f(x)=2cos²x+2√3sinx*cosx-1=cos(2x)+√3sin(2x)=2sin(2x+π/6)x∈[0,π/2],∴π/6≤2x+π/6≤7π/6-1/2≤sin(2x+π/6)≤1-1≤2sin(2x+π/6)≤2∴f(x)的值域为:[-1,2].从倒数第三行就看不懂了 sinπ/6=1/2 怎么会出现-1/2呢?
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f(x)=2cos²x+2√3sinx*cosx-1=cos(2x)+√3sin(2x)=2sin(2x+π/6)x∈[0,π/2],∴π/6≤2x+π/6≤7π/6-1/2≤sin(2x+
f(x)=2cos²x+2√3sinx*cosx-1=cos(2x)+√3sin(2x)=2sin(2x+π/6)x∈[0,π/2],∴π/6≤2x+π/6≤7π/6-1/2≤sin(2x+π/6)≤1-1≤2sin(2x+π/6)≤2∴f(x)的值域为:[-1,2].从倒数第三行就看不懂了 sinπ/6=1/2 怎么会出现-1/2呢?
f(x)=2cos²x+2√3sinx*cosx-1
=cos(2x)+√3sin(2x)
=2sin(2x+π/6)
x∈[0,π/2],
∴π/6≤2x+π/6≤7π/6
-1/2≤sin(2x+π/6)≤1
-1≤2sin(2x+π/6)≤2
∴f(x)的值域为:[-1,2].
从倒数第三行就看不懂了 sinπ/6=1/2 怎么会出现-1/2呢?
f(x)=2cos²x+2√3sinx*cosx-1=cos(2x)+√3sin(2x)=2sin(2x+π/6)x∈[0,π/2],∴π/6≤2x+π/6≤7π/6-1/2≤sin(2x+π/6)≤1-1≤2sin(2x+π/6)≤2∴f(x)的值域为:[-1,2].从倒数第三行就看不懂了 sinπ/6=1/2 怎么会出现-1/2呢?
sin7π/6=? 是不是-1/2