题目

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?

提问时间：2021-02-03

答案

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?
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