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

max=3/2,min=1 解: f(x) =sin²x+√3sinxcosx =(1-cos2x)/2+(√3/2)sin2x =sin(2x-π/6)+1/2 ∵x∈[π/4,π/2] ∴2x-π/6∈[π/3,5π/6] ∴ f(x)_max=1+1/2=3/2 f(x)_min=1/2+1/2=1

求采纳,

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(...

解: f(x)=sin²x-cos²x-2√3sinxcosx =-(cos²x-sin²x)-√3(2sinxcosx) =-cos2x-√3sin2x =-2[(√3/2)sin2x+(1/2)cos2x] =-2sin(2x+π/6) (1) f(2x/3)=-2sin[2(2x/3)+ π/6]=-2sin(4x/3 +π/6) (2) 最小正周期T=2π/2=π 2kπ+ π/2≤2x+...

、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)=2sinx(sinx+cosx) =2(sinx)^2+2sinxcosx =1-cos2x+sin2x =√2sin(2x-π/4)+1 所以函数fx的最小正周期是T=2π/2=π 如果不懂,请追问,祝学习愉快!

f(x)=(sinx)^2+2√3sinxcosx+sin(x+π/4)sin(x-π/4) =1/2(1-cos2x)+√3sin2x+sin(x+π/4)sin[-π/2+(x+π/4)] =1/2-1/2cos2x+√3sin2x-sin(x+π/4)cos(x+π/4) =1/2-1/2cos2x+√3sin2x-1/2sin(2x+π/2) =1/2-1/2cos2x+√3sin2x-1/2cos2x =√3sin2x-cos2x+1/2 ...

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

fx=1+根号2cos(2x-派/4)/sin(x+派/2) =1+(cos2x+sin2x)/cosx =1+cosx+sin^x/cosx+2sinx =1+cosx+1/cosx-cosx+2sinx =1+1/cosx+2sinx

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