高中数学三角函数化简题已知 fα=sin(π-α)cos(2π-α)cos(-α+3/2π)/cos(π/2-α)sin(-π-α)已知fα=sin(π-α)cos(2π-α)cos(-α+3/2π)/cos(π/2-α)sin(-π-α)化简若α第三象限,且cos(α-3/2π)=1/5,求fα若α=-31/3π,求f
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高中数学三角函数化简题已知 fα=sin(π-α)cos(2π-α)cos(-α+3/2π)/cos(π/2-α)sin(-π-α)已知fα=sin(π-α)cos(2π-α)cos(-α+3/2π)/cos(π/2-α)sin(-π-α)化简若α第三象限,且cos(α-3/2π)=1/5,求fα若α=-31/3π,求f
高中数学三角函数化简题已知 fα=sin(π-α)cos(2π-α)cos(-α+3/2π)/cos(π/2-α)sin(-π-α)
已知fα=sin(π-α)cos(2π-α)cos(-α+3/2π)/cos(π/2-α)sin(-π-α)
化简
若α第三象限,且cos(α-3/2π)=1/5,求fα
若α=-31/3π,求fα
高中数学三角函数化简题已知 fα=sin(π-α)cos(2π-α)cos(-α+3/2π)/cos(π/2-α)sin(-π-α)已知fα=sin(π-α)cos(2π-α)cos(-α+3/2π)/cos(π/2-α)sin(-π-α)化简若α第三象限,且cos(α-3/2π)=1/5,求fα若α=-31/3π,求f
∵fα=sin(π-α)cos(2π-α)cos(-α+3/2π)/cos(π/2-α)sin(-π-α)
∴f(a) = -sinacosasina/(sinasina)=-cosa
∵cos(a-3/2π) = -sina = 1/5 又α第三象限
∴cosa = - 2*(根号6)/5
∴f(a) = -cosa = 2*(根号6)/5
(2)∵α=-31/3π
f(a) = -cos(-31/3π) = -cos(31/3π) = -cos(10π + 1/3π) = -cos(1/3π)= -1/2