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已知函数F(x)=√3sinxCosx-Cos²x-½(x∈R)...

f(x)=1/(3/2*sin2x-cos2x-2)(3/2)^2+1^2再开方;既是根号13比上2;以下用a代替;f(x)=asin(2x-b)分之1(b为角);最小周期就是1/π;最小值就是sin值为-1的时候,即为-1/a;

解:f(x)=2[(√3/2)sinxcosx-(1+cos2x)/2-1/2. =(√3/2)sin2x-[(1/2)cos2x-1/2-1/2. =sin(2x-π/6)-1.当sin(2x-π/6)=-1, f(x)min=-2. ----函数f(x) 的最小值.T=2π/2=π . 所求f(x)的最小正

①f(x)=(√3)sinxcosx+1/2 =[(√3)/2]sin2xcosx+1/2 =[(√3)/4][sin(2x+x)+sin(2x-x)]+1/2 =[(√3)/4](sin3x+sinx)+1/2∵ sin3x、sinx的周期分别为2π/3、2π∴f(x)=(√3)sinxcosx

(1)f(x)=√3sinxcosx-cosx+1/2=(√3/2)sin2x-(1+cos2x)/2+1/2=sin(2x-π/6).∴2x-π/6=2kπ+π/2,即x=kπ+π/3时,所求最大值为:1.(2)f(A)=1→A=π/3=60°.故依余弦定理得b=√(a+c-2acco

f(x)=2√3sinxcosx-3sinx-cosx+2=√3sin2x-3(1-cosx)-cosx+2=√3sin2x-3+3cosx-cosx+2=√3sin2x+2cosx-1=√3sin2x+cos2x=2(√3/2*sin2x+1/2*cos2x)=2sin(2x+π/6)∵x∈[

f(x)=√3sinxcosx+cosx-1/2=(√3/2)sin2x+(1+cos2x)/2-1/2=(√3/2)sin2x+(1/2)cos2x=sin(2x+π/6)递减区间是:[kπ+π/6,kπ+2π/3](k∈Z)对称轴方程是:x=kπ+π/6(k∈Z)对称中心是(kπ-π/12,0)(k∈Z)y=sinx向左移π/6后,再纵坐标不变,横坐标变为原来的1/2,就可以得到f(x)=sin(2x+π/6)

∵ cos2x=2cosx-1 ∴ cosx=(cos2x+1)/2 ∵ sin2x=2sinxcosx ∴ f(x)=√3sinxcosx-cosx+1/2 =√3/2sin2x-1/2cos2x-1/2+1/2 =√3/2sin2x-1/2cos2x =sin2xcosπ/6-sinπ/6cos2x =sin(2x-π/6)(1) 函数f(x)的最小正周期为:2π/2=π(2) 单调增:2kπ-π/2<2

解:f(x)=√3sinxcosx-cosx-1/2 =√3/2(2sinxcosx)-1/2(1 cos2x)-1/2 =√3/2sin2x-1/2cos2x-1 =sin2xcosπ/6-cos2xsinπ/6-1 =sin(2x-π/6)-1 故 f(x)的最小正周期是 π,最小值是 -2.

f(x)=√3sinxcosx+cosx=(√3/2)2sinxcosx+(2cosx-1)+=(√3/2)sin2x+cos2x+=sin(2x+π/6) + 最小正周期T=2π/2=π

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