作业帮求y=sinx+2sinxcosx+cosx-4的值域
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求y=sinx+2sinxcosx+cosx-4的值域
求y=sinx+2sinxcosx+cosx-4的值域
求y=sinx+2sinxcosx+cosx-4的值域
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]
y=sinx+2sinxcosx+cosx-4
=1+2sinxcosx+sinx+cosx-5
=(sinx+cosx)^2+sinx+cosx-5
=(sinx+cosx)^2+(sinx+cosx)+1/4-1/4-5
=(sinx+cosx+1/2)^2-21/4
=[√2sin(x+π/4)+1/2]^2-21/4
-1<=sin(x+π/...
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y=sinx+2sinxcosx+cosx-4
=1+2sinxcosx+sinx+cosx-5
=(sinx+cosx)^2+sinx+cosx-5
=(sinx+cosx)^2+(sinx+cosx)+1/4-1/4-5
=(sinx+cosx+1/2)^2-21/4
=[√2sin(x+π/4)+1/2]^2-21/4
-1<=sin(x+π/4)<=1
-√2<=√2sin(x+π/4)<=√2
-√2+1/2<=√2sin(x+π/4)+1/2<=√2+1/2
0<=[√2sin(x+π/4)+1/2]^2<=9/4+√2
-21/4<=[√2sin(x+π/4)+1/2]^2-21/4<=9/4+√2-21/4
-21/4<=[√2sin(x+π/4)+1/2]^2-21/4<=√2-3
所以y=sinx+2sinxcosx+cosx-4的值域为：[-21/4,√2-3]
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