已知x属于 (派/12,派/3) ,cos(2x+派/3)=-5/13 求sin2x
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![已知x属于 (派/12,派/3) ,cos(2x+派/3)=-5/13 求sin2x](/uploads/image/z/2832906-66-6.jpg?t=%E5%B7%B2%E7%9F%A5x%E5%B1%9E%E4%BA%8E+%28%E6%B4%BE%2F12%2C%E6%B4%BE%2F3%29+%2Ccos%EF%BC%882x%2B%E6%B4%BE%2F3%EF%BC%89%3D-5%2F13+%E6%B1%82sin2x)
已知x属于 (派/12,派/3) ,cos(2x+派/3)=-5/13 求sin2x
已知x属于 (派/12,派/3) ,cos(2x+派/3)=-5/13 求sin2x
已知x属于 (派/12,派/3) ,cos(2x+派/3)=-5/13 求sin2x
x∈(π/12,π/3)
所以2x+π/3∈(π/2,π)
所以sin(2x+π/3)=12/13
sin2x=[(2x+π/3)-π/3]
=sin(2x+π/3)×cosπ/3-cos(2x+π/3)×sinπ/3
=12/13×(1/2)-(-5/13)×(√3/2)
=(12+5√3)/26