*Notes to a video lecture on http://www.unizor.com*

__Trigonometry+ 04__

*Problem A*

Find the sums

Σ

_{k∈[0,n−1]}

**sin²(x+k·π/n)***Solution A*

First, convert

*sin²(...)*into

*cos(...)*by using the identity

*cos(2φ) = cos²(φ) − sin²(φ) =*

= 1 − 2sin²(φ)

= 1 − 2sin²(φ)

from which follows

*sin²(φ) = ½(1 − cos(2φ))*

Now our sum looks like this

Σ

_{k∈[0,n−1]}

*Σ*

**½(1−cos(2x+2k·π/n)) =**

= n/2 −

− ½= n/2 −

− ½

_{k∈[0,n−1]}

**cos(2x+2k·π/n)**To calculate Σ

_{k∈[0,n−1]}above, let's use the result of the previous lecture

*Trigonometry 03*that proved the following

Σ

_{k∈[0,n]}

*[*

**cos(x+k·y) =**

==

*]*

**sin(x+(2n+1)·y/2) −**

− sin(x−y/2)− sin(x−y/2)

*[*

**/**

//

*]*

**2·sin(y/2)**Since in our sum we have

*n*terms instead of

*n+1*, as in the formula above, to use this formula in our case we have to use

*n−1*instead of

*n*, which will result in

Σ

_{k∈[0,n−1]}

*[*

**cos(x+k·y) =**

==

*]*

**sin(x+(2n−1)·y/2) −**

− sin(x−y/2)− sin(x−y/2)

*[*

**/**

//

*]*

**2·sin(y/2)**To use this formula in our case, we have to substitute

*2x*instead of

*x*and

*2π/n*instead of

*y*.

In this case the numerator in the formula above would be

**sin(x+(2n−1)·π/n) − sin(x−π/n)**The first term in the above expression for the numerator equals to

**sin(x+(2n−1)·π/n) =**

= sin(x+2n·π/n−π/n) =

= sin(x−π/n+2π) = sin(x−π/n)= sin(x+2n·π/n−π/n) =

= sin(x−π/n+2π) = sin(x−π/n)

(since the value

*2π*is a period of function

*sin*)

which makes the first term exactly the same as the second term and they will cancel each other resulting in zero in the numerator.

That leaves the sum in our case to be equal to

Σ

_{k∈[0,n−1]}

**½(1−cos(2x+2k·π/n)) = n/2***Answer A*

Σ

_{k∈[0,n−1]}

**sin²(x+k·π/n) = n/2***Problem B*

Given a regular

*n*-sided polygon with vertices

*,*

**A**_{1}*,...,*

**A**_{2}*inscribed into a circle of radius*

**A**_{n}*with a center at point*

**R***and any point*

**O***on this circle.*

**P**Calculate the sum of squares of distances from point

*to all vertices*

**P**

**A**_{k}*Σ*

**S =**_{k∈[1,n]}

*.*

**(PA**_{k})²*Hint*:

Let ∠

*.*

**POA**_{k}= φ_{k}Then

*.*

**φ**_{k}= φ_{1}+ (k−1)·2π/nUse the results of

*Problem A*.

*Answer*

**S = 2nR²**
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