Trigonometry?

Question

Find the sum to n terms of the series
Cos^2 a + Cos^2 2a+Cos^2 3a+………..

Find Sina - Sin2a + Sin3a - Sin4a ………. n terms.

Answer 1

cos^2(a)+sin^2(a)=1
cos^2(a)-sin^2(a)=cos(2a)
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cos^(2a)+sin^2(2a)=1
cos^2(2a)-sin^2(2a)=cos(4a)
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cos^(3a)+sin^2(3a)=1
cos^2(3a)-sin^2(3a)=cos(6a)
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.....
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cos^(na)+sin^2(na)=1
cos^2(na)-sin^2(na)=cos(2na)


--> 2·[ Cos^2 a + Cos^2 2a+Cos^2 3a+………..] =
n+ [Cos(2a) + Cos(4a)+.... + Cos(2na)]

--> 2·[ Cos^2 a + Cos^2 2a+Cos^2 3a+………..] =
(n-1)+ [1+ Cos(2a) + Cos(4a)+.... + Cos(2na)]

--> [ Cos^2 a + Cos^2 2a+Cos^2 3a+………..] =
(n-1)/2 + 1/2 · [1+ Cos(2a) + Cos(4a)+.... + Cos(2na)] =
(n-1)/2 + 1/2 · Real [ e^(0ai)+e^(2ai)+e^(4ai)+....+e^(2nai)] =
(n-1)/2 + 1/2 · Real { [e^(2nai+2ai)-1]/[e^(2ai)-1]} =
(n-1)/2 + 1/2· { [cos(2·(n+1)·a)-1] · [cos(2a)-1] + [sin(2·(n+1)·a)·sin(2a)]} / { [cos(2a)-1]^2+ sin^2(2a)] =

(n-1)/2+1/2 · { cos(2na) - cos(2na+2a) - cos(2a)+1} / {2-2cos(2a)} =

(n-1)/2 +1/4 · { cos(2na) - 2·cos[(n+2)a] · cos(na) +1} /{1-cos(2a)}

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Sina - Sin2a + Sin3a - Sin4a ………. =
Im{ e^(ai)-e^(2ai)+.....+(-1)^(n+1)·e^(nai)}=
=Im { [(-1)^n· e^(nai+eai) - e^(ai)] /[ -e^(2ai)-1]}

saludos.

Answer 2

For the second
sin pa= imag (e^ia)^p
so the sum of sines is the imaginary part of the sum of a geometric progression with r = e^ia

S= imag[( e^ia)^(n+1)-e ^ia ]/(e^ia -1)
The next steps are simple calculations with complex numbers
remember e^(ia)^p = e^iap = cosap +isinap