Demonstrate that 1+cos2pie/5+cos4pie/5+cos6pie/5+cos8pie/5=0 without calculating.use demoivre

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You should consider the following summation `S_2 = sin((2pi)/5) + sin((4pi)/5) + sin((6pi)/5) + sin((8pi)/5)`  if `S_1 = 1 + cos((2pi)/5) + cos((4pi)/5) + cos((6pi)/5) + cos((8pi)/5).`

You need to perform the following addition such that:

`S_1 + i*S_2 = 1 + (cos((2pi)/5) + i*sin((2pi)/5)) + (cos((4pi)/5) + i*sin((4pi)/5)) + (cos((6pi)/5) + i*sin((6pi)/5)) + (cos((8pi)/5) + i*sin((8pi)/5))`

You should notice that `(cos((2pi)/5) + i*sin((2pi)/5)) ` represents the trigonometric form of complex number`z`  such that:

`z = (cos((2pi)/5) + i*sin((2pi)/5))`

You may use De Moivre's identity such that:

`z = (cos alpha + i*sin alpha)^n = cos n*alpha + i*sin n*alpha`

Reasoning by analogy yields:

`(cos((2pi)/5) + i*sin((2pi)/5))^2 = (cos((4pi)/5) + i*sin((4pi)/5)) = z^2`

`(cos((2pi)/5) + i*sin((2pi)/5))^3 = (cos((6pi)/5) + i*sin((6pi)/5)) = z^3`

`(cos((2pi)/5) + i*sin((2pi)/5))^4 = (cos((8pi)/5) + i*sin((8pi)/5)) = z^4`

`S_1 + i*S_2 = 1 + z + z^2 + z^3 + z^4`

You need to use the following identity, such that:

`a^n - b^n = (a - b)(a^(n-1) + a^(n-2)*b + ... + a*b^(n-2) + b^(n-1))`

Reasoning by analogy yields:

`z^5 - 1 = (z - 1)(z^4+ z^3+ z^2+ z + 1)`

`z^4 + z^3 + z^2 + z + 1 = (z^5 - 1)/(z - 1)`

`S_1 + i*S_2 = (z^5 - 1)/(z - 1)`

Substituting back `(cos((2pi)/5) + i*sin((2pi)/5))`  for`z`  and using De Moivre's identity yields:

`z^5 =(cos((2pi)/5) + i*sin((2pi)/5))^5 = cos(2pi) + i*sin(2pi)`

Since `cos 2pi = 1`  and `sin 2pi = 0`  yields:

`z^5 = 1 => z^5 - 1 = 1 - 1 = 0`

`S_1 + i*S_2 = (z^5 - 1)/(z - 1) = 0/(z-1) = 0 => {(S_1 = 0),(S_2 = 0)}.`

Since `S_1 = 1 + cos((2pi)/5) + cos((4pi)/5) + cos((6pi)/5) + cos((8pi)/5) => 1 + cos((2pi)/5) + cos((4pi)/5) + cos((6pi)/5) + cos((8pi)/5) = 0.`

Hence, checking if the expression `S_1 = 1 + cos((2pi)/5) + cos((4pi)/5) + cos((6pi)/5) + cos((8pi)/5) = 0` , without calculating it, yields that the identity `S_1 = 0`  holds.

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