만약 cos(pi/4-a)cos(pi/4+a)=(√2)/6(0<a<pi/2)이면 sin2a=?

만약 cos(pi/4-a)cos(pi/4+a)=(√2)/6(0<a<pi/2)이면 sin2a=?

(π/4-a)cos(π/4+a)=cos(a-π/4)cos(a+π/4)=근호 2/6Cos(π/4-a)cos(a+π/4)-sin(π/4-a)sin(π/4+a)=근호 2/6Cos(π/4-a+a+π/4)=cosπ/2=0 그래서 sin(π/4-a)sin(π/4-a)sin(π/4+a)=근호 2/4+a)=근호 2/6 그래서 sin(a-π/4)sin(π/4)sin(π/4+a+a+a+a+a+a)=-근호 2/4+코스...