求值域y=-2cos^2x+2sinx+3/2,x∈[-π/4,π/4]
2个回答
y=-2cos^2x+2sinx+3/2
=-2(1-sin^2x)+2sinx+3/2
=-2sin^2x+2sin^2x-1/2
=-2(sinx-1/2)^2
-π/4
相关问题
y=cos^2 x+sinx,x∈[-π/4,π/4]求值域
求值域y=2sinx+cos^2x,x∈[π/6,2π/3)
函数y=cos2x-2sinx(-π/4≤x≤π/4)的值域是
求函数y=2cos²x+5sinx-4(π/3≤x≤5π/6)的值域
cos[x+(π/4)]=(√2)/10,x∈(π/2,3π/4),求sinx和sin[2x+(π/3)].
求下列函数的值域:(1)y=4-3sin(x-π/3) (2)y=cos^2x-sinx
求y=1/2sin2x+cos^2x,x∈[-π/4,π/4]的值域
求函数y=sinx[2cos(x+π/2)-1],x属于[-6/π,2/π)的值域
求函数y=3sin2x-4sinx+1,x∈[π/3,π]的值域
求函数y=cos^2x+3cosx+2,x属于(3π/4,π)的值域