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F x sin 2x sinx 1

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)=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)

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)当a=2时,f(x)=sin2x-4(sinx+cosx)+4=2sinxcosx-4(sinx+cosx)+4,令sinx+cosx=t(-2≤t≤2),则2sinxcosx=t2-1,∴g(t)=t2-4t+3=(t-2)2-1,∵g(t)在[-2,2]上单调递减,∴g(t)min=g(2)=5-42.(2)令sinx+cosx=t(-2≤t≤2),∵h...

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

(1)令 x= 9π 4 ,得 2 a+4+9=13-9 2 ,得a=-9.(2) f(x+π)=-9(|sin(x+π|+|cos(x+π)|)+4sin2(x+π)+9 =-9(|sinx|+|cosx|)+4sin2x+9=f(x) 所以,f(x)的最小正周期为π.(3)不存在n满足题意. 当 x∈[0, π 2 ] 时,f(x)=-9(sinx+cosx)+4...

已知函数f(x)=2sinxsin(π/2+x)-2sin²x+1(x∈R)(1)求函数f(x)的最小正周期及函数f(x)的单调递增区;(2)若f(xo/2)=(√2)/3,xo∈(-π/4,π/4),求cos(2xo)的值。 解:(1) f(x)=2sinxcosx-(1-2sin²x)=sin2x-cos2x=(√2)[sin2xcos(π/4)-co...

(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

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