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已知函数F(x)=√3/2sin2x%Cos^2x%1/2,x∈R

解:先用降幂公式把函数化为:f(x)=√3/2sin2x-1/2cos2x-1=sin(2x-π/6)-1 (1)最小值为-2,最小正周期为π (2)由f(C)=0知sin(2C-π/6)=1,从而可得C=π/3,再由余弦定理知:c^2=a^2+b^2-2abcosC 3=a^2+4a^2-2a*2acosπ/3,解得a=1,故b=2

解:f(x)=(√3/2)sin2x-(1+cos2x)/2-1/2. =(√3/2)sin2x-(1/2)cos2x-1/2-1/2. ∴f(x)=sin(2x-π/6)-1, (x∈R). (1) 当x∈[-π/12,5π/12]时,求函数f)x)的最小值和最大值。 ∵-π/12≤x≤5π/12. -π/6≤2x≤5π/6. -π/6-π/6≤2x-π/6≤5π/6-π/6. ∴ -π/3≤(2x-π/6)≤2π...

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fx=1/2sin2x-√3cos²x =1/2sin2x+√3/2cos2x-√3/2 =sin2xcosπ/3+cos2xsinπ/3-√3/2 =sin(2x+π/3)-√3/2 最小正周期=2π/2=π 最小值=-1-√3/2

(1)f(x)=√3/2sin2x-cos∧2x-1/2 =√3/2sin2x-1/2cos2x-1 =sin(2x-π/6)-1 最小值=-1-1=-2 最小正周期T=2π/2=π (2)f(C)=sin(2C-π/6)-1 =0 2C--π/6=π/2 C=π/3 m,n共线,则 sinB=2sinA 即b=2a 余弦定理 c^2=a^2+b^2-2ab*cosC 3=a^2+4a^2-2a^2 所以 a^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(...

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见图 解:(I)f(x)==sin2x+cos2x=sin(2x+). 令 2kπ-≤(2x+)≤2kπ+,可得 kπ-≤x≤kπ+,k∈z. 即f(x)的单调递增区间为[kπ-,kπ+],k∈z. (II)在△ABC中,由,可得sin(2A+)=,∵<2A+<2π+, ∴<2A+= 或,∴A= (或A=0 舍去). ∵b,a,c成...

f(x)=√3sinxcosx+cos²x =(√3/2)·2sinxcosx+½(2cos²x-1)+½ =(√3/2)sin2x+½cos2x+½ =sin(2x+π/6) +½ 最小正周期T=2π/2=π

f(x)=1/2sin2x-cos²(x+π/4) =1/2sin2x-1/2cos(2x+π/2)-1/2 =1/2sin2x+1/2sin2x-1/2 =sin2x-1/2 f(B/2)=sinB-1/2=0 sinB=1/2 锐角三角形 B=30° sinB=1/2 A+C=180°-B=150° C=150°-A属于(0°,90°) A属于(60°,90°) b=1 a=bsinA/sinB =2sinA c=...

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