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F(x)=sinx^4+2倍根号3 sinxCosx%Cosx^4 求(1)函...

f(x)=√3sinxcosx-cos²x-1/2 =(√3/2)sin2x-(1+cos2x)/2-1/2 =(√3/2)sin2x-(1/2)cos2x-1 =sin(2x-π/6)-1. 早小正周期T=2π/2=π. sin(2x-π/6)=-1, 即x=kπ-π/6时, 最小值f(x)|min=-2。

解: (1) f(x)=cos²x-√3sinxcosx+½ =½[1+cos(2x)]-(√3/2)sin(2x)+½ =½cos(2x)-(√3/2)sin(2x)+1 =cos(2x+π/3)+1 最小正周期T=2π/2=π cos(2x+π/3)=1时,f(x)取得最大值f(x)max=1+1=2 cos(2x+π/3)=-1时,f(x)取得最小值f(...

、先将f(x)=m*n化成一个函数式子,f(x)=m*n=√3sinxcosx/16+cosx/4,根据二倍角公式,等于三十二分之根号三倍的sin2x加上八分之cos2x加上1,。并进行提取公因式得到十六分之一倍的括号里二分之根号三倍的sin2x加上二分之cos2x括号结束再加上一,...

f(x)=3cosx+4sinx =5×(3/5×cosx+4/5×sinx) =5×(sina×cosx+cosa×sinx),sina=3/5,cosa=4/5 =5sin(x+a) ∵-1≤sin(x+a)≤1 ∴-5≤sin(x+a)≤5 故,函数的值域为[-5,5]。 答题不易,望采纳~~~

(1)f(x)=4sinx[sin(π/4+x/2)]^2+2(cosx)^2+1+a =2sinx[1-cos(π/2+x)]+2+cos2x+a =2sinx(1+sinx)+cos2x+2+a =2sinx+2(sinx)^2+cos2x+2+a =2sinx+3+a, x∈R是一个奇函数, ∴f(0)=3+a=0,a=-3. (2)f(x)=2sinx, f(2x)=2sin2x>=-√3, ∴sin2x>=-√3/2, ∴...

f(x)=根号3sin2x+cos2x=2sin(2x+π/6) 1) T=2π/2=π x∈[0,π/2] 2x+π/6[π/6,7π/6] f(x)小=2sin(π/6)=1 f(x)大=2sin(π/2)=2 2) sin(2x0+π/6)=3/5 co2x0=cos(2x0+π/6-π/6)=(3-4根号3)/10 字限制

解: f(x)=2√3sinxcosx+1 -2sin²x =√3sin(2x)+cos(2x) =2[(√3/2)sin(2x)+(1/2)cos(2x)] =2sin(2x+π/6) -1≤sin(2x+π/6)≤1 -2≤2sin(2x+π/6)≤2 -2≤f(x)≤2 函数的值域为[-2,2]

你好,你可以试着先在一个平面坐标系画2个图像(即y=sinx和y=cosx的图像),是否发现有两个交点?那两个交点对应的x轴坐标就是π/4和5π/4。推导过程: 因为是交点,即sinx=cosx可得sinx/cosx=1,即tanx=1,可得x=π/4和5π/4(0-2π范围内) 至于范围,...

求采纳,

(1) f(x)=2√3sinxcosx+2cos²x-1=√3sin2x+cos2x=2(√3/2sin2x+1/2cos2x)=2sin(2x+π/6) 函数的最小正周期T=π (2) f(x)=2sin(2x+π/6)=6/5 sin(2x+π/6)=3/5 x∈[π/4,π/2], cos(2x+π/6)=-4/5 cos2x=cos(2x+π/6-π/6)=cos(2x+π/6)cosπ/6...

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