中三數學,三角學的應用,數學高手請進

1. sin82度/tan8度-cos8度tan82度 (ans: 0)
= sin82度*(cos8度/sin8度)-cos8度*(sin82度/cos82度)
= sin82度*(sin82度/cos82度)-sin82度*(sin82度/cos82度)
=0

2. (sinΘ+cosΘ)^2-1
=sin^2Θ+2sincosΘ+cos^2Θ-1
=2sinΘcosΘ

3. 若sinΘ=1/2, 求(sinΘ+cosΘ)/(2sinΘ+cosΘ)的值
sinΘ=1/2
(1) cosΘ=sqrt(3)/2
(sinΘ+cosΘ)/(2sinΘ+cosΘ)=(1/2+sqrt(3)/2)/(1+sqrt(3)/2)=(1+sqrt(3)/(2+sqrt(3))
(2) cosΘ=-sqrt(3)/2
(sinΘ+cosΘ)/(2sinΘ+cosΘ)=(1/2-sqrt(3)/2)/(1-sqrt(3)/2)=(1-sqrt(3)/(2-sqrt(3))
(sinΘ+cosΘ)/(2sinΘ+cosΘ)= (1+sqrt(3)/(2+sqrt(3))or =(1-sqrt(3)/(2-sqrt(3))


2009-05-25 03:14:22 補充：
4. 若cosΘ=7/25, 求(sinΘ-2cosΘ)/(sinΘ+3cosΘ)的值
cosΘ=7/25
(1) sinΘ=24/25
(sinΘ-2cosΘ)/(sinΘ+3cosΘ)
=(24/25-2*7/25)/(24/25+3*7/25) =(24-14)/(24+21)=10/45=2/9
(2) sinΘ=-24/25
(sinΘ-2cosΘ)/(sinΘ+3cosΘ)
=(-24/25-2*7/25)/(-24/25+3*7/25) =(-24-14)/(-24+21)=-38/-3=38/3
(sinΘ-2cosΘ)/(sinΘ+3cosΘ) =2/9 or =38/3

2009-05-25 03:14:43 補充：
5. 若tanΘ=1, 求(3cosΘ-sinΘ)/(3sinΘ+cosΘ)的值
(3cosΘ-sinΘ)/(3sinΘ+cosΘ)
=(3cosΘ/cosΘ-sinΘ/cosΘ)/ (3sinΘ/cosΘ)+cosΘ/cosΘ)
=(3-tanΘ)/(3tanΘ+1)=(3-1)/(3+1)=1/2

6. cos30tan30/sin^2 30
=(sqrt(3)/2)*(1/sqrt(3))/(1/2)^2=(1/2)/(1/4)=2

To chilok1999220:

乜你 copy and paste 第一個答案可以幫到人架咩？

1.
sin82o/tan8o - cos8otan82o
= sin(90o - 8o)/tan8o - cos8otan(90o - 8o)
= cos8o/tan8o - cos8o/tan8o
= 0

2.
(sinq + cosq)2 -1
= (sin2q + 2sinqcosq + cos2q) - 1
= (sin2q + cos2q) + 2sinqcosq - 1
= 1 + 2sinqcosq - 1
= 2sinqcosnq

3.
sinq = 1/2

cosq
= √(1 - sin2q)
= √[1 - (1/2)2]
= (√3)/2

(sinq + cosq)/(2sinq + cosq)
= [1/2 + (√3)/2]/[2(1/2) + (√3/2)]
= 2[1/2 + (√3)/2]/2[2(1/2) + (√3/2)]
= (1 + √3)/(2 + √3)

4.
cosq = 7/25

sinq
= √[1 - (7/25)2]
= 24/25

(sinq - 2cosq)/(sinq + 3cosq)
= [24/25 - 2(7/25)]/[24/25 + 3(7/25)]
= [10/25]/[45/25]
= 10/45
= 2/9
(The given answer may be incorrect or the question is mistyped or misunderstood.)

5.
tanq = 1
sinq/cosq = 1
所以 sinq = cosq

(3cosq - sinq)/(3sinq + cosq)
= (3cosq - cosq)/(3cosq + cosq)
= (2cosq)/(4cosq)
= 1/2

6.
cos30otan30o/sin230o
= (√3/2)(1/√3)/(1/2)2
= (1/2)/(1/4)
= (1/2) x 4
= 2

1.
sin82o/tan8o - cos8otan82o
= sin(90o - 8o)/tan8o - cos8otan(90o - 8o)
= cos8o/tan8o - cos8o/tan8o
= 0


2.
(sinq + cosq)2 -1
= (sin2q + 2sinqcosq + cos2q) - 1
= (sin2q + cos2q) + 2sinqcosq - 1
= 1 + 2sinqcosq - 1
= 2sinqcosnq


3.
sinq = 1/2

cosq
= √(1 - sin2q)
= √[1 - (1/2)2]
= (√3)/2

(sinq + cosq)/(2sinq + cosq)
= [1/2 + (√3)/2]/[2(1/2) + (√3/2)]
= 2[1/2 + (√3)/2]/2[2(1/2) + (√3/2)]
= (1 + √3)/(2 + √3)


4.
cosq = 7/25

sinq
= √[1 - (7/25)2]
= 24/25

(sinq - 2cosq)/(sinq + 3cosq)
= [24/25 - 2(7/25)]/[24/25 + 3(7/25)]
= [10/25]/[45/25]
= 10/45
= 2/9
(The given answer may be incorrect or the question is mistyped or misunderstood.)

5.
tanq = 1
sinq/cosq = 1
所以 sinq = cosq

(3cosq - sinq)/(3sinq + cosq)
= (3cosq - cosq)/(3cosq + cosq)
= (2cosq)/(4cosq)
= 1/2


6.
cos30otan30o/sin230o
= (√3/2)(1/√3)/(1/2)2
= (1/2)/(1/4)
= (1/2) x 4
= 2
=