Summation of series

Question

!!sum_(k=1)^10 sin((2kpi)/11)!! is:

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Solution

There a few methods to solve this question, firstly:

Using the forumla !!sin(x)+sin(2x)+..+sin(nx)=(sin(((n+1)x)/2)sin((nx)/2))/sin(x/2)!!

!!sum_(k=1)^10 sin((2kpi)/11)!!

!!=(sin((2xx11xxpi/11)/2)sin((10xx2pi/11)/2))/sin((pi/11)/2)!!

!!0!!

Secondly

We know that the 11th root of unity are of the form !!e^((ixx2kpi)/11)!! for !!kin0"to"10!!

These are the roots of the equation !!z^11=1!!

So the sum of roots of this equation are zero.

So !!sum_(k=0)^10 e^((ixx2kpi)/11)=0!!

!!=>sum_(k=0)^10 cos((2kpi)/11)+isum_(k=0)^10 sin((2kpi)/11)=0!!

!!=>sum_(k=0)^10 sin((2kpi)/11)=0!!

!!=>sum_(k=1)^10 sin((2kpi)/11)=0!!

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