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已知sinα=5分之3,α∈(2分之π,π),求tAnα,Cosα.

答: 因为:π/2

sinα=3/5 =>cosα=4/5 or -4/5 sin2α =2sinα.cosα =24/25 or -24/25

α∈(3π/2,2π),则cosα>0 sinα=-5/13 cosα=√(1-sin²α)=√[1-(-5/13)²]=12/13 tanα=sinα/cosα=(-5/13)/(12/13)=-5/12 tan2α =2tanα/(1-tan²α) =2·(-5/12)/[1-(-5/12)²] =-120/119 tan2α的值为-120/119

(sina-cosa)²=1-2sinacosa=1-2×5/2=-4

tanα=2tanπ/5 cos(α-3π/10)/sin(α-π/5) =(cosαcos3π/10+sinαsin3π/10) / (sinαcosπ/5-cosαsinπ/5) =(cos3π/10+tanαsin3π/10) / (tanαcosπ/5-sinπ/5) =(cos3π/10+2tanπ/5sin3π/10) / (2tanπ/5cosπ/5-sinπ/5) ={sin(π/2-3π/10)+2tanπ/5cos(π/2-3π/...

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

角在第三象限,所以sina小于0 不懂请追问

公式:tan(a-b)=(tana-tanb)/(1+tanatanb) 解:因为sin2α=3/5(π/2

解答:解∵sinθ+cosθ=15,θ∈(0,π ),∴(sinθ+cosθ )2=125=1+2sinθ cosθ,∴sinθ cosθ=-1225<0.由根与系数的关系知,sinθ,cosθ 是方程x2-15 x-1225=0的两根,解方程得x1=45,x2=-35.∵sinθ>0,cosθ>0,∴sinθ=45,cosθ=-35.则tanθ=-43; s...

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