Question

By considering `1+ cistheta + cis2theta + cis3theta + .... + cisntheta` as a geometric series, can you find: `sum_(r=0)^n cosrtheta`?

Answer 1

`(sin((n+1)theta/2)cos(ntheta/2))/(sin(theta/2))`

Explanation:

Rewording the question, we are asked to find an equation for `1+costheta+cos2theta+...+cosntheta`

To solve this you need De Moirve's theorem: `(cistheta)^n=cisntheta=cosntheta+isinntheta`

Notice that the sum `1+cistheta+cis2theta+...+cisntheta` is the same as `1+cistheta+(cistheta)^2+...+(cistheta)^n`; a geometric series.
Seeing that the first term is `1`, the ratio `cistheta` and the number of terms is `n+1`, we can represent this neatly:
`1+cistheta+(cistheta)^2+...+(cistheta)^n=1((cistheta)^(n+1)-1)/(cistheta-1)`
`1+cistheta+(cistheta)^2+...+(cistheta)^n=((cis(n+1)theta)-1)/(cistheta-1)`

I am going to use a trick when simplifying `cistheta-1` here:
`cisntheta-1=c`
`=1-2sin^2(ntheta/2)+i2sin(ntheta/2)cos(ntheta/2)-1`
`=-2sin(ntheta/2)(sin(ntheta/2)+icos(ntheta/2))`
`=-2sin(ntheta/2)(cos(90-ntheta/2)+isin(90-ntheta/2))`
`=-2sin(ntheta/2)cis(90-ntheta/2)`

Notice that the left hand side can be expanded to:
`1+costheta+isintheta+cos2theta+isin2theta+...+cosntheta+isinntheta`, and that the real part of this is what we want. Thus, we need to find the real component of `((cis(n+1)theta)-1)/(cistheta-1)`

`((cis(n+1)theta)-1)/(cistheta-1)`
`(-2sin((n+1)theta/2)cis(90-(n+1)theta/2))/(-2sin(theta/2)cis(90-theta/2))`
`(sin((n+1)theta/2)cis(90-(n+1)theta/2-(90-theta/2)))/(sin(theta/2))`
`(sin((n+1)theta/2)cis(-ntheta/2))/(sin(theta/2))`
Getting the real part, we see get
`(sin((n+1)theta/2)cos(-ntheta/2))/(sin(theta/2))`
`(sin((n+1)theta/2)cos(ntheta/2))/(sin(theta/2))`
and we're done

Answer 2

`sum_(k=0)^n cos(kx) = 1/2(cos(n x) + sin(n x)(cos(x/2)/sin(x/2))+1)`

Explanation:

`sum_(k=0)^n e^(ik x) = sum_(k=0)^n cos(kx)+i sum_(k=0)^nsin(kx)`
`sum_(k=0)^n e^(-ik x) = sum_(k=0)^n cos(kx)-i sum_(k=0)^nsin(kx)`

so

`sum_(k=0)^n cos(kx) = 1/2(sum_(k=0)^n e^(ik x)+sum_(k=0)^n e^(-ik x) )`

but

`sum_(k=0)^n e^(ik x)=(e^(i(n+1)x)-1)/(e^(ix)-1) = (cos((n+1)x)-1+isin((n+1)x))/(cosx-1+isinx)`

and

`sum_(k=0)^n e^(-ik x)=(e^(-i(n+1)x)-1)/(e^(-ix)-1) = (cos((n+1)x)-1-isin((n+1)x))/(cosx-1-isinx)`

so after simplifications

`sum_(k=0)^n e^(ik x)+sum_(k=0)^n e^(-ik x) =cos(n x) + sin(n x)(cos(x/2)/sin(x/2))+1`

and finally

`sum_(k=0)^n cos(kx) = 1/2(cos(n x) + sin(n x)(cos(x/2)/sin(x/2))+1)`