Question

y=根号6cosx-根号2sinx的值域

要过程 急

Answer 1

y=√6cosx-√2sinx
=√2(√3cosx-sinx)
=2√2(√3/2*cosx-1/2*sinx)
=2√2(cosπ/6*cosx-sinπ/6*sinx)
=2√2cos(π/6+x)
-1<=cos(π/6+x)<=1
-2√2<=2√2cos(π/6+x)<=2√2
y∈[-2√2,2√2]

Answer 2

解；
y=√6cosx-√2sinx
=√(6+2)[cosx*(√3/2)-sinx*(1/2)]
=2√2(cosxsinπ/3-sinxcosπ/3)
=2√2sin(π/3-x)
=-2√2sin(x-π/3)
-1<=sin(π/3-x)<=1
所以函数值域是[-2√2,2√2]

Answer 3

y=√6cosx-√2sinx
y=√8[(√6/√8)cosx-(√2√8)sinx ]
y=2√2[(√3/2) cosx-(1/2)sinx]
y=2√2[(√3/2) cosx-(1/2)sinx]
令a=2kπ+π/3
y=2√2[sinacosx+cosasinx]
y=2√2[sin(x+a)]
所以y=√6cosx-√2sinx 的值域 是 y∈(-2√2，2√2)

Answer 4

哥们，思路比结果重要结果是[-2√2,2√2]
楼上用的都是asinx+bcosx=√(a^2+b^2)[sinx*a/√(a^2+b^2)+cosx*b/√(a^2+b^2)]=√(a^2+b^2)sin(x+t)
cost=a/√(a^2+b^2),sint=b/√(a^2+b^2)