若sin(Q+π\4)=-1\3且Q属于(3π\4,π),则cos2Q

若sin(Q+π\4)=-1\3且Q属于(3π\4,π),则cos2Q
若sin(Q+π\4)=-1\3且Q属于(3π\4,π),则cos2Q

若sin(Q+π\4)=-1\3且Q属于(3π\4,π),则cos2Q
已知 Q属于(3π/4,π),那么：
Q+π/4 属于(π,5π/4)
若sin(Q+π\4)=-1\3,则有：
cos(Q+π/4)=-根号[1-sin平方(Q+π/4)]=-2(根号2)/3
所以：cos2Q=sin(2Q+ π/2)
=2sin(Q+π/4)*cos(Q+π/4)
=2*(-1/3)*[-2(根号2)/3]
=4(根号2)/9