a.maths三角函數

2009-04-21 3:36 am
解下列方程,其中x大過或等於0゚但小於360゚
1+cos2x/2cosx=sin2x/1-cos2x

回答 (2)

2009-04-21 5:14 am
✔ 最佳答案
(1+cos2x)/(2cosx)=sin2x/(1-cos2x )
(2cos^2x)/(2cosx)=2sinxcosx/(2sin^2x)
(2cos^2x)/(2cosx)-2sinxcosx/(2sin^2x)=0
(cos^2x)/(cosx)-sinxcosx/(sin^2x)=0
(sin^2xcos^2x-sinxcos^2x)/(sin^2xcosx)=0
(sin^2xcos^2x-sinxcos^2x)/(sin^2xcosx)=0
(sinx)(cos^2x)(sinx-1)/cosxsin^2x=0
(sinx-1)=0 或 sinx=0(rejected) 或 cosx=0(rejected) <<因cosx,sinx非0
sinx=1
x=90゚
2009-04-21 3:56 am

圖片參考:http://g.imagehost.org/0982/ScreenHunter_10_Apr_20_08_05.gif

1-cos22x=2sin2xcosx
sin22x=2sin2xcosx
sin22x-2sin2xcosx=0
sin2x(sin2x-cosx)=0
sin2x=0或sinx=1/2
2x=180o或540o或x=30o或150o
x=30o或90o或150o或270o


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