f(x)=5√3cos²x+√3sin²x-4sinxcosx(π/4≤x≤7π/24)的最小值,并求其单调区间 啊··
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![f(x)=5√3cos²x+√3sin²x-4sinxcosx(π/4≤x≤7π/24)的最小值,并求其单调区间 啊··](/uploads/image/z/5259791-47-1.jpg?t=f%28x%29%3D5%E2%88%9A3cos%26sup2%3Bx%2B%E2%88%9A3sin%26sup2%3Bx-4sinxcosx%EF%BC%88%CF%80%2F4%E2%89%A4x%E2%89%A47%CF%80%2F24%EF%BC%89%E7%9A%84%E6%9C%80%E5%B0%8F%E5%80%BC%2C%E5%B9%B6%E6%B1%82%E5%85%B6%E5%8D%95%E8%B0%83%E5%8C%BA%E9%97%B4+%E5%95%8A%C2%B7%C2%B7)
f(x)=5√3cos²x+√3sin²x-4sinxcosx(π/4≤x≤7π/24)的最小值,并求其单调区间 啊··
f(x)=5√3cos²x+√3sin²x-4sinxcosx(π/4≤x≤7π/24)的最小值,并求其单调区间 啊··
f(x)=5√3cos²x+√3sin²x-4sinxcosx(π/4≤x≤7π/24)的最小值,并求其单调区间 啊··
f(x)=5√3cos²x+√3sin²x-4sinxcosx
=4√3cos²x+√3-4sinxcosx
=2√3(2cos²x-1)+3√3-2*2sinxcosx
=2√3cos2x-2sin2x+3√3
=4(√3/2*cos2x-1/2*sin2x)+3√3
=4(cos2x*cosπ/6-sin2x*sinπ/6)+3√3
=4cos(2x+π/6)+3√3.
因为π/4≤x≤7π/24,
所以2π/3≤2x+π/6≤3π/4,
-√2/2
自己算
(x)=5√3cos²x+√3sin²x-4sinxcosx
=4√3cos²x+√3-4sinxcosx
=2√3(2cos²x-1)+3√3-2*2sinxcosx
=2√3cos2x-2sin2x+3√3
=4(√3/2*cos2x-1/2*sin2x)+3√3
=4(co...
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(x)=5√3cos²x+√3sin²x-4sinxcosx
=4√3cos²x+√3-4sinxcosx
=2√3(2cos²x-1)+3√3-2*2sinxcosx
=2√3cos2x-2sin2x+3√3
=4(√3/2*cos2x-1/2*sin2x)+3√3
=4(cos2x*cosπ/6-sin2x*sinπ/6)+3√3
=4cos(2x+π/6)+3√3.
因为π/4≤x≤7π/24,
所以2π/3≤2x+π/6≤3π/4,
-√2/2<=cos(2x+π/6)<=-1/2,
故f(x)的最小值: 3√3-2√2。
由2kπ<=2x+π/6<=2kπ+π,得:
kπ-π/12<=x<=kπ+5π/12,
所以f(x)在[kπ-π/12,kπ+5π/12]单调递减;(k为整数)
由2kπ-π<=2x+π/6<=2kπ,得:
kπ-7π/12<=x<=kπ-π/12,
所以f(x)在[kπ-7π/12,kπ-π/12]单调递增。(k为整数)
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