三角函数问题
已知f(x)=sin(sinx-cosx),求f(x)的周期和单调区间.
已知f(x)=sin(sinx-cosx),求f(x)的周期和单调区间 f(x)=sin(sinx-cosx)=sin[√2sin(x-π/4)] 显然有:f(2kπ+x)=f(x)---->f(x)的周期是2kπ,k∈Z ∵y=sint在[-π/2,π/2]上单调增,同时t=√2sin(x-π/4)∈[-√2,√2]包含于[-π/2,π/2] ∴y=f(t)的单调性与t=sin(x-π/4)的单调性相同,即: x∈[2kπ-π/2,2kπ+π/2]时,f(x)单调增;x∈[2kπ+π/2,2kπ+3π/2]时,f(x)单调减
T=2pi
答:已知函数f(x)=ax+bsin^3 x+1(a、b为常数),且f(5)=7,则f(-5)=________ 令:g(x)=f(x)-1=ax+bsin^3 x...详情>>
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