函数f(x)=cos(x+π)(x∈【π/12,5π/6】)的值域为多少?

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函数f(x)=cos(x+π)(x∈【π/12,5π/6】)的值域为多少?函数f(x)=cos(x+π)(x∈【π/12,5π/6】)的值域为多少?函数f(x)=cos(x+π)(x∈【π/12,5π

函数f(x)=cos(x+π)(x∈【π/12,5π/6】)的值域为多少?
函数f(x)=cos(x+π)(x∈【π/12,5π/6】)的值域为多少?

函数f(x)=cos(x+π)(x∈【π/12,5π/6】)的值域为多少?
x∈[π/12,5π/6]
x+π∈[π/12+π,5π/6+π]
-cos5π/6≤f(x)=cos(x+π)≤-cosπ/12

f(x)=cos(x+π)=-cosx
增区间为[0,π]
(x∈[π/12,5π/6])
f(x)∈(-(√6+√2)/4,√3/2)

解 函数f(x)=cos(x+π)在【0,π】上为增函数,所以函数f(x)=cos(x+π)在【π/12,5π/6】为增函数 cos(π/12+π)