设锐角三角形ABC中,2sin^2A-cos^2A=2.(1)求
问题描述:
设锐角三角形ABC中,2sin^2A-cos^2A=2.(1)求
答
2sin^2A-cos^2A=23sin^2A-(sin^2A+cos^2A)=23sin^2A=2+(sin^2A+cos^2A)=2+1=3sinA=1(1),∠A=90°(2),y=2sin^2B+sin(2B+π/6)=1-cos(2B)+(0.5√3)*sin(2B)+0.5cos(2B)=1+(0.5√3)*sin(2B)-0.5cos(2B)=1+sin(2B-π/6)=...