若cos(π/4-a)cos(π/4+a)=)=(√2)/6 (0<a<π/2)则sin2a=?

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若cos(π/4-a)cos(π/4+a)=)=(√2)/6 (0<a<π/2)则sin2a=?
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若cos(π/4-a)cos(π/4+a)=)=(√2)/6 (0<a<π/2)则sin2a=?
若cos(π/4-a)cos(π/4+a)=)=(√2)/6 (0<a<π/2)则sin2a=?

若cos(π/4-a)cos(π/4+a)=)=(√2)/6 (0<a<π/2)则sin2a=?
Cos(π/4-a)cos(π/4+a)=cos(a-π/4)cos(a+π/4)=根号2/6
Cos(π/4-a)cos(a+π/4)-sin(π/4-a)sin(π/4+a)=cos(π/4-a+a+π/4)=cosπ/2=0
所以 sin(π/4-a)sin(π/4+a)=根号2/6
所以sin(a-π/4)sin(π/4+a)=-根号2/6
Cos2a=cos(a-π/4+a+π/4)=cos(a-π/4)cos(a+π/4)-sin(a-π/4)sin(a+π/4)=根号2/3
0<a<π/2
所以sin2a=根号7/3