Question

求y=sinx+2sinxcosx+cosx-4的值域

Answer 1

令u=sinx+cosx = √2 sin(x+π/4) ∈[-√2,√2]

u² = sin²x+cos²x+2sinxcosx = 1+2sinxcosx

∴4sinxcosx = 2(u²-1)y = u - 2(u²-1) + 1= -2u² + u +1= - 2 (u - 1/4)² + 9/8u∈[-√2,√2]当u=1/4时取最大值 9/8当u= -√2时取最小值 -3-√2故原函数的值域 [ -3-√2,9/8]