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sin& 4+x

原式=cos[π/2-(π/4-x)]sin(π/4+x) =1/2[2cos(π/4+x)sin(π/4+x))] =1/2*sin(π/2+2x) =1/2*cos2x

由sin(派/4+x)=二分之根号二(sinx+cosx)=1/3,即sinx+cosx=三分之根号二,两边同时平方得,1+2sinxcosx=2/9,所以sin2x=2sinxcosx=2/9-1=-7/9

f(x)=4sinxsin²(π/4+x/2)+cos2x-1 =4sinxsin²[(π/2+x)/2]+cos2x-1 =4sinx[1-cos(π/2+x]/2+cos2x-1 =2sinx+2sin²x+1-2sin²x-1 =2sinx ∴f(ωx)=2sinωx x∈[-π/2,2π/3]是增函数 f'(ωx)=2ωcosωx>0 ∵ω>0 ∴cosωx>0 ω·2π/3≤π/2→ω≤3/4...

cos(π/4+x)*1 复合求导,高考中重点的重点,先对SIN求导,在对π/4+x求导,注意象限,注意符号。

tanx = 1/2 2sinx= cosx 2sinxcosx = (cosx)^2 sin2x = 1/(secx)^2 = 1/[( tanx)^2 +1] =1/(1+ 1/4) = 4/5 [sin(π/4+x)]^2 =[(√2/2)(sinx+cosx) ]^2 =(1/2)(1+sin2x) =(1/2)(1+ 4/5) = 9/10

sinx(x+4分之派)=sinxcos四分之派+ cosxsin四分之派 cos四分之派 =sin四分之派 =二分之根号二 结果为二分之根号二倍的(sinx+cosx)

cos(π/4+x)=cosπ/4cosx-sinπ/4sinx=(1/√2)(cosx-sinx)=3/5, cosx-sinx=3√2/5,两边平方得,cos²‎x+sin²‎x-2cos‎xsinx=18/25, 1-sin2x=18/25,sin2x=7/18 打字不易,如满意,望采纳。

sin(π/4+X) =sin[π/2-(π/4-X)]=cos(π/4-X) 因为0

解.f(x)=2sinx[1-cos(x+π/2)]+1-2sin²x=2sinx(1+sinx)+1-2sin²x=2sinx+1 (1)y=f(wx)=2sinwx+1 因在区间[-π/2,2π/3]上是增函数,所以最小正同期T=2π/w≥2(π/2+2π/3) 即0

已知0<x<π/4,sin(π/4-x)=5/13,求cos2x/cos(π/4+x)的值 已知0<x<π/4,则sinx>0,cosx>0 sin(π/4-x)=sinπ/4·cosx-cosπ/4·sinx=√2/2(cosx-sinx)=5/13 则cosx-sinx=5√2/13,又因为sin²x+cos²x=1 2cosx·sinx=-(cosx-sinx)²+sin²...

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