几道高中数学三角函数题

几道高中数学三角函数题

1.证明tan²β-sin²β=tan²βsin²β
证明:∵(tan²β-sin²β)/(tan²βsinvβ)=1/sin²β-1/tan²β=csc²β-cot²β=1
∴tan²β-sin²β=tan²βsin²β
2.求证(1-2sinxcosx)/(cos²x-sin²x)=(1-tanx)/(1+tanx)
证明:左边=(cosx-sinx)²/(cosx+sinx)(cosx-sinx)=(cosx-sinx)/(cosx+sinx)
=(1-sinx/cosx)/(1+sinx/cosx)=(1-tanx)/(1+tanx)=右边.
3.求证 tanxsinx/(tanx-sinx)=(tanx+sinx)/tanxsinx
证明:左边=sin²x/(sinx-sinxcosx)=sinx/(1-cosx)=cot(x/2)
\右边=(sinx+sinxcosx)/sin²x=(1+cosx)/sinx=cot(x/2)
故左边=右边.
4.求证(1-sin⁶x-cos⁶x)/(1-sin⁴x-cos⁴x)=3/2
证明:左边=[1-(sin²x+cos²x)(sin⁴x-sin²xcos²x+cos⁴x)]/[1-(sin²x+cos²x)²+2(sin²xcos²x)]
=[1-(sin²x+cos²x)²+3sin²xcos²x]/(2sin²xcos²x)=3sin²xcos²x/2sin²xcos²x=3/2=右边
5,已知1/sinx,1/cosx是二次方程x²-(√2)ax+a=0的两个根,求tanx及a的值.
1/sinx+1/cox=(sinx+cosx)/sinxcosx=(√2)a.(1)
1/sinxcosx=a.(2)
(2)代入(1)式得 (sinx+cosx)a=(√2)a
故 (√2/2)(sinx+cosx)=1
sinxcos(π/4)+cosxsin(π/4)=sin(x+π/4)=1
故 x+π/4=π/2+2kπ,x=π/4+2kπ
∴tanx=tan(π/4+2kπ)=tan(π/4)=1
a=1/sin(π/4)cos(π/4)=1/(2/4)=2.
6.在△ABC中,(√2)sinA=√(3cosA),求∠A的弧度数.
将原式两边平方之,得 2sin²A=3cosA
2(1-cos²A)=3cosA,2cos²A+3cosA-2=(2cosA-1)(cosA+2)=0
∴cosA=1/2,A是锐角,故A=π/3.
7,sinx-cosx=(1-√3)/2,x∈(0,π),求sinx,cosx.
(√2)[sinxcos(π/4)-cosxsin(π/4)]=(1-√3)/2
故 (√2)sin(x-π/4)=(1-√3)/2=(1/2)-(√3/2)=sin(π/6)-sin(π/3)=2cos(π/2)sin(-π/6)=0
∴sin(x-π/4)=0,x-π/4=0,x=π/4
∴ sinx=sin(π/4)=(√2)/2
cosx=cos(π/4)=(√2)/2