cos(π+α)=1/2,α=?

cos(π+α)=1/2,α=?
cos(π+α)=1/2,α=?

cos(π+α)=1/2,α=?
cos(π+α)＝1/2
→π+α＝2kπ±π/3
∴α＝2kπ-2π/3或α＝2kπ-4π/3.
(其中k为整数)

cos(π+α)=1/2
π+α=2kπ±π/3
α=2kπ-π±π/3

由sin（π＋α）=-1/2得, -sinα= -1/2，所以sinα=1/2
因此有 (1)sin（5π－α）=sin(π－α)=sinα = 1/2
(2) sin(π/2+α) =cosα =√（1－sin^2 α) = √[1^2-(1/2)^2] = ±√3/2
　　　(3) cos( α－3π/2）=cos( 3π/2－α)=cos(π/...

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由sin（π＋α）=-1/2得, -sinα= -1/2，所以sinα=1/2
因此有 (1)sin（5π－α）=sin(π－α)=sinα = 1/2
(2) sin(π/2+α) =cosα =√（1－sin^2 α) = √[1^2-(1/2)^2] = ±√3/2
　　　(3) cos( α－3π/2）=cos( 3π/2－α)=cos(π/2－α)=sinα= 1/2
(4) tan(π/2－α) = cotα= cosα/sinα= ±√3
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