F.2數學(三角比的關係)20分..急.......

1.化簡:
sinθcosθtanθ-1
=sinθcosθ(sinθ/cosθ)-1
=sin2θ-1
=cos2θ


2.如果tanθ=3，求7cosθ-sinθ/2cosθ的值。
(7cosθ-sinθ)/2cosθ
=(7-tanθ)/2
=(7-3)/2
=2

3.如果sinθ=1/5，求4cos2θ+3sin2θ的值。
4cos2θ+3sin2θ
=4(1-sin2θ)+3sin2θ
=4-4sin2θ+3sin2θ
=4-sin2θ
=4-(1/5)2
=99/25

4.如果sinA=2/3，求(cosA/tanA)+sinA的值。
(cosA/tanA)+sinA
=cosA/(sinA/cosA)+sinA
=cos2A/sinA +sinA
=(cos2A+sin2A)/sinA
=1/sinA
=1/(2/3)
=3/2

第一條
sinθcosθtanθ-1
= sinθcosθ(sinθ / cosθ ) -1
= sin2次θ - 1
= cos2次θ //

第二條
tanθ = sinθ / cosθ = 3

7cosθ-sinθ/2cosθ
= 7cosθ-1/2 (tanθ)
= 7cosθ - 1/2 (3)
= 7cosθ - 3/2
= [2(7cosθ)-3 ]/ 2
= (14cosθ-3 )/ 2


第三條

sinθ = 1/5

sin^2θ+cos^2θ = 1
cos^2θ = 1-(1/5)^2
cos^2θ = 24/25

4cos^2θ+3sin^2θ
= 4(24/25) + 3(1/25)
= 96/25 + 3/25
= 99/25 //

第4條

sinA=2/3

sin^2A+cos^2A=1
cos^2A = 1-(2/3)^2
cos^2A = 5/9
cosA = 開方5/9

sinA / cosA = tanA
2/3 / 開方5/9 = tanA
tanA = 2/3*開方9/5

(cosA/tanA) + sinA
= (開方5/9 / 2/3*開方9/5) +2/3
= 開方5/9 * 3/2 * 開方5/9 +2/3
= 5/9 * 3/2 + 2/3
= 5/6 * 1/2 + 2/3
=5/6+ 2/3
=5/6 + 4/6
=9/6
=3/2 //

2008-05-14 14:01:46 補充：
第二條錯了 ,正確的如下 .

第二條
tanθ = sinθ / cosθ = 3

7cosθ-sinθ/2cosθ
= 7cosθ / 2cosθ - sinθ/2cosθ
= 7/2 - (1/2)tanθ
= 7/2 - 3/2
= 4/2
= 1/2 //