求值域y=-2cos^2x+2sinx+3/2,x∈[-π/4,π/4]_数学解答

求值域y=-2cos^2x+2sinx+3/2,x∈[-π/4,π/4]

人气：499 ℃　时间：2020-06-19 18:30:40

解答

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