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设函数Fx等于4sinx(Cos%sinx)加3 1.当x属于(0,派...

解:(1)最简单的方法是用“积化和差”公式 2sinαcosβ = sin(α+β) + sin(α-β) 原式=2×2sinxcos(x+π/3) =2 [ sin(x+x+π/3) + sin(x-x-π/3) ] = 2 [ sin(2x + π/3) + sin (-π/3) ] = 2sin(2x + π/3) - √3 显然 T= 2π/ω = 2π/2 = π ※ 此法虽然简单,但...

参考

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...

f(x)=4√3sinxcosx-4cos²x+1 =2√3sin(2x)-2[cos(2x)+1]+1 =2√3sin(2x)-2cos(2x) -1 =4[sin(2x)(√3/2)-cos(2x)(1/2)] -1 =4sin(2x -π/6) -1 x∈[0,π/2] 0-π/6≤2x-π/6≤2·(π/2) -π/6 -π/6≤2x-π/6≤5π/6 -1/2≤sin(2x-π/6)≤1 4×(-1/2)-1≤4sin(2x -...

f(x)=2sinxcosx-sin²x+cos²x =sin2x+cos2x =√2sin(2x+π/4) 在[0, π]上,2x+π/4∈[π/4, 2π+π/4] 其中单调增区间为: 2x+π/4 ∈[π/4, π/2]U[3π/2,9π/4] 单调减区间: 2x+π/4 ∈[π/2, 3π/2] 即f(x)的单调增区间为:[0, π/8]U[5π/8, π] f(x)...

f(x)=4sinxcos(x+π/3)+根号3 =4sinx(1/2cosx-√3/2sinx)+√3 =2sinxcosx-2√3sin²x+√3 =sin2x-2√3sin²x+√3(sin²x+cos²x) =sin2x+√3cos²x-√3sin²x =sin2x+√3(cos²x-sin²x) =sin2x+√3con2x =2sin(2x+π/3) ...

lim(x->0) [1-cos(1-cosx)]/x^4 (0/0) =lim(x->0) [sin(1-cosx)] .(sinx) /(4x^3) =lim(x->0) [sin(1-cosx)] /(4x^2) ( x->0 sinx~x) (0/0) =lim(x->0) [cos(1-cosx)].( sinx) /(8x) =lim(x->0) [cos(1-cosx)] /8 ( x->0 sinx~x) =1/8 希望能解...

f(x)=a(1+cosx+sinx)+b=2asin(x+π4)+a+b,(1)当a=-1时,由2kπ+π2≤x+π4≤2kπ+32π,得2kπ+π4≤x≤2kπ+54π,∴f(x)的单调增区间为[2kπ+π4,2kπ+54π](k∈Z);(2)∵0≤x≤π,∴π4≤x+π4≤54π,∴-22≤sin(x+π4)≤1,依题意知a≠0,分两种情况考虑:1...

f(x)= - √3sin^2(x)+sinxcosx = -√3[(1-cos2x)/2]+(1/2)sinxcosx =(1/2)sin2x+(√3/2)cos2x-(√3/2) =sin2xcos60º+cos2xsin60º - (√3/2) f(x)=sin(2x+60º)-(√3/2) f(x/2)=sin[2(x/2)+60º] - (√3/2) =sin(x+60º) - (√3/2)...

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