# Show 1+ cos(2pie/5)+cos(4pie/5)+cos(6pie/5)+cos(8pie/5)=0?

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### 1 Answer

You should come up with the following notation for the summation to the left side, such that:

`S = 1 + cos((2pi)/5) + cos((4pi)/5) + cos((6pi)/5) + cos((8pi)/5)`

You should test if `S = 0` , hence, you may perform the following multiplication, such that:

`S*sin(pi/5) = sin(pi/5) + cos((2pi)/5)*sin(pi/5) + cos((4pi)/5)*sin(pi/5) + cos((6pi)/5)*sin(pi/5) + cos((8pi)/5)*sin(pi/5)`

Converting the products into summations, yields:

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

`cos((2pi)/5)*sin(pi/5) = (sin((3pi)/5) - sin(pi/5))/2`

`cos((4pi)/5)*sin(pi/5) = (sin((4pi)/5 + pi/5) - sin((4pi)/5 - pi/5))/2`

`cos((4pi)/5)*sin(pi/5) = (sin(pi) - sin((3pi)/5))/2`

`cos((6pi)/5)*sin(pi/5) = (sin((6pi)/5 + pi/5) - sin((6pi)/5 - pi/5))/2`

`cos((6pi)/5)*sin(pi/5) = (sin((7pi)/5) - sin pi)/2`

`cos((8pi)/5)*sin(pi/5) = (sin((8pi)/5 + pi/5) - sin((8pi)/5 - pi/5))/2`

`cos((6pi)/5)*sin(pi/5) = (sin((9pi)/5) - sin((7pi)/5))/2`

`S*sin(pi/5) = sin(pi/5) + (sin((3pi)/5) - sin(pi/5) + sin(pi) - sin((3pi)/5) + sin((7pi)/5) - sin pi + sin((9pi)/5) - sin((7pi)/5)/2`

Reducing duplicate members yields:

`S*sin(pi/5) = sin(pi/5) + (sin((9pi)/5) - sin(pi/5))/2`

`S*sin(pi/5) = (2 sin(pi/5) + sin((9pi)/5) - sin(pi/5))/2`

`S*sin(pi/5) = (sin((9pi)/5) + sin(pi/5))/2`

Converting back `sin((9pi)/5) + sin(pi/5)` into a product, yields:

`S*sin(pi/5) = (2sin ((9pi)/5 + pi/5)/2*cos ((9pi)/5 - pi/5)/2)/2`

`S*sin(pi/5) = sin ((2pi)/2)*cos((4pi)/5)`

`S*sin(pi/5) = sin pi*cos((4pi)/5)`

Since `sin pi = 0` yields:

`S*sin(pi/5) = 0 => {(sin(pi/5) != 0),(S = 0):}`

**Hence, testing if the given summation is equal to zero yields that `1 + cos((2pi)/5) + cos((4pi)/5) + cos((6pi)/5) + cos((8pi)/5) = 0` , holds.**

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