唔識做依題數呀,,,urgent plz

2011-06-04 8:45 am

a)solve the equation tan3x=cot2x for 0<x<90
b)it is given that (tan3x)=(3tanx-tan^3x)/(1-3tan^2x)
express tan3x=cot2x in the form atan^4x+btan2x+c=0
c)prove that tan18=開方[(5-2開方5)/5]
只係唔識c(應該用result of a and b)

回答 (1)

螞蟻雄兵

2011-06-04 11:56 am

✔ 最佳答案

a)solve the equation Tan3x=Cot2x for0<x<90degree
Tan3x=Cot2x
Tan3x=Tan(90－2x)
3x=90degree 3x=270degree 3x=450degree
x=18degreor x=54degree or x=90degree(不合)

b)it is given that (Tan3x)=(3Tanx－Tan^3 x)/(1－3Tan^2 x)
express Tan3x=Cot2x in the form aTan^4x+bTan2x+c=0
Sol
Cot2x=1/Tan2x=(1－Tan^2 x)/(2Tanx)
So
(3Tanx－Tan^3 x)/(1－3Tan^2 x)=(1－Tan^2 x)/(2Tanx)
(3Tanx－Tan^3 x)(2Tanx)=(1－3Tan^2 x)(1－Tan^2 x)
6Tan^2 x－2Tan^4 x=1－Tan^2 x－3Tan^2 x+3Tan^4 x
5Tan^4x－10Tan^2 x+1= 0
c)prove that Tan18degree=√[(5－2√5)/5]
Sol
Set w=Tan18degree<1
from a) x=18degree is a solution
5w^4－10w^2+1=0
w^2=(10－√(100-20))/10 (+ reject since w^2>1)
=(10－4√5)/10
=(5－2√5)/5
w=√[(5－2√5)/5]

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