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已知函数Fxsin2x

f(x)=cosxsin2x=2sinxcos2x=2sinx(1-sin2x)=2sinx-2sin3x,对于A:∵f(π-x)+f(π+x)=(2sinx-2sin3x)-(2sinx-2sin3x)=0,∴y=f(x)的图象关于(π,0)中心对称,即A正确;对于B:∵f(π-x)=2sinx-2sin3x=f(x),∴y=f(x)的图象关于x=...

f(x)=(sin2x-2sin²x)/sinx =(2sinxcosⅹ-2sin²x)/sinx =2(cosx-sinx) =2√2[cosxcos(π/4)-sinxsin(π/4)] =2√2cos(x+π/4) 显然,定义域为x∈R,即(-∞,+∞). cos(x+π/4)=1, 即x=2kπ-π/4时, 最大值f(x)|max=2√2. 0

k=2 。 这是一个分段函数。首先一个函数在一点处联系的充分必要条件是:1、在这一点首先要有定义。2、左右极限相等且等于该点的函数值。 该函数在零处地左极限lim (sin2x)/x=2.有极限limf(x)=k. x->0- x->0+ 有f(0) =k.所以k=2

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

亲,网友,您说的是不是下面的问题: 已知函数f x=sin(兀/2-x)sinx-根号3cos^2x,求周期、最值。 f(x)=1/2 sin2x-√3/2(cos2x+1) =sin(2x-π/3)-√3/2 T=2π/2=π。 f max=1-√3/2, f min=-1-√3/2. 送您 2015 夏祺 凉快

f(x)=sin2x-2sin²x =sin2x-(1-cos2x) =sin2x+cos2x-1 =√2sin(2x+π/4)-1 T=2π/2=π 当2x+π/4=2kπ+π/2时函数取到最大值 此时x=kπ+π/8 f(x)max=√2-1

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

(1) f(x)=sin(2(x-π/12)) x∈[0,π]时,2(x-π/12)∈[-π/6,11π/6] 单调递减区间是2(x-π/12)∈[π/2,3π/2] 即x-π/12∈[π/4,3π/4] 则x∈[π/3,5π/6] (2)x∈[-π/12,π/2]时, 2(x-π/12)∈[-π/3,5π/6] 而当2(x-π/12)∈[-π/3,π/3]时,sin(2(x-π/12))∈[sin(-π/3),...

(1)T=2π2=π.(2)由2kπ-π2≤2x+π6≤2kπ+π2,得kπ-π3≤x≤kπ+π6,k∈Z,∴函数的单调增区间为[kπ-π3,kπ+π6](k∈Z).(3)∵x∈[0,π2],∴2x+π6∈[π6,7π6],∴-12≤sin(2x+π6)≤1,∴当2x+π6=π2,即x=π6时函数有最大值1,当2x+π6=7π6时,即x=π2,函数有...

f(x)=sin2x-2√3sin^2x+√3+1 =sin2x+√3(-2sin^2x+1)+1 =(sin2x+√3cos2x)+1 =(sin2xcos(π/3)+cos2xsin(π/3))*2+1 =2sin(2x+π/3)+1 最小正周期=π -π/2+2kπ

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