maths questions

[匿名]

2012-07-30 7:22 am

1.If tanθ=4.find the values of
a) 2cosθ/3sinθ
b)3sinθ-4cosθ/2cosθ

2.Find the value of θ in each of the following equations
a)√3 sinθ =cosθ
b)2cos 3θ=1

3.If sinθ =2/3, find cosθ and tanθ (leav your answers in surd form)

4.Prove the following indentities
a) 1-(cosθ-sinθ)^2=2sinθcosθ
b) cos^2θ/1+cosθ + cos^2θ/1-cosθ = 2/tan^2θ

5.sinθ cosθ and tanθ are also defined for θ>90°
Using tan (θ-180°)=tanθ, solve 3 - √3 tan 3θ = 0 for 0°≤ θ ≤ 90°

6.solve tanθ cos θ=0 for 0°≤ θ ≤ 90°

回答 (1)

PeterChow

2012-07-31 4:45 pm

✔ 最佳答案

1.
a)2cosθ/3sinθ=2/3tanθ=2/(3×4)=1/6
b)3sinθ-4cosθ/2cosθ=3sinθ/2cosθ-4cosθ/2cosθ=3tanθ/2-2=3(4)/2-2=4
2.
a)√3 sinθ=cosθ
tanθ=1/√3
θ=30 or 210
b)2cos 3θ=1
cos 3θ=1/2
3θ=60 or 300
θ=20 or 100
3.
cosθ=√(1-sin^2θ)=√(1-(2/3)^2)=√5/3
tanθ=sinθ/cosθ=(2/3)/(√5/3)=2√5/5
4.
a) L.H.S.=1-(cosθ-sinθ)^2
=1-(cos^2θ-2cosθsinθ+sin^2θ)
=1-(1-2cosθsinθ)
=2cosθsinθ=R.H.S.
b)L.H.S.=cos^2θ/1+cosθ + cos^2θ/1-cosθ
=(cos^2θ-cos^3θ+cos^2θ+cos^3θ)/(1-cos^2θ)
=2cos^2θ/sin^2θ
=2/tan^2θ=R.H.S.
5.
3-√3 tan 3θ= 0
tan3θ=3/√3
tan3θ=√3
3θ=60
θ=20
我諗唔到點用 tan (θ-180°)=tanθ，但係上面嗰個都work
6.
tanθcosθ=0
sinθ=0
θ=0

參考: me

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