Question

已知sinα+sinβ=1,cosα+cosβ=0,求cos2α+cos2β

Answer 1

sinα+sinβ=1,两边平方得(sinα)^2+(sinβ)^2+2sinα*sinβ=1,
(1)
cosα+cosβ=0,两边平方得(cosα)^2+(cosβ)^2+2cosα*cosβ=0,
(2)
(1)+(2)得2+2cos(α-β)=1,则cos(α-β)=-1/2,(3)
由(3)得α-β=(2k+1)P-P/3,或者α-β=(2k+1)P+P/3,P=Pai
(1)-(2)得cos2α+cos2β+2cos(α+β)=-1,
(4)
当α-β=(2k+1)P-P/3时,
代入sinα+sinβ=1,得....代入cosα+cosβ=0得...,两式平方得sin2β,进而得cos2β.
cos(α+β)=cos(2β+2/3P)可求出,代入(4)即得cos2α+cos2β
实在很难算，算不下去了，不好意思

Answer 2

sina=1-sinb.....(1)
cosa=-cosb.......(2)
将两式平方,得
(sina)^2+(cosa)^2=(1-sinb)^2+(-cosb)^2=1
整理得,sinb=1/2
sina=1/2
cos2a=1-2(sina)^2=1/2
cos2b=1-2(sinb)^2=1/2
Cos2α+Cos2β=1