trigonometry 2

1.
θ lies in the 4th quadrant.


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2.
θ lies in the 3rd quadrant.


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3.
θ lies in the 2nd quadrant.


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4.
tanθ = 1/3
sinθ/cosθ = 1/3
cosθ = 3sinθ ...... (1)

sin²θ + cos²θ = 1
sin²θ + (3sinθ)² = 1
10sin²θ = 1
sin²θ = 1/10
sinθ = ±1/√10

Put sinθ = ±1/√10 into (1)
cosθ = ±3/√10

Since tanθ > 0, θ lies inthe 1st quadrant or the 3rd quadrant.

When θ lies in the 1st quadrant,sinθ = 1/√10 and cosθ = 3/√10
sinθcosθ = (1/√10) * (3/√10) =3/10

When θ lies in the 3rd quadrant,sinθ = -1/√10 and cosθ = -3/√10
sinθcosθ = (-1/√10) * (-3/√10)= 3/10

Hence, sinθcosθ= 3/10

2011-08-26 03:47:52 補充：
發問時點數已經扣除，刪除問題只會損人不利己。

Tanθ=1/3
Sinθ=+/-1/√10
Sin^2 θ=1/10
Sinθ/Cosθ=1/3
Cosθ=3Sinθ
SinθCosθ=Sinθ*3Sinθ=3Sin^2 θ=3/10