sinx cosx=cos2x
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(Ⅰ)∵f(x)=sin2x+cos2x∴f(π4)=sinπ2+cosπ2=1(Ⅱ)f(α2)=cosα+sinα=22∴sin(α+π4)=12,cos(α+π4)=±32.sinα=sin(α+
y=7-4sinxcosx+4cos2x-4cos4x=7-2sin2x+4cos2x(1-cos2x)=7-2sin2x+4cos2xsin2x=7-2sin2x+sin22x=(1-sin2x)2
派再问:求过程再答:再答:给好评再问:三角形3边ABC满足B^2=AC求f(B的取值范围。。)再答:先给好评,立马帮你解决再答:应该是abc吧?再问:嗯。再问:?
sin2x=2sinxcosxcos2x=(cosx)^2-(sinx)^2sin2x+sinxcosx-cos2x=3sinxcosx-(cosx)^2+(sinx)^2由于sinx-2cosx=0
y=√3cos2x+sin2x=√[1²+(√3)²]sin(2x+z)=2sin(2x+z)其中tanz=√3/1=√3所以最大=2,最小=-2T=2π/2=π
(Ⅰ)f(x)=32sin2x-(cos2xcosπ3-sin2xsinπ3)-cos2x+12=3sin2x-cos2x-12=2sin(2x-π6)-12,∴f(x)的最小正周期T=π.当2kπ-
sinx/cosx=tanx=2sinx=2cosx带入恒等式sin²x+cos²x=1cos²x=1/5sinxcosx=(2cosx)cosx=2cos²x
f(x)=(√3)sinxcosx+cos2x+1f(x)=(√3)(2sinxcosx)/2+cos2x+1f(x)=(√3/2)sin2x+cos2x+1f(x)=(√7/2)[(√3/2)(2/
f(x)=2√3sinxcosx-cos2x=√3sin2x-cos2x=2(sin2x*√3/2-cos2x*1/2)=2sin(2x-π/6)x=π/12;函数f(x)的图象可以由函数y(x)=2
(1)∵f(x)=3sinxcosx-cos2x+12=3sinxcosx-2cos2x−12=32sin2x-12cos2x=sin(2x-π6),∴T=2π2=π.(2)设由y=f(x)(0≤x≤
f(x)=cos2x-sin2x+2√3sinxcosx=cos2x-sin2x+√3sin2x=cos2x+(-1+√3)sin2x=√(5-2√3)sin(2x+φ)=(√5-√3)si(2x+φ
分两部分求2sin2x=4sinxcosx注:sin2x=2sinxcosx=4sinxcosx/{(cosx)^2+(sinx)^2}注:{(cosx)^2+(sinx)^2=1=4tanx/{1+
tanx=3cosx^2=1/(1+tanx^2)=1/10sinxcosx+cos2x=tanx*cosx^2+2cosx^2-1=3/10+2/10-1=-1/2
f(x)=2sinxcosx+cos2x=sin2x+cos2x=√2sin(2x+π/4)2x+π/4=π/2+2kπ时f(x)有最大值f(x)=√2x=π/8+kπ2x+π/4=3π/2+2kπ时
f(x)=sin2x+cos2x=√2sin(2x+π/4)f(π/4)=√2sin(2*π/4+π/4)=√2*√2/2=10
f(x)=sin2x+cos2x=√2sin(2x+π/4)所以T=2π/2=π最大值=√2f(θ+π/8)=√2sin(2θ+π/4+π/4)=√2cos2θ=√2/3cos2θ=1/3θ锐角则si
f(x)=√3sinxcosx+cos2x+1=(√3/2)sin2x+cos2x+1=[(√7)/2][(√3/√7)sin2x+(2/√7)cos2x]+1=[(√7)/2]sin(2x+α)+1
cos2x-2根号3sinxcosx=cos2x-根号3(2sinxcosx)运用倍角公式得=cos2x-根号3sin2x运用辅助角公式得=-2sin(2x-六分之π)由ω=2,T=二派除以ω,所以周
解:原式=√3sin2x+cos2x+1=2(√3/2sin2x+1/2cos2x+1=2cos(2x-pai/3)+1.
不对,因为f(x)=cos2x-2√3sinxcosx=cos2x-√3sin2x=2[sinπ/6cos2x-cosπ/6sin2x]=2sin(π/6-2x)左移5π/12,f(x)=2sin【π