20分 請解答amaths題目（詳細步驟）解trigo有秘訣嗎？？？

(1)
LHS
= (sec x - cos x)(1 + cot x + tan x)
= (1/cos x - cos x)(1 + cos x/sin x + sin x/cos x)
= (1 - cos² x)/cos x * (sin x cos x + cos² x + sin² x)/(sin x cos x)
= sin² x (sin x cos x + 1)/(sin x cos² x)
= sin x (sin x cos x + 1)/cos² x

RHS
= sec² x/csc x + sec x/csc² x
= (1/cos² x)/(1/sin x) + (1/cos x)/(1/sin² x)
= sin x/cos² x + sin² x/cos x
= (sin x + sin²x cos x)/cos² x
= sin x (1 + sin x cos x)/cos² x

So, LHS = RHS.


(2)
LHS
= [sec(-x) + sin(-x - 90°)]/[csc(540° - x) - cos(270° + x)]
= [1/cos(-x) - sin(90° + x)]/[1/sin(540° - x) - sin x]
= [1/cos x - cos x]/[1/sin(180° - x) - sin x]
= [(1 - cos² x)/cos x]/[1/sin x - sin x]
= [sin² x/cos x]/[(1 - sin² x)/sin x]
= [sin² x/cos x]/[cos² x/sin x]
= sin³ x/cos³ x
= tan³ x
= RHS


(3)
sec x (sec x - tan x) = x
1/cos x (1/cos x - sin x/cos x) = x
1/cos x * (1 - sin x)/cos x = x
(1 - sin x)/cos² x = x
(1 - sin x)/(1 - sin² x) = x
(1 - sin x)/[(1 - sin x)(1 + sin x)] = x
1/(1 + sin x) = x
1 = x + x sin x
1 - x = x sin x
sin x = (1 - x)/x


(4)
Since sin² x + sin x - 1 = 0
So, sin² x = 1 - sin x -------------(*)

(a)
cos^4 x + cos² x - 1
= (1 - sin² x)² + (1 - sin² x) - 1
= 1 - 2 sin² x + sin^4 x - sin² x
= 1 - 3 sin² x + (sin² x)²
= 1 - 3 (1 - sin x) + (1 - sin x)²【由(*)】
= 1 - 3 + 3 sin x + (1 - 2 sin x + sin² x)
= -1 + sin x + sin² x
= -1 + sin x + (1 - sin x)【By (*)】
= 0

(b)
cos^8 x + cos^6 x + cos² x - 1
= cos^8 x + cos^6 x - cos^4 x + cos^4 x + cos² x - 1【補上cos^4 x項】
= cos^4 x (cos^4 x + cos² x - 1) + (cos^4 x + cos² x - 1)
= cos^4 x (0) + (0)【由(a)】
= 0


(5)
LHS
= (cos² x - sin x cos x + tan x)/(cos² x + sin x cos x - tan x)
= (cos² x - sin x cos x + sin x/cos x)/(cos² x + sin x cos x - sin x/cos x)
= [(cos³ x - sin x cos² x + sin x)/cos x]/[(cos³ x + sin x cos² x - sin x)/cos x]
= (cos³ x - sin x cos² x + sin x)/(cos³ x + sin x cos² x - sin x)
= (1 - tan x + tan x/cos² x)/(1 + tan x - tan x/cos² x)【分子分母除以cos³ x】
= (1 - tan x + tan x sec² x)/(1 + tan x - tan x sec² x)
= [1 - tan x + tan x (1 + tan² x)]/[1 + tan x - tan x (1 + tan² x)]
= [1 + tan³ x]/[1 - tan³ x]
= RHS


(6)
Product of roots = cos x csc x = cos x/sin x = 1/tan x
On the other hand, product of roots = (-1)/3 = -1/3
So, 1/tan x = -1/3
tan x = -3

Since π/2 < x < π, x-coordinate = -1, y-coordinate = 3
sin x = 3/√(3² + 1²) = 3/√10
cos x = -1/√10

Sum of roots = cos x + csc x = cos x + 1/sin x
On the other hand, sum of roots = (-h)/3 = -h/3
So, -h/3 = cos x + 1/sin x
h = -3 [-1/√10 + 1/(3/√10)]
h = 3/√10 + (-3) √10/3
h = 3/√10 - √10
h = 3√10 /10 - √10
h = (3√10 - 10√10)/10
h = -7√10 /10


(7)
Sum of roots = sin x + cos x = -(-4)/k = 4/k --------------(#)

Since sin x and cos x are roots of kx² - 4x + 3 = 0
k sin² x - 4 sin x + 3 = 0 ----------(1)
k cos² x - 4 cos x + 3 = 0 ----------(2)

(1) + (2):
k (sin² x + cos² x) - 4 (sin x + cos x) + 6 = 0
k - 4 (4/k) + 6 = 0
k - 16/k + 6 = 0
k² - 16 + 6k = 0
k² + 6k - 16 = 0
(k + 8)(k - 2) = 0
k = -8　or 　k = 2


Hope it helps! ^^

2006-12-11 00:44:04 補充：
做三角問題最常用的方法是將tan,sec,csc,cot變為sin,cos再作簡化，因為sin,cos有更多公式可用。

1. Prove (secX - cosX)(1 + cotX + tanX) = sec^2X/cscX + secX/csc^2X
LHS=
(secX - cosX)(1 + cotX + tanX)
=(1/cosX - cosX)(1 + cosX/sinX + sinX/cosX)

=(1-cos^2X)/cosX * (cosXsinX + cos^2X + sin^2X)/cosXsinX

=(sin^2X)/cosX * (cosXsinX + 1)/cosXsinX.....(cos^2X + sin^2X=1)

=(sin^2X)*(cosXsinX + 1)/cos^2XsinX
=(sinX)*(cosXsinX + 1)/cos^2X.........(divided by sinX)
=(sin^2X*cosX + sinX)/cos^2X
=sin^2X/cosX + sinX / cos^2X
=secX/csc^2X + sec^2X/cscX

2. Prove [sec(-X) + sin(-X-90°)]/[csc(540°-X) - cos(270°+X)] = tan^3X
LHS
=[secX - sin(90+X)] / [csc(360°+180°-X) - cos(360°-270°-X)]
=[secX - sin(180-90-X)] / [cscX - cos(90°-X)]
=[secX-cosX]/[1/sinX - sinX]
=[(1-cos^2X)/cosX] / (1-sin^2X)/sinX
=sin^2X/cosx * sinX / cos^2X
=sin^3X/cos^3X
=tan^3X
3. If secX(secX - tanX) = x , express sinX in terms of x.
secX(secX - tanX) = x
1/cosX * [1/cosX - sinX/cosX]=x
1/cos^2X * [1-sinX]=x

[1-sinX] / (1-sin^2X)=x

[1-sinX] / (1-sinX)(1+sinX)=x
if [1-sinX]=0, then sinX=1, cosX=0. Then, secX is undefined. We assume 1-sinX not equal to zero.
1 / (1+sinX) = x.....(cancel out [1-sinX])
1/x = 1+sinX
sinX=(1/x) -1
4. It is given that sin^2X + sinX - 1 = 0 . Without solving for the values of X, prove that:
(a). cos^4X + cos^2X - 1 = 0

sin^2X + sinX - 1 = 0
sinX = 1 - sin^2X = cos^2X

So, LHS=cos^4X + cos^2X - 1=[cos^2X]^2 + [cos^2X] -1 = [sinX]^2 + sinX-1 = 1-1=0
(b). cos^8X + cos^6X + cos^2X - 1 = 0

cos^4X + cos^2X - 1=0

cos^4X = 1- cos^2X =sin^2X
cos^8X = sin^4X

cos^6X =(sin^2X)*cos^2X =sin^3X
LHS= cos^8X + cos^6X + cos^2X - 1
= sin^4X + sin^3X + sinX - 1
= (1-cos^2X)*(1-cos^2X)+sinX*sin^2X + sinX -1
=(1-sinX)(1-sinX) + sinX*(1-cos^2X) + sinX -1
=1-2sinX +sin^2X + sinX*(1-sinX) + sinX -1
= sin^2X - 2sinX + 1 - sinX - sin^2X + sinX -1
=0
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