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已知θ是第四象限角,sin(θ+π/4)=3/5,tAn(θ%π/4...

-4/3 解: ∵ θ是第四象限角 ∴ θ+π/4是第四象限或第一象限角 又∵ sin(θ+π/4)=3/5>0 ∴ θ+π/4是第一象限角 ∴ cos(θ+π/4)=√[1-sin²(θ+π/4)]=4/5 ∴ cot(θ+π/4)=4/3 tan(θ-π/4) =tan[(θ+π/4)-π/2] =-tan[π/2-(θ+π/4)] =-cot(θ+π/4) =-4/3

-4/3 解析: tan(θ-π/4) =-tan(π/4-θ) =-cot[π/2-(π/4-θ)] =-cot(π/4+θ) =-cos(π/4+θ)/sin(π/4+θ) =-4/3

∵θ是第二象限角, sinθ= 4 5 ,∴cosθ=- 3 5 ,tan θ=- 4 3 ,∴ tan( 5π 4 -θ) =tan( π 4 -θ )= tan π 4 -tanθ 1+tan π 4 tanθ = 1+ 4 3 1- 4 3 =-7.故选D.

因为sinx+cosx= 1 5 ,而sin 2 x+cos 2 x=1 即(sinx+cosx) 2 -2sinxcosx=1,所以 1 25 -2sinxcosx=1 所以2sinxcosx=- 24 25 又因为sin 2 x+cos 2 x=(sinx-cosx) 2 +2sinxcosx=1 所以(sinx-cosx) 2 - 24 25 =1,所以(sinx-cosx) 2 = 49 ...

已知sin(π/4-α)=3/5,且α∈(0,π/4),那么: -α∈(-π/4,0),即有:π/4 - α∈(0,π/4) 所以:cos(π/4-α)=根号[1-sin²(π/4-α)]=根号(1- 9/25)=4/5 所以:sinα=sin[π/4 - (π/4-α)] =sin(π/4)*cos(π/4-α)-cos(π/4)*sin(π/4-α) =(根号2...

已知α为第二象限角 且sinα=3/5 则tan(π+α)的值是 tan(π+α)=(tanπ+tanα)/(1-tanπtanα) =(0+tanα)/1 =tanα =sinα/cosα =3/5÷(-4/5) =-3/4

(1) f(α)=[sinαtan(-α)/tan(-α)]*sinα =sin²α (2) 若sinα=-3/5 f(α)=sin²α=(-3/5)²=9/25 (3) 若α=330° 那么sinα=sin330°=sin(-30°)=-1/2 所以f(α)=sin²α=(-1/2)²=1/4

因为θ∈(π/2,π),所以cosθ=- 4/5,tanθ=-3/4,cos(θ+π/4)=负的十分之七倍的根号2

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