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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

解:3π

sin(π-θ)=4/5(π/2<θ<π) sinθ=4/5 cos2θ=1-(4/5)^2=9/25 tanθ=-4/3 tan(θ-π/4)=-4/3-1/1-4/3=-7 cos(2θ-π/3)=9/25*1/2+24/25*√3/2=9+24√3/50

∵sin(π/2+θ)=-3/5 ∴-cosθ=-3/5 ∴cosα=3/5 ∵π/2<θ<π ∴sinθ=4/5 ∴tan(π-θ)=tan(-θ)=-tanθ=-sinθ/cosθ=-(4/5)/(3/5)=-4/3

依题意知α为第四象限角,则θ也是第四象限角,∴sinθ<0,cosθ>0,tanθ<0,∴y=-1+1-1=-1,故选B.

∵tan(θ+ π 4 )= tanθ+1 1-tanθ = 1 7 ,即7tanθ+7=1-tanθ,∴tanθ=- 3 4 ,又0<θ<π,tanθ= sinθ cosθ <0,∴sinθ>0,cosθ<0,∴cosθ=- 1 1 +tan 2 θ =- 4 5 ,sinθ= 1 -cos 2 θ = 3 5 ,则sinθ+cosθ=- 1 5 .故选A

sinα+cosα=4/3 ∴(sinα+cosα)²=1+2sinacosa=16/9 ∴2sinacosa=7/9 ∵(0<α<π/4) ∴sina-cosa<0 ∴sina-cosa=-√(sina-cosa)²=-√(1-7/9)=-√2/3 明教为您解答, 如若满意,请点击[满意答案];如若您有不满意之处,请指出,我一定改正! 希望...

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