Question

代数的题，这个怎么做，在线等，急

设a,x∈R,n∈N 计算以下总和：
1+acos⁡x+a^2 cos⁡2x+⋯+a^n cos⁡nx
asin⁡x-a^2 sin⁡2x+⋯+(-1)^(n-1) a^n sin⁡nx
sin^2⁡x+sin^2⁡2x+⋯+sin^2⁡nx
cos^2⁡x+cos^2⁡2x+⋯+cos^2⁡nx

Answer 1

(1) 等比复数级数
1 + a(cosx+isinx) + a^2(cosx+isinx)^2 + ...... + a^n(cosx+isinx)^n
= [a^(n+1)(cosx+isinx)^(n+1) - 1]/[a(cosx+isinx)-1]
= {a^(n+1)[cos(n+1)x+isin(n+1)x] - 1}/[(acosx-1)+isinx]
= {a^(n+1)[cos(n+1)x+isin(n+1)x] - 1}[(acosx-1)-isinx]/[(acosx-1)^2+(sinx)^2]
取实部， 得
1+acos⁡x+a^2cos⁡2x+⋯+a^ncos⁡nx
= {[a^(n+1)cos(n+1)x-1](acosx-1)+sin(n+1)xsinx}/[(acosx-1)^2+(sinx)^2]
(2) 等比复数级数
a(cosx+isinx) - a^2(cosx+isinx)^2 + ...... + (-1)^(n-1)a^n(cosx+isinx)^n
= a(cosx+isinx)[1-(-1)^n a^n(cosx+isinx)^n]/[1+a(cosx+isinx)]
= a(cosx+isinx)[1-(-1)^n a^n(cosnx+isinnx)]/[(1+acosx)+iasinx]

= a[(1+acosx)-iasinx](cosx+isinx)[1-(-1)^n a^n(cosnx+isinnx)]
/[(1+acosx)^2+(asinx)^2]
= a[(1+acosx)cosx+a(sinx)^2+i[sinx(1+acosx)-asinxcosx]
·[1-(-1)^n a^n(cosnx+isinnx)]/[(1+acosx)^2+(asinx)^2]
= a(1+a+isinx)·[1-(-1)^na^ncosnx - i(-1)^na^nsinnx)]/[(1+acosx)^2+(asinx)^2]
取虚部，得
asinx - a^2(sin2x) + ...... + (-1)^(n-1)a^n(sinnx)
= a{(-1)^(n+1)(1+a)a^nsinnx)+sinx[1-(-1)^na^ncosnx]}/[(1+acosx)^2+(asinx)^2]
(3) 由 (1), 取 a = 1， x = 2u，得
cos⁡2u+cos⁡4u+⋯+cos⁡2nu
= -1+{[cos2(n+1)u-1](cos2u-1)+sin2(n+1)usin2u}/[(cos2u-1)^2+(sin2u)^2]
即 cos⁡2x+cos⁡4x+⋯+cos⁡2nx
= -1+{[cos2(n+1)x-1](cos2x-1)+sin2(n+1)xsin2x}/[(cos2x-1)^2+(sin2x)^2]
则 sin^2⁡x+sin^2⁡(2x)+⋯+sin^2⁡(nx)
= (1/2)[1-cos2x+1-cos4x + ...... + 1-cos2nx]
= (1/2)[n-(cos2x+cos4x + ...... + cos2nx)]
= (1/2)【n+1-{[cos2(n+1)x-1](cos2x-1)+sin2(n+1)xsin2x}/[(cos2x-1)^2+(sin2x)^2]】
（4） 由 (3), cos^2⁡x+cos^2⁡(2x)+⋯+cos^2⁡(nx)
= 1-sin^2⁡x+1-sin^2⁡(2x)+⋯+1-sin^2⁡(nx)
= n-[sin^2⁡x+sin^2⁡(2x)+⋯+sin^2⁡(nx)]
= (1/2)【n-1+{[cos2(n+1)x-1](cos2x-1)+sin2(n+1)xsin2x}/[(cos2x-1)^2+(sin2x)^2]】