sin(-14兀/3)+cos(-20兀/3)+cot(-53兀/6)=?

sin(-14兀/3)+cos(-20兀/3)+cot(-53兀/6)=-sin(4π+2兀/3)+cos(6兀+2兀/3)-cot(8兀+5兀/6)=-sin(2兀/3)+cos(2兀/3)-cot(5兀/6)=-sin(兀-兀/3)+cos(兀-兀/3)-cot(兀-兀/6)=-sin(兀/3)-cos(兀/3)+cot(兀/6)=-√3/2-1/2+√3/3 =-(3+√3)/6 ...

=-√3sin(2π/3)cos(π/3)-cos(3π/4)tan(-π/4)=-√3×√3/2×1/2-√2/2×1 =3/4-√2/2

回答：根据三角函数的定义,在直角三角型中,正切tanA是对边比邻边,余切cotA是邻边比对边。正弦sinA是对边比斜边,余弦CcosA是邻边比斜边 画两个直角三角形,3.,4,5与5,12,13 再根据sin2a=2sinacosa cos2b=cosb^2-sinb^2计算 sina,cosb的值根据所画的直角三角形得来 具体过程如下:

应用公式：sin（2kπ+α）=sinα cos（2kπ+α) =cosα tan ( kπ+α) = tan α cot ( kπ+α) = cot α (1)sin(-25/3π)= sin（-8π- π/3)= sin(-π/3)= -sin(π/3)= - √3/2 (2) tan13/3π = tan(4π+π/3)= tan (π/3)= √3 ...

解如下图所示

解：原式=sinθ*(-sinθ)*(-cotθ)/tanθsecθsecθ=sinθ^4cotθ^2

sin0=0 sinπ/6=0.5 sinπ/4=0.7071=二分之根号2 sinπ/3=0.8660=二分之根号3 sinπ/2=1 cos0=1 cosπ/6=0.866025404=二分之根号3 cosπ/40.707106781=二分之根号2 cosπ/3=0.5 cosπ/2=0 tan0=0 tanπ/6=0.577350269=三分之根号3 tanπ/4=1 tanπ/3=1.732050808=...

解：（1）f(α)=[sin(π+α)cos(2π-α)tan(-α+3π/2)tan(-α-π)]/sin(π-α)=[(-sinα)*cosα*cotα*(-tanα)]/sinα =cosα；（2）∵cos(α-3π/2)=3/5 ==>cos(3π/2-α)=3/5 ==>-sinα=3/5 ∴sinα=-3/5 ∵α是第三象限角 ∴cosα<0 ∴cosα=-...

cosθcotθtan(-θ)/sinθ(-cotθ)=-cosθ(-1)/sinθ(-cotθ)=1

不正确，因为：{sin(-α-3/2π）sin（3/2π-α)tan^2(2π-α)} / {cos（π/2-α)cos(π/2+α)cot(π-α)} = {sin(π/2-α）sin（3/2π-α)tan^2(2π-α)} / {cos（π/2-α)cos(π/2+α)cot(π-α)} = {cosα(-cosα)tan²α)} / {sinαsinα(-...