Question

高三数学三角函数求解答

高三数学三角函数求解答已知sinα=cos2π/5,0<α＜π,则α的取值集合为

Answer 1

sinα=cos2π/5
= sin(π/10) or sin(9π/10)
α = π/10 or 9π/10

Answer 2

sinα=cos(2π/5)=sin(π/2 -2π/5)=sin(π/10)
α=2kπ+ π/10或α=(2k+1)π- π/10，(k∈Z)
α=kπ+(-1)ᵏ π/10，(k∈Z)
α的取值集合为{α|α=kπ+(-1)ᵏ π/10，k∈Z}

追问

角有范围呀

追答

哦，抱歉
sinα=cos(2π/5)=sin(π/2 -2π/5)=sin(π/10)
0<α<π
α=π/10或α=9π/10

追问

谢谢你！