Question

几道高中数学三角函数题

Answer 1

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.
解: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.
解: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.
解:(√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

Answer 2

楼上的证明都很对，就是第七题好像不太合适。
7,sinx-cosx=(1-√3)/2, x∈(0, π),求sinx, cosx.
解：（sinx-cosx）^2=1- √3/2
（sinx+cosx）^2=1+ √3/2=（1+√3）^2 /4
∴ sinx+cosx=(1+√3)/2
∴ sinx=1/2
cosx=√3/2

Answer 3

1.左边=[(sinβ)^2/(cosβ)^2] - (sinβ)^2
=[(sinβ)^2 - (sinβ)^2(cosβ)^2] / (cosβ)^2
={(sinβ)^2*[1-(cosβ)^2]} / (cosβ)^2
=(sinβ)^2*(sinβ)^2 / (cosβ)^2
=(tanβ)^2*(sinβ)^2=右边
2.左边=(sinx^2+cosx^2-2sinxcosx) / (cosx+sinx)(cosx-sinx)
=(cosx-sinx)^2 / (cosx+sinx)(cosx-sinx)
=(cosx-sinx) / (cosx+sinx) , (分子分母同时除以cosx)
=(1-tanx) / (1+tanx) =右边
3.左边=[(sinx/cosx)*sinx] / [(sinx/cosx) - sinx] = [(sinx)^2/cosx] / [(sinx - sinxcosx)/cosx]
=(sinx)^2 / (sinx)*(1-cosx) =(sinx) / (1-cosx) =[(sinx)*(1+cosx)] / [1-(cosx)^2]
=(1+cosx) / (sinx) , (分子分母同时乘以tanx)
=(tanx + sinx) / tanxsinx =右边
4. 分子=1 - [(sinx)^6 + (cosx)^6] = 1 - [(sinx^2)^3 + (cosx^2)^3]
=1 - {[(sinx)^2 + (cosx)^2]*[(sinx)^4 - (sinxcosx)^2 + (cosx)^4]}
=1 - (sinx)^4 + (sinxcosx)^2 - (cosx)^4
=1 - (sinx)^4 - (cosx)^4 + (sinxcosx)^2
代回后=1 + {(sinxcosx)^2/[1 - (sinx)^4 - (cosx)^4]}
=1 + {(sinxcosx)^2/[(1+sinx^2)(1-sinx^2) - (cosx)^4]}
=1 + {(sinxcosx)^2/[(1+sinx^2)*(cosx)^2 - (cosx)^4]}
=1 + {(sinxcosx)^2/(cosx)^2*[1+(sinx)^2 - (cosx)^2]}
=1 + {(sinxcosx)^2/(cosx)^2*[2(sinx)^2]}
=1 + 1/2 = 3/2=右边
5.由韦达定理： 1/sinx + 1/cosx =(cosx+sinx)/(sinxcosx)=√2a ，
(1/sinx)*(1/cosx)=1/(sinxcosx)=a
可得：sinx+cosx=√2
移项后，两边平方：(sinx)^2 = 2 - 2√2cosx + (cosx)^2
1 - (cosx)^2 = 2 - 2√2cosx + (cosx)^2
2(cosx)^2 - 2√2cosx + 1=0
(√2cosx - 1)^2=0
cosx=√2/2 ，则sinx=√2/2
tanx=sinx/cosx=1
a=1/sinxcosx=2
6. 两边平方：(√2sinA)^2 = (√3cosA)^2
2(sinA)^2 = 3cosA
2[1-(cosA)^2] = 3cosA
(2cosA-1)(cosA+2)=0
cosA=1/2 ， cosA=-2(无意义，舍)
所以，在三角形ABC中，∠A=π/3
7. 两边平方：(sinx+cosx)^2 = (1-√3)^2/4
1+sin2x=1-(√3)/2
sin2x=-(√3)/2
(sinx-cosx)^2=1-sin2x=(2+√3)/2
∵sinx+cosx=(1-√3)/2＜0 ,且x∈(0, π),
∴sinx＞0，cosx＜0
∴sinx-cosx=(1+√3)/2
由已知sinx+cosx=(1-√3)/2
解得：sinx=1/2 ，cosx=-(√3)/2