求最值 (1)y=√(1-1/2sinx) (2)cos²+sinx-2

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求最值 (1)y=√(1-1/2sinx) (2)cos²+sinx-2

求最值 (1)y=√(1-1/2sinx) (2)cos²+sinx-2
求最值 (1)y=√(1-1/2sinx) (2)cos²+sinx-2

求最值 (1)y=√(1-1/2sinx) (2)cos²+sinx-2
y=根号(1-1/2sinx)
-1/2 ≤ -1/2sinx ≤ 1/2
1/2 ≤ 1-1/2sinx ≤ 3/2
根号2/2 ≤ 根号(1-1/2sinx)≤ 根号6/2
最小值:根号2 /2
最大值:根号6 /2
cos²+sinx-2
= 1-sin^2x+sinx-2
= - sin^2x+sinx-1/4+1/4+1-2
= - (sinx-1/2)^2 -3/4 ≤ -4/3
-1≤sinx≤1
0 ≤ (sinx-1/2)^2 ≤ 9/4
-9/4 ≤ -(sinx-1/2)^2 ≤ 0
-3 ≤ - (sinx-1/2)^2 -3/4 ≤ -3/4
最小值:-3
最大值:-3/4