help.........

Chun Hei

2007-03-15 7:52 am

By using the trigonometric identities

1.tan^2θxcosθ+cosθ
2.1/(1-cosθ)-(1+cosθ)/sin^2θ
3.[1-(1)/cosθ][1/cosθ+1]
4.If sinθcosθ=k,express tanθ+1/tanθ in terms of k
5.θ is represents an acute angle,given that sinθ =square root2/10 and tanθ=1/7,find the value of cosθ
6.Given that sinθ=7/8,find the value of cosθ and tanθ

回答 (1)

[匿名]

2007-03-15 9:03 am

✔ 最佳答案

1. tan^2θxcosθ+cosθ
=(sin^2θ/cos^2θ) xcosθ +cosθ
=(sin^2θ+cos^2θ)/cosθ
=1/cosθ

2. 1/(1-cosθ)-(1+cosθ)/sin^2θ
=[sin^2θ-(1-cosθ)(1+cosθ)]/(1-cosθ)(sin^2θ)
=[sin^2θ-1+cos^2θ]/(1-cosθ)(sin^2θ)
=[1-1]/(1-cosθ)(sin^2θ)
=0

3..[1-(1)/cosθ][1/cosθ+1]
=[(cosθ-1)]/cosθ] [(1+cosθ)/cosθ]
=[cos^2θ-1]/cos^2θ
=[cos^2θ-cos^2θ-sin^2θ]/cos^2θ
= -sin^2θ/cos^2θ
= -tan^2θ

4. tanθ+1/tanθ
=(sinθ/cosθ) + (cosθ/sinθ)
=(sin^2θ+cos^2θ)/cosθsinθ
=1/k

5. from (4), tanθ+1/tanθ = 1/k = 1/(sinθcosθ)
hence,
(1/7)+7=1/[(square root2/10)(cosθ)]
50/7 = 1/[(square root2/10)(cosθ)]
cosθ=(7/50)(sq. root 5)
cosθ=7/(10√5)
cosθ=0.313049516

6. sinθ=7/8
cosθ=√15/8
tanθ=7/√15

參考: me

👍 0 👎 0