F.4 Amaths trigo(15分)

tan 3θ
=tan(2θ+θ)
=(tan2θ+tanθ)/(1-tan2θ tanθ)



as we all know ,if tan2θ = (tan3θ-tan2θ-tanθ)/(tan3θtanθ) , the statement is ture

then tan2θtan3θtanθ= tan3θ-tan2θ-tanθ)

==> tan2θtan3θtanθ/ tan3θ-tan2θ-tanθ=1

so , we can prove tan2θtan3θtanθ/ tan3θ-tan2θ-tanθ = 1 is right

then tan2θ = (tan3θ-tan2θ-tanθ)/(tan3θtanθ) is true...

tan2θtan3θtanθ/ tan3θ-tan2θ-tanθ
=[(tan2θ+tanθ)tan2θtanθ/(1-tan2θ tanθ)]/[(tan2θ+tanθ)/(1-tan2θ tanθ)-tan2θ-tanθ)]

=[(tan2θ+tanθ)tan2θ tanθ ]/ [tan2θ+tanθ-tan2θ(1-tan2θ tanθ)-tanθ(1-tan2θ tanθ)

=(tan2θ+tanθ)tan2θ tanθ /[tan2θ+tanθ-tan2θ+tan2θ(tan2θ tanθ)-tanθ+(tanθ)(tan2θ tanθ)]
=(tan2θ+tanθ)tan2θ tanθ /[tan2θ(tan2θ tanθ)+(tanθ)(tan2θ tanθ)]]

=(tan2θ+tanθ)tan2θ tanθ/(tan2θ+tanθ)tan2θ tanθ

=1//

so tan2θ = (tan3θ-tan2θ-tanθ)/(tan3θtanθ)

tan3θ
=tan(2θ+θ)
=[tan2θ+tanθ]/[1-tan2θtanθ]

RHS=(tan3θ-tan2θ-tanθ)/(tan3θtanθ)
*=([tan2θ+tanθ]/[1-tan2θtanθ]-tan2θ-tanθ)/([tan2θ+tanθ]tanθ/[1-tan2θtanθ])
*=[(tan2θ+tanθ-tan2θ+tan^22θtanθ-tanθ+tan^2θtanθ)/(1-tan2θtanθ)]/
([tan2θ+tanθ]tanθ/[1-tan2θtanθ])
*=(tan^22θtanθ+tan2θtan^2θ)/(tan2θtanθ+tan^2θ)
*=[tan2θ(tan2θtanθ+tan^2θ)]/(tan2θtanθ+tan^2θ)
*=tan2θ
LHS=RHS

Note:
1.*代tan3θ個條式入去,因為之前計左
2.*唔同分母既通分母
3.*之前tan2θ-tan2θ
tanθ-tanθ
1-tan2θtanθ約左
所以冇左咁多野
4.*抽左tan2θ出黎
5.*約左(tan2θtanθ+tan^2θ)
註:tan^22θ= (tan2θ)^2
tan^2θ =(tanθ)^2
因為再用括號會好亂!!sor!!