Question

若是X一个三角形的最小内角,则函数y=sinx-cosx的值域是

Answer 1

sinx-cosx = 根号2sin(x-π/4)
x最小角，0< x<=π/3
-π/4 < x-π/4 <= π/12
-1 < 根号2sin(x-π/4) < = 根号2sin π/12 = 根号2 * (根号6-根号2)/4 = (根号3-1)/2

Answer 2

y=sinx-cosx=√2sin(x-π/4)，x是三角形的最小内角，所以0<x≤π/3，-π/4<x-π/4≤π/12，
而正弦函数y=sinx在(-π/4，π/12]上单调递增，所以-√2/2<sin(x-π/4)≤(√6-√2)/4
故y的值域为(-1,√3/2-1/2]

Answer 3

三角形的最小内角，所以x的范围是(0,π/3]，sinx-cosx=根号2*sin(x-π/4)，x-π/4的范围是(-π/4，π/12],sin(x-π/4)的范围是(-根号2/2,sinπ/12]，所以y的值域是(-1,(根号3-1)/2]。

Answer 4

y=根号2 SIN(X-π/4)因为0<x<π/3所以-π/4<X-π/4<π/12-1<Y<（根号6-根号2）/4

Answer 5

负1到2分之根3减1（1在分数线上）