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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

①2 ②sin(2α+π/2)=4/5 =cos2α sin2α=3/5 答案3/10十(4根号3)/10

由θ为第四象限角,且 cosθ= 3 5 得 sinθ=- 4 5 (2分)又 sin2θ=2sinθcosθ=2?(- 4 5 )? 3 5 =- 24 25 (2分)∴ cos2θ=co s 2 θ-si n 2 θ=( 3 5 ) 2 -(- 4 5 ) 2 =- 7 25 (2分) ∴ sin(2θ+ π 3 ) = sin2θ?cos π 3 +cos2θ?sin π 3 =(- 24 25 )? ...

解: α是第二象限角,则tanα

tanα=4√3 sinα=4√3/7 cosα=1/7 cos(α+β)=-11/14,且α、β均为锐角 0

(1)f(α)=sin(-α-π)cos(5π-α)tan(2π-α)/cos(π/2-α)tan(-π-α) =(-sinα)*(-cosα)*(-tanα)/cosαtanα =-sinαcosαtanα/cosαtanα =-sinα (2)由tan(π-α)=-2得:tanα=2,即:sinα/cosα=2,所以sinα=2cosα, 代入(sinα)^2+(cosα)^2=1的:(c...

解: sinθ+cosθ=√2/3 (sinθ+cosθ)^2=(√2/3)^2. sin^2θ+2sinθcosθ+cos^2θ=2/9. 1+2sinθcosθ=2/9. 2sinθcosθ=2/9-1. =-7/9. (sinθ-cosθ)^2=(sinθ+cosθ)^2-4sinθcosθ. =2/9-2*(-7/9). =2/9+14/9. =16/9. sinθ-cosθ=±4/3 ∵π/21 【这是不可能的】 ∵si...

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