trigo problem

用戶117714

2007-03-03 9:07 am

(A)Prove that cos3θ=4cos^3θ-3cosθ
(B)then , by puttingx=2cosθ,show that the equation x^3-3x-1=0 can be transformed to cos3θ=1/2
(C)Hence find correct to two decimal places,the three roots of the equations in (B)

回答 (1)

chaoseng2000

2007-03-03 11:04 am

✔ 最佳答案

(A)cos3θ
=cos(2θ+θ)
=cos2θcosθ-sin2θsinθ
=[2(cosθ)^2-1]cosθ-2sinθcosθsinθ
=2(cosθ)^3-cosθ-2[1-(cosθ)^2]cosθ<-------------(sinθ)^2=1-(cosθ)^2
=2(cosθ)^3-cosθ-2cosθ+2(cosθ)^3
=4(cosθ)^3-3cosθ

(B)
(2cosθ)^3-3*2cosθ-1=0
8(cosθ)^3-6cosθ=1
2[4(cosθ)^3-3cosθ]=1<-----------by(A)
cos3θ=1/2

(C)by(B)
cos3θ=1/2
θ=20度
x=2cos20度=1.88
因(x-1.88)為其中一個因式
則(x^3-3x-1)/(x-1.88)=x^2+1.88x+0.5344是因式
即x^2+1.88x+0.5344=0
用求根公式
x=[-b+-(b^2-4ac)^1/2]/2a得
x=-0.36
x=-1.54
x=1.88or-0.36or-1.54

2007-03-03 03:11:19 補充：
x^2+1.88x+0.5344=0用求根公式x=[-b+-(b^2-4ac)^1/2]/2a得x=-0.35x=-1.53x=1.88or-0.35or-1.53

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