求解几道三角函数题目！

Question

1. cos2a/sin(a-π/4)=-√2/2,那么sina+cosa的值为？2.在三角形ABC中，tanA=1/2,cosB=3√10/10,则tanC的值是？3. 要使得sina-√3cosa9=(4m-6)/(4-m)有意义，则m的取值范围是？

Answer 1

1. cos2a/sin(a-π/4)=-√2/2,那么sina+cosa的值为？
cos2a=(-√2/2)(sinacosπ/4-cosasinπ/4)=(-sina-cosa)/2
cos2a=cos^2a-sin^2a=(-sina-cosa)/2
cos^2a-cosa/2=sin^2a-sina/2
(cosa-1/4)^2=(sina-1/4)^2
所以cosa=sina=√2/2,cosa+sina=√2或cosa-1/4=1/4-sina,cosa+sina=1/2

2.在三角形ABC中，tanA=1/2,cosB=3√10/10,则tanC的值是？
tanA=1/2,2sinA=cosA,4sin^2A=cos^2A,sinA=√5/5,cosA=2√5/5
cosB=3√10/10,sinB=√10/10
sinC=sin(B+A)=sinBcosA+sinAcosB=√2/2
cosC=-cos(B+A)=-(cosBcosA-sinBsinA)=-√2/2
tanC=-1

3. 要使得sina-√3cosa9=(4m-6)/(4-m)有意义，则m的取值范围是？
sina-√3cosa9???sina-√3cosa?
-2<=sina-√3cosa=2(sina/2-√3cosa/2)=2sin(a-π/3)<=2
要使得sina-√3cosa=(4m-6)/(4-m)有意义,则
-2<=(4m-6)/(4-m)<=2,(m<>4)
-2<=(4m-6)/(4-m)=[-4(4-m)+16-6]/(4-m)=-4+10/(4-m)<=2
2<=10/(4-m)<=6
1/5<=1/(4-m)<=3/5
5>=4-m>=5/3
1>=-m>=-7/3
-1<=m<=7/3