若a∈(0,π/2),cos(a+π/6)=-4/5,sina,

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若a∈(0,π/2),cos(a+π/6)=-4/5,sina,

cos(a+π/6)=-4/5
cosacos(π/6)-sinasin(π/6)=-4/5
(cosa)(√3)/2-(sina)/2=-4/5
(√3)cosa-sina=-8/5
(√3)cosa=sina-8/5
[(√3)cosa]²=(sina-8/5)²
3cos²a=sin²a-16sina/5+64/25…………(1)
又:cos²a+sin²a=1
cos²a=1-sin²a
代入(1),有:
3(1-sin²a)=sin²a-16sina/5+64/25
3-3sin²a=sin²a-16sina/5+64/25
4sin²a-16sina/5-11/25=0
sin²a-(4/5)sina-11/100=0
(sina-2/5)²=27/100
sina=(4±3√3)/10
因为:a∈(0,π/2)
所以:sina=(4+3√3)/10