五條附加數問題! - Ans.fyi 參考答案

五條附加數問題!

ka-kin

2007-01-29 7:16 pm

1.slove 2cosA-cotA=0-for0<360
2.slove (tan^2)A+4(sec^2)A-9=0 for 0<360.
3.If the quadratic equation
(3sinA)x^2-(4cosA)x+2=0
in x has equal roots,find the value (s) of A where 0<360
4.Find the maximum and minimum value of y=1-2sin^2+4cosx
and the corresponding values of x for 0<360 at which such extreme values are obtained.
5.If cosA and cosescA are the two roots of of the equation 3x^2+hx-1=0,where 180<360,find the value(s) of h and leave your answer in surd form.

更新1:

5.If cosA and cosescA are the two roots of of the equation 3x^2+hx-1=0,where 180

更新2:

5.If cosA and cosescA are the two roots of of the equation 3x^2+hx-1=0,where 180

回答 (1)

myisland8132

2007-01-29 7:56 pm

✔ 最佳答案

1

2cosA-cotA=0

2cosA=cosA/sinA

2cosAsinA=cosA

cosA(2sinA-1)=0

cosA=0 or sinA=1/2

A=90, 270, 30, 150

2

(tan^2)A+4(sec^2)A-9=0

(tan^2)A+4(1+tan^2)A-9=0

5tann^2A+4tanA-9=0

(5tanA+9)(tanA-1)=0

tanA=-9/5 or 1

A=119.05 or 299.05 or 45 or 225

3

discriminant=0

(4cosA)^2-24sinA=0

[16(1-sin^2A)]-24sinA=0

16-16sin^2A-24sinA=0

16sin^2A+24sinA-16=0

2sin^2A+3sinA-2=0

(2sinA-1)(sinA+2)=0

sinA=1/2 or sinA=-2 (rejected)

A=30 or 150

y

=1-2sin^2+4cosx

=1-2(1-cos^2x)+4cosx

=2cos^2x+4cosx-1

=2(cosx+1)^2-2-1

=2(cosx+1)^2-3

So when cosx=-1 (x=180), y gets the minimum value -3

when cosx=1 (x=0), y gets the maximum value 5

5

cosA +cscA=-h/3 ...(1)

(cosA)(cscA)=-1/3...(2)

from (2)

cotA=-1/3

cot^2A=1/9

1+cot^2A=csc^2A=10/9

cscA=-√10/3

cosA=-1/√10 or 1/√10

sub into (1)

when cosA=1/√10

cosA +cscA=-h/3

1/√10-√10/3=-h/3

3-10=-√10h

h=7/√10

when cosA=-1/√10

cosA +cscA=-h/3

-1/√10-√10/3=-h/3

-3-10=-√10h

h=13/√10

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