作业帮只求6 7题
来源：学生作业帮 编辑：作业帮 分类：数学作业 时间：2024/11/05 18:56:37
只求6 7题
解析：6,1)2sinAsinB=1+cosC=1-cos(A+B)=1-cosAcosB+sinAsinB,
cosAcosB+sinAsinB=1
cos(A-B)=1,
A-B=0,
A=B,
ABC为等腰三角形.
2）cosA/cos(B-C)=-cos(B+C)/cos(B-C)=-(cosBcosC-sinBsinC)/(cosBcosC+sinBsinC)
=(tanBtanC-1)/(1+tanBtanC)=(1/5-1)/(1+1/5)=-2/3,
7,1),θ∈(π/2,π),θ-π/3∈(π/6,2π/3)
cos(θ-π/3)=1/5＞0,得,θ-π/3∈(π/6,π/2),则sin(,θ-π/3)=2√6/5,
cosθ=cos[(θ-π/3)+π/3]=cos(θ-π/3)cosπ/3-sin(θ-π/3)sinπ/3=1/5*1/2-2√6/5*√3/2=1/10-6√2/10,
看不清了求什么
cosAcosB+sinAsinB=1
cos(A-B)=1,
A-B=0,
A=B,
ABC为等腰三角形.
2）cosA/cos(B-C)=-cos(B+C)/cos(B-C)=-(cosBcosC-sinBsinC)/(cosBcosC+sinBsinC)
=(tanBtanC-1)/(1+tanBtanC)=(1/5-1)/(1+1/5)=-2/3,
7,1),θ∈(π/2,π),θ-π/3∈(π/6,2π/3)
cos(θ-π/3)=1/5＞0,得,θ-π/3∈(π/6,π/2),则sin(,θ-π/3)=2√6/5,
cosθ=cos[(θ-π/3)+π/3]=cos(θ-π/3)cosπ/3-sin(θ-π/3)sinπ/3=1/5*1/2-2√6/5*√3/2=1/10-6√2/10,
看不清了求什么