函数f(x)=cos(-1/2)+sin(π-x/2).x∈R,⑴求f(x)周期,⑵求f(x)在[函数f(x)=cos(-1/2)+sin(π-x/2).x∈R,⑴求f(x)周期,⑵求f(x)在[0,π]上的减区间

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函数f(x)=cos(-1/2)+sin(π-x/2).x∈R,⑴求f(x)周期,⑵求f(x)在[函数f(x)=cos(-1/2)+sin(π-x/2).x∈R,⑴求f(x)周期,⑵求f(x)在[0,π]上的减区间
函数f(x)=cos(-1/2)+sin(π-x/2).x∈R,⑴求f(x)周期,⑵求f(x)在[
函数f(x)=cos(-1/2)+sin(π-x/2).x∈R,⑴求f(x)周期,⑵求f(x)在[0,π]上的减区间

函数f(x)=cos(-1/2)+sin(π-x/2).x∈R,⑴求f(x)周期,⑵求f(x)在[函数f(x)=cos(-1/2)+sin(π-x/2).x∈R,⑴求f(x)周期,⑵求f(x)在[0,π]上的减区间
f(x)=cos(-x/2)+sin(π-x/2)=cos(x/2)+sin(x/2)=√2/2sin(x/2+π/4)
周期T=2π*2=4π
π/2+2kπ≤x/2+π/4≤3π/2+2kπ
解得π/2+4kπ≤x≤5π/2+4kπ
即f(x)在[0,π]上的减区间为[π/2,π]