已知函数f(x)=sin(π-x)sin(π/2-x)+cos平方x,当x∈【-π/8,3π/8】时,求函数的单调区间

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已知函数f(x)=sin(π-x)sin(π/2-x)+cos平方x,当x∈【-π/8,3π/8】时,求函数的单调区间已知函数f(x)=sin(π-x)sin(π/2-x)+cos平方x,当x∈【-π

已知函数f(x)=sin(π-x)sin(π/2-x)+cos平方x,当x∈【-π/8,3π/8】时,求函数的单调区间
已知函数f(x)=sin(π-x)sin(π/2-x)+cos平方x,当x∈【-π/8,3π/8】时,求函数的单调区间

已知函数f(x)=sin(π-x)sin(π/2-x)+cos平方x,当x∈【-π/8,3π/8】时,求函数的单调区间
答案请看图

f(x)=sin(π-x)sin(π/2-x)+cos² x
=sinxcosx+cos² x
=1/2*sin2x+(1+cos2x)/2
=1/2*sin2x+1/2*cos2x+1/2
=√2(√2/2*sin2x+√2/2*cos2x)+1/2
=√2(sin2xcosπ/4+cos2xsinπ/4)+1/2
=√2sin...

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f(x)=sin(π-x)sin(π/2-x)+cos² x
=sinxcosx+cos² x
=1/2*sin2x+(1+cos2x)/2
=1/2*sin2x+1/2*cos2x+1/2
=√2(√2/2*sin2x+√2/2*cos2x)+1/2
=√2(sin2xcosπ/4+cos2xsinπ/4)+1/2
=√2sin(2x+π/4)+1/2
x∈[-π/8,,3π/8]
2x∈[-π/4,,3π/4]
2x+π/4∈[0,,π]
2x+π/4∈[0,,π/2]递增
2x∈[-π/4,,π/4]
x∈[-π/8,,π/8]

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