已知cosα=√3/3,化簡(jiǎn)并求值:(1+(tan2α)^2)[cos(2α+π/3)+cos(2α-π/3)]

數(shù)學(xué)人氣：387 ℃時(shí)間：2020-10-02 03:12:11

優(yōu)質(zhì)解答

(1+(tan2α)^2)[cos(2α+π/3)+cos(2α-π/3)]=[1+(2tanα/(1-tan^2α)^2][cos2αcos-sin2αsinπ/3+cos2αcos+sin2αsinπ/3]=[(1+tan^2α)/(1-tan^2α)]^2(2cos2αcosπ/3)=[1/(cos^2α-sin^2α)]^2(2cos2αcosπ/...=2cosπ/3/[2(cosα)^2-1]到這一步之前都是對(duì)的，不過(guò)最后答案好像算錯(cuò)哩，是-3吧2cosπ/3=2*1/2=1(cosα)^2=1/3[2(cosα)^2-1]=2*1/3-1=2/3-1=-1/32cosπ/3/[2(cosα)^2-1]=1/(-1/3)=-3原來(lái)錯(cuò)了