁡  (  2  n  x  ) and   cot  ⁡  x

Selena Pratt · Accepted Answer

From the formula      ∑          k      =      1              n        sin  ⁡  (  2  k  θ  )  =      1    2    cot  ⁡  θ  −      1    2              cos      ⁡      (      2      n      +      1      )      θ              sin      ⁡      θ        ,it seems that the upper envelope is given by       1    2    (  cot  ⁡  θ  +      |    csc  ⁡  θ      |    ) and likewise the lower envelope is given by       1    2    (  cot  ⁡  θ  −      |    csc  ⁡  θ      |    ). When   0  &amp;lt;  θ  &amp;lt;  π, we can simplify them as      1    2    (  cot  ⁡  θ  +  csc  ⁡  θ  )  =      1    2    cot  ⁡      θ    2      and        1    2    (  cot  ⁡  θ  −  csc  ⁡  θ  )  =  −      1    2    tan  ⁡      θ    2    .

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