已知sin(a+π/3)=3/5 a属于(π/6,2π/3) 则cosa=?
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![已知sin(a+π/3)=3/5 a属于(π/6,2π/3) 则cosa=?](/uploads/image/z/8979400-64-0.jpg?t=%E5%B7%B2%E7%9F%A5sin%28a%2B%CF%80%2F3%EF%BC%89%3D3%2F5+a%E5%B1%9E%E4%BA%8E%EF%BC%88%CF%80%2F6%2C2%CF%80%2F3%EF%BC%89+%E5%88%99cosa%3D%3F)
已知sin(a+π/3)=3/5 a属于(π/6,2π/3) 则cosa=?
已知sin(a+π/3)=3/5 a属于(π/6,2π/3) 则cosa=?
已知sin(a+π/3)=3/5 a属于(π/6,2π/3) 则cosa=?
sin(a+π/3)=3/5
a∈(π/6,2π/3)
a+π/3∈(π/2,π)
所以cos(a+π/3)<0
cos(a+π/3)=-4/5
cosa=cos(a+π/3-π/3)
=1/2cos(a+π/3)+√3/2sin(a+π/3)
=(3√3-4)/10