F.4 TRIGONOMETRY M2
2011-06-15 3:02 am
回答 (2)
2011-06-15 4:52 am
2011-06-15 3:12 am
(a)
Sin2θ=2Tanθ/(1+Tan^2 θ)
Sin(π/6)=1/2=2Tan(π/12)/[1+Tan^2(π/12)]
4Tan(π/12)=1+Tan^2(π/12)
Tan^2 (π/12)－4Tan(π/12)+1=0
(b)
Tan^2 (π/12)－4Tan(π/12)+1=0
Tan((π/12)=(4+/-√(16-4))/2=2+/-√3
Since Tan(π/12)<Tan(π/14)=1
So
Tan((π/12)=2－√3
Sin2θ=2Tanθ/(1+Tan^2 θ)
Sin(π/6)=1/2=2Tan(π/12)/[1+Tan^2(π/12)]
4Tan(π/12)=1+Tan^2(π/12)
Tan^2 (π/12)－4Tan(π/12)+1=0
(b)
Tan^2 (π/12)－4Tan(π/12)+1=0
Tan((π/12)=(4+/-√(16-4))/2=2+/-√3
Since Tan(π/12)<Tan(π/14)=1
So
Tan((π/12)=2－√3
收錄日期: 2021-04-30 15:52:53
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20110614000051KK00839