三角比 求人急救 ! ( 20 點 ) !!!

5. 已知tanθ=2 搵 3sinθ- 4cosθ/3sinθ+ 4cosθ的值

5. 3sinθ=3X(2/√5),4cosθ=4X(1/ √5)
3sinθ+ 4cosθ=3X(2/√5)+4X(1/ √5)

3X(2/√5)-4X(1/ √5) / 3X(2/√5))+4X(1/ √5)

=2/5 /2
=4/5

1.√sin^2 60-cos^2 60)sin30
=√(sin^2 60-(1-sin^2 60)sin30
=√(2sin^2 60-1)sin30
=√(0.5)0.5
=0.5 or 1/2

2.tan60-tan30+tan60tan30 ( I think)
= √3 -1/√3 +1
=(9-1+√3)/√3
=(8√3+9)/3

3.tanθ=sinθ/cosθ
1/√3=sinθ/(√3/2)
sinθ=2/3

4. 1+ sinθtanθ/ cosθ
= 1+sin^2 θ
= 1+(1-cos^2 θ)
= 1+0.01
=1.01

5.吾識
6.sinθ(sin^2θ+cos^2θ-1)
=0

7.cosθsinθ(1/tan^2θ+1)
= cos^3θ/sinθ+cosθsinθ
= 1/sinθ(cos^3θ+cosθ+sin^2θ)
= 1/sinθ(cosθ(cos^2θ+sin^2θ))
=1/tanθ

8.((1+sinθ)+(1-sinθ))/((1-sinθ)(1+sinθ))
= 2/(1-sin^2θ)
=2/cos^2θ

9.吾識
10. tanθ+(cos(90-θ)/sin(90-θ))
=2tanθ

11.4(cos(90-θ)/sin(90-θ))-sinθ/cosθ
=3tanθ

12.sin^2θ(cos^2x+sin^2x)+cos^2θ
=sin^2θ(1)+cos^2θ
=1

13.(cos^2θ+sin^2θ)(cos^2θ-sin^2θ)
=(1)(cos^2θ-(1-cos^2θ)
=2cos^2θ-1

14.吾識 sor..

2009-03-08 00:01:09 補充：
sor...
應該係
1.√sin^2 60-cos^2 60)sin30
=√(sin^2 60-(1-sin^2 60)sin30
=√(2sin^2 60-1)sin30
=√(0.5)0.5
=0.5 or 1/2
2.tan60-tan30+tan60tan30 ( I think)
= √3 -1/√3 +1
=(9-1+√3)/√3
=(8√3+9)/3
3.tanθ=sinθ/cosθ
1/√3=sinθ/(√3/2)
sinθ=2/3
4. 1+ sinθtanθ/ cosθ
= 1+sin^2 θ
= 1+(1-cos^2 θ)
= 1+0.01
=1.01
6.sinθ(sin^2θ+cos^2θ-1)
=0

2009-03-08 00:01:51 補充：
7.cosθsinθ(1/tan^2θ+1)
= cos^3θ/sinθ+cosθsinθ
= 1/sinθ(cos^3θ+cosθ+sin^2θ)
= 1/sinθ(cosθ(cos^2θ+sin^2θ))
=1/tanθ
8.((1+sinθ)+(1-sinθ))/((1-sinθ)(1+sinθ))
= 2/(1-sin^2θ)
=2/cos^2θ
10. tanθ+(cos(90-θ)/sin(90-θ))
=2tanθ
11.4(cos(90-θ)/sin(90-θ))-sinθ/cosθ
=3tanθ
12.sin^2θ(cos^2x+sin^2x)+cos^2θ
=sin^2θ(1)+cos^2θ
=1

2009-03-08 00:02:04 補充：
13.(cos^2θ+sin^2θ)(cos^2θ-sin^2θ)
=(1)(cos^2θ-(1-cos^2θ)
=2cos^2θ-1

2009-03-08 00:11:32 補充：
5.tanθ=2
sinθ/cosθ=2/1
sinθ=2 and cosθ=1
3sinθ-4cosθ/3sinθ+4cosθ
=28/3 (guess)

2009-03-08 00:17:51 補充：
14.√1-cosθ/1+cosθ
=√((1-cosθ/1+cosθ)*(1+cosθ)/(1+cosθ))
=√((1-cos^2θ)/(1+cosθ)^2)
=sinθ/(1+cosθ)