急求助!!F.4 A.maths

2007-02-07 5:06 am
Solve the following equations for 0 ≦ x ≦ 360(Q1-Q4)
1. ( 1 - tanx)(1 + sin2x)=1+tanx
2.sin^4x + cos ^4x = 1/2sin2x
3.cos4x + sin2x = 0
4.Suppose y =(sec^2x csc^2x) - 2
a. Show that y =( 4/sin^2 2x ) -2
5 Show that sin3x/sinx + cos3x/cosx = 4cos2x
6 Let f(x) = sin^6x + cos^6 x Espressf(x) in the form A + Bcos4x where A and B are constants Hence find the maximun and minimun values of f (x)
solve the equation sec^2x = (2 - cosx - sinx)/1 - sinx for 0 ≦ x ≦ 360

回答 (1)

2007-02-07 9:59 am
✔ 最佳答案
1) (1 - tanx)(1 + sin2x)=1+ tanx
1+ 2sinx cosx - tanx - 2sinx cosx tanx = 1+ tanx
2sinx cosx-2(sinx)^2=2 tanx
sinx(cosx - sinx)=sinx/cosx
sinx=0 or cosx - sinx=1/cosx
sinx=0 or (cosx)^2 - sinxcosx=1
sinx=0 or 1 - (cosx)^2 + sinxcosx = 0
sinx=0 or (sinx)^2 + sinxcosx = 0
sinx=0 or sinx(sinx+cosx) = 0
sinx=0 or tanx = -1
x=0, 180 or -135, -315

2)(sinx)^4 + (cosx)^4 = 1/2sin2x
((sinx)^2 + (cosx)^2)^2 - 2(sinxcosx)^2 = 1/2sin2x
(1)^2 - 0.5(sin2x)^2 = 1/2sin2x
sin2x - 0.5(sin2x)^3 - 0.5 = 0
(sin2x)^3 -2 sin2x + 1 = 0
Let y = sin2x
y^3 -2y + 1 = 0
(y-1)(y^2+y-1)=0
y=1 or (-1+(root5))/2 or (-1-(root5))/2
sin2x = 1 or (-1+(root5))/2 or (-1-(root5))/2
2x = 90,450 or 38.173, 141.827, 398.173, 501.827 or no solution
x = 45 / 225 / 19.086 / 70.914 / 199.086 / 250.914

3) cos4x + sin2x = 0
1-2(sin2x)^2+sin2x=0
2(sin2x)^2 - sin2x - 1=0
(sin2x - 1)(2sin2x + 1)=0
sin2x=1 or -0.5
2x=90, 450 or 210, 330, 570, 690
x=45, 225 or 105, 165, 285, 345

4)Suppose y =(secx)^2 (cscx)^2 - 2
(a) y =(secx)^2 (cscx)^2 - 2 = 1/(cosxsinx)^2-2
=1/(sin2x/2)^2-2 = 4/(sin2x)^2-2

5) sin3x/sinx + cos3x/cosx
= (sinxcos2x+cosxsin2x)/sinx + (cosxcos2x-sinxsin2x)/cosx
=cos2x+2(cosx)^2 + cos2x - 2(sinx)^2
=2cos2x+2((cosx)^2-(sinx)^2) = 2cos2x+2cos2x = 4cos2x

6) f(x) = (sinx)^6 + (cosx)^6
=((sinx)^2 + (cosx)^2)^3 - 3(sinx)^2(cosx)^2((sinx)^2+(cosx)^2)
=(1)^3 - 3(sinx)^2(1-(sinx)^2)(1)
=1 - 3(sinx)^2 + 3(sinx)^4
=cos2x - (sinx)^2 + 3(sinx)^4
=cos2x - 0.5 + 0.5cos2x + 3/4 * (1 - 2cos2x+(cos2x)^2)
=cos2x - 0.5 + 0.5cos2x + 3/4 - 1.5cos2x + 3/4 * (cos2x)^2
=0.25 + 0.75* (0.5+0.5cos4x)
=0.25 + 0.385 + 0.385cos4x = 5/8 + 3/8 * cos4x
max value=5/8+3/8*(1)=1, min value=5/8+3/8*(-1)=1/4

7)(secx)^2 = (2 - cosx - sinx)/(1 - sinx)
1+(tanx)^2 = (1-cosx)/(1-sinx)+1
(tanx)^2 = (1-cosx)/(1-sinx)
(1-sinx)(tanx)^2 = (1-cosx)
(sinx)^2-(sinx)^3 = (cosx)^2 - (cosx)^3
(sinx)^2-(cosx)^2 = (sinx)^3 - (cosx)^3
(sinx+cosx)(sinx-cosx) = (sinx-cosx)((sinx)^2+sinxcosx+(cosx)^2)
sinx-cosx=0 or sinx+cosx=1+sinxcosx
tanx=1 or 1+sinxcosx-sinx-cosx = 0
tanx=1 or 1-sinx+cosx(sinx-1) = 0
tanx=1 or (1-cosx)(1-sinx)= 0
tanx=1 or sinx=1 or cosx=1
x=0,45,90,225
sec 90 is undefined, thus x is not 90
Thus, x=0, 45, 225

2007-02-07 17:59:09 補充:
1)x=0, 180, 360 or -135, -3157)x=0, 45, 225, 360


收錄日期: 2021-04-15 21:01:30
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20070206000051KK03771

檢視 Wayback Machine 備份