Question

高一三角函数题，求各位详细的解答过程，非常感谢

高一三角函数题，求各位详细的解答过程，非常感谢谢谢

Answer 1

sinA-cosA=2/3(sinA-cosA)^2=(cosA)^2 -2cosAsinA+(sinA)^2=4/9
=>((cosA)^2 +(sinA)^2)-2cosAsinA=4/9
=>1-2cosAsinA=4/9
=>-2cosAsinA=4/9-1=-5/9
=>2cosAsinA=5/9
=>sin2A=5/9
(cos2A)^2+(sin2A)^2=1
=>(cos2A)^2=1-(sin2A)^2=1-25/81=56/81
=>cos2A=±根号(56)/9