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F x sin 2x sinx 1

f(x)=sin²x+sinxcosx+1 =(1-cos2x)/2+sin2x/2+1 =½(sin2x-cos2x)+3/2 =√2/2sin(2x-π/4)+3/2 ∴最小正周期是π 单调递增区间2x-π/4∈(2kπ-π/2,2kπ+π/2)→x∈(kπ-π/8,kπ+3π/8) 单调递减区间2x-π/4∈(2kπ+π/2,2kπ+3π/2)→x∈(kπ+3π/8,kπ+7π/8)

首先是奇函数,但是它一定有最大值 所以没有正确的选项 你的题意不明,1/2是什么的系数?

解答如下: f(x)=(1+sinx+cosx+sin2x)/(1+sinx+cosx); =(sin^2x+cos^2x+sinx+cosx+sin2x)/(1+sinx+cosx); =(sin^2x+2sinxcosx+cos^2x+sinx+cosx)/(1+sinx+cosx); =[(sinx+cosx)^2+sinx+cosx]/(1+sinx+cosx); =(sinx+cosx)*(...

f'(x)=1/2cos2x*(2x)]+cosx =cos2x+cosx 显然这是偶函数 cos2x+cosx =2cos²x-1+cosx =2(cosx+1/4)²-9/8 -1

f(x)=sin2xcosx/(1-sinx)=2sinx*(cosx*cosx)/(1-sinx) =2sinx*(1-sinx)*(1+sinx)/(1-sinx) =2sinx*(1+sinx),(sinx不等于1) =2*(sinx+sinx*sinx) =2*[(sinx+1/2)^2-1/4] 因为-1

设sinx=u x=arcsinu f(u^2)=arcsinu/u f(x)=arcsinx/√x ∫{[√x f(x)] / √(1-x) }dx =∫{(√x*arcsinx/√x) / √(1-x) }dx =∫{arcsinx/√(1-x)}dx 分部积分 设u=arcsinx dv=1/√(1-x)dx du=1/√(1-x^2)dx v=-2√(1-x) =-2arcsinx√(1-x)+∫{2√(1-x)/√(1-x^2)...

f(x)=sin2x/sinx+2sinx=2cosx+2sinx=2√2sin(x+π/4) 定义域R和最小正周期T=2π f(a)=2 , 2=2√2sin(a+π/4) , sin(a+π/4)=√2/2 , a=π/2 f(a+π/12)=2√2sin(a+π/12+π/4)=2√2sin(a+π/3)=2√2sin(π/2+π/3)=2√2cosπ/3=√2

(1)∵cos2x = 1-2sin²x ∴f(x) = sin2x +cos2x -1=√2[﹙√2/2﹚sin2x +﹙√2/2﹚cos2x] -1 =√2﹙sin2xcosπ/4+cos2xsinπ/4﹚=√2sin﹙2x+π/4) ∵T=2π/2=π (2)令u=2x+π/4 ∵u=π/2 +2kπ(k∈Z) 时,f(x)取最大值 ∴2x+π/4=π/2 +2kπ(k∈Z) 解得...

f(x)=sin2x-2√3sin^2x+√3+1 =sin2x+√3(-2sin^2x+1)+1 =(sin2x+√3cos2x)+1 =(sin2xcos(π/3)+cos2xsin(π/3))*2+1 =2sin(2x+π/3)+1 最小正周期=π -π/2+2kπ

用升幂公式:cos2x=(cosx)^2-(sinx)^2=2(cosx)^2-1 f(x)=[2*(cos^2)+8*(sin^2)]/(2sinx*cosx)=(1+4tan^2)/tan =(1/tanx)+4*tanx>=2*根号[(1/tanx)*4tanx]=4 所以最小值为4 追问: 麻烦问下2*根号[(1/tanx)*4tanx]怎么推出来呢啊?

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