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求解这几道高中数学题!急需啊!

被浏览: 0次 2023年02月09日 06:01
热门回答(4个)
游客1

sin70=cos20 sin50=cos40
cos10+√3sin10
=2(1/2cos10+√3/2sin10)
=2(sin30cos30+sin10+cos30)
=2sin40

cos10=sin80=2cos40sin40
(cos20)^2=(1+cos40)/2

[cos40+sin50(1+√3*tan10)]/[sin70*√(1+cos40)]
=[sin50+cos40(1+√3sin10/cos10)]/(cos20)^2 (分子分母同*cos10 )
=[sin50cos10+cos40(cos10+√3sin10)]/(cos20)^2cos10
=[sin50cos10+2cos40sin40]/(cos20)^2cos10
=[sin50cos10+cos10]/(cos20)^2cos10
=[cos40+1]/(cos20)^2
=2

2.
π/2<β<α<3π/4,第二象限
π<α+β<3π/2第三象限
0<α-β<π/4第一象限

cos( α-β)=12/13==> sin( α-β)=5/13,
sin(α+β)=-3/5==>cos(α+β)=-4/5

cos2β
=cos(α+β - α-β)
=cos(α+β)cos(α-β)+sin(α+β)sin(α-β)
=-4/5*12/13 -3/5*5/13
=-63/65

3.
x∈(π , 17π/12].
sinx<0
cosx<0

g(x)=cosx*f(sinx)+sinx*f(cosx),
=cosx*√[(1-sinx)/(1+sinx)]+sinx*√[(1-cosx)/(1+cosx)]
=cosx*√[(1-sinx)^2/(1-sin^2x)]+sinx*√[(1-cosx)^2/(1-cos^2x)]
=cosx*√[(1-sinx)^2/(cos^2x)]+sinx*√[(1-cosx)^2/(sin^2x)]
=cosx*(1-sinx)/(-cosx)+sinx*(1-cosx)/(-sinx)
=-(1-sinx)-(1-cosx)
=sinx+cosx-2
=√2(√2/2*sinx+√2/2*cosx)-2
=√2(sinxcosψ+cosxsinψ)-2(ψ=π/4)
=√2sin(x+ψ)-2

-1<=sin(x+ψ)<=1
-√2-2<=g(x)=√2-2

游客2

1、原式=[cos40°+sin50°(1+√3*sin10°/cos10°)]/(√2*(cos20)^2)
=[cos40°+2(sin50°/cos10°)*(1/2cos10°+√3/2sin10°)]/
√2*(cos20)^2)
=[cos40°+2(sin50°/cos10°)*(sin40°)]/√2*(cos20)^2)
=[cos40°+sin80°/cos10°]/(√2*(cos20)^2)
=[cos40°+1]/(√2*(1+cos40°)/2)
=√2
2、讲讲思路吧!cos2α=cos[(α+β)+(α-β)]

游客3

1) 1+ 根3 ×tan10°=(cos10+根3*sin10)/cos10
=2*(1/2*cos10+1/(根3)*sin10)/cos10=2*sin40/cos10
=2*sin40/sin80=1/cos40=1/sin50

cos40°+sin50°(1+ 根3 ×tan10°)=1+cos40

sin70°×根(1+cos40°)=sin70*根2*cos20=根2*(sin70)^2
=根2/2*(1-cos140)=根2/2*(1+cos40)

[cos40°+sin50°(1+ 根3 ×tan10°)]÷[sin70°×根(1+cos40°)]=根2

2、π<α+β<3π/2,0<α-β<π/4

coc(α+β)=-4/5,sin(α-β)=5/13

cos2α=cos[(α+β)+(α-β)]=coc(α+β)*cos(α-β)-sin(α+β)*sin(α-β)=33/65

3、f(sinx)=(sin(x/2)-cos(x/2))/(sin(x/2)+cos(x/2))

f(cosx)=sin(x/2)/cos(x/2)

g(x)=-(sin(x/2)-cos(x/2))^2+2*(sin(x/2))^2=sinx-cosx=根2*sin(x+7π/4)

[-根2/2,1/2)

游客4

1、[cos40°+sin50°*2(cos10°sin30°/cos10°+ cos30° ×tan10)]÷[sin70°×根(2cos20°^2)]
=[cos40°+sin50°*2*(sin40°/cos10°)]÷[sin70°×cos20°根(2)]
=[cos40°+cos40°*2*(sin40°/cos10°)]÷[sin70°×cos20°根(2)]
=[cos40°+(sin80°/cos10°)]÷[sin70°×cos20°根(2)]
=[cos40°+1]÷[cos20°×cos20°根(2)]
=(2COS20^2)]÷[cos20°×cos20°
=根(2)
2、π/2<β<α<3π/4 π<α+β<3π/2 sin(α+β)=-3/5 cos(α+β)=-4/5
0<α-β<π/2cos(α-β)=12/13 sin(α-β)=5/13
cos2α=cos(α+β+α-β)=cos(α+β)cos(α-β)-sin(α+β)sin(α-β)=-33/65
3、f(t)=(1-t/1+t)开根 x∈(π , 17π/12]. cosx<0
sinx<0
f(sinx)=(1-sinx/1+sinx)开根=(分子分母同时乘1-sinx)=sinx-1/cosx
f(cosx)=(1-cosx/1+cosx)开根=分子分母同时乘1-cosx=1-cosx/-sinx
g(x)=cosx·f(sinx)+sinx·f(cosx)=sinx-1-1+cosx=sinx+cosx-2=根2*sin(X+ ψ)-2
sinψ=根2/2 cosψ=根2/2 ψ=π/4 A=根2 ω=1
(2)x∈(π , 17π/12].x+ψ∈(5π/4 , 5π/3].
在x+ψ=5π/3取最大 x+ψ= 3π/2取最小
max=2-根3 /2 min=-3
g(x) 值域是闭区间【-3,2-根3/2】