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已知角θ∈(0,π2),且满足条件sinθ+Cosθ=3+12,sin...

(Ⅰ)sinθ1?1tanθ+cosθ1?tanθ=sinθsinθ?cosθsinθ+cosθcosθ?sinθcosθ=sin2θsinθ?cosθ-cos2θcosθ?sinθ =sin2θ?cos2θsinθ?cosθ=sinθ+cosθ=3+12.(Ⅱ)把sinθ+cosθ=3+12 平方可得 1+2sinθ cosθ=4+234,∴sinθ cosθ=34,∴m2=34,∴m=32. 此时,sinθ 和...

解: 因为π/2

原不等式等价于(3+2sinθcosθ-asinθ-acosθ)2≥14,θ∈[0,π2]①,由①得a≥3+2sinθcosθ+12sinθ+cosθ②,或a≤3+2sinθcosθ?12sinθ+cosθ③,在②中,1≤sinθ+cosθ≤2,3+2sinθcosθ+12sinθ+cosθ=(sinθ+cosθ)+52(sinθ+cosθ),显然当1≤x≤2时,f(x)=x+52x为...

∵ α∈( 3π 4 ,π) ,∴ α+ π 4 ∈(π, 5π 4 ),∴sinα+cosα= 2 sin( α+ π 4 )<0,又 sinα?cosα=- 12 25 ,则sinα+cosα=- 1+2sinαcosα = 1- 24 25 =- 1 5 .故选B

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

∵cos(π2?α)=12,∴sinα=12∵α是第二象限角,∴α=2kπ+5π6(k∈Z)∴2α=4kπ+5π3(k∈Z)∴sin2α=?32故答案为:?32

sin(π/3+α)=12/13,α∈(π/6,2π/3), π/3+α∈(π/2,π) 所以 cos(π/3+α)=-5/13 cosα=cos(π/3+α-π/3) =cos(π/3+α)cosπ/3+sin(π/3+α)sinπ/3 =-5/13×1/2+12/13×√3/2 =(-5+12√3)/26

f(x)=sin(x+π/2)=cosx f(α)=3/5,f(β)=12/13 得cosα=3/5,cosβ=12/13 又α,β∈(0,π/2) 所以sinα=4/5 sinβ=5/13 (根据sina*sina+cosa*cosa=1) f(α-β)=cos(α-β)=cosα cosβ+sinα sinβ=36/65+20/65=56/65

解:θ属于(0,π) 且sin(θ+π/3)>0 所以cos(θ+π/3)即可能大于0.也可能小于0,根据两者平方和等于1,得cos(θ+π/3)=±2√2/3 cos(θ+π/12) =cos(θ+π/3-π/4) =cos(θ+π/3)cos(π/4)+sin(θ+π/3)sin(π/4) =√2/2cos(θ+π/3)+√2/6 把cos(θ+π/3)=±2√2/3...

(1)f(x)=2sin(ωx-π6)sin(ωx+π3)=2sin(ωx-π6)sin[(ωx-π6)+π2]=2sin(ωx-π6)cos(ωx-π6)=sin(2ωx-π3),∵T=π,∴ω=1,∴f(x)=sin(2x-π3),∵x∈[π8,5π12],∴2x-π3∈[-π12,π2],根据正弦函数在此区间单调递增,得到:f(x)min=sin...

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