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已知函数F(x)=23sinxCosx%2sin2x+1(x∈R).(1...

解(1)∵f(x)=2sin2x+23sinxcosx-1=3sin2x-cos2x=2(sin2xcosπ6-cos2xsinπ6),∴f(x)=2sin(2x-π6).∴函数f(x)的图象可由y=sinx的图象按如下方式变换得到:①将函数的y=sinx图象向右平移π6个单位,得到函数y=sin(x-π6)的图象;②将函数y=...

(Ⅰ)∵f(x)=2sinxcosx-sin2x+1=2sinxcosx+cos2x=sin2x+cos2x=2(22sin2x+22cos2x)=2sin(2x+π4)---(2分)∴f(x)的最小正周期为π;--------------------(3分)∵-π2+2kπ≤2x+π4≤π2+2kπ(k∈Z),∴-3π8+kπ≤x≤π8+kπ(k∈Z),∴f(x)的增区间为...

y=sin2x+sinx+cosx+2 y=2sinxcosx+1+sinx+cosx+1 y=[2sinxcosx+sin^2(x)+cos^2(x)]+sinx+cosx+1 y=(sinx+cosx)^2+(sinx+cosx)+1 y=(sinx+cosx+1/2)^2+3/4 又因为:负根号2

(sinx-cosx)sinx =(sinx)^2-sinxcosx =[-1+2(sinx)^2]/2+1/2-(sin2x)/2 =(-1/2)(cos2x+sin2x)+1/2 =(-√2/2)sin(2x+φ)+1/2 =(-√2/2)sin[2(x+φ/2)]+1/2 ∴最小正周期是(2π)/2=π

(1)由于f(x)=-2sin2x+23sinxcosx+5=?2×1?cos2x2+3sin2x+5=cos2x?3sin2x+4=2sin(2x+π6)+4则函数f(x)的最小正周期T=2π2=π(2)由(1)知f(x)=2sin(2x+π6)+4,∵x∈[π2,π],∴7π6≤2x+π6≤13π6∴sin(2x+π6)∈[?1,12],故f(x)∈[2,5]∴f(x)的...

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

(I)f(x)=sin2x-3cos2x+1=2sin(2x-π3)+1,∵ω=2,∴T=π;(II)∵x∈[π4,π2],2x-π3∈[π6,2π3],∴sin(2x-π3)∈[12,1],即2sin(2x-π3)+1∈[2,3],则函数f(x)的最大值为3,最小值为2.

(I)f(x)=1?cos2x2+32sin2x+(1+cos2x)=32sin2x+12cos2x+32=sin(2x+π6)+32.∴f(x)的最小正周期T=2π2=π.由题意得2kπ?π2≤2x+π6≤2kπ+π2,k∈Z,即kπ?π3≤x≤kπ+π6,k∈Z.∴f(x)的单调增区间为[kπ?π3,kπ+π6],k∈Z.(II)先把y=sin2x图象上所有...

f(x)=2√3sinxcosx-2sin²x+1 =√3sin2x+cos2x =2sin(2x+π/6) (2) 2kπ-π/2

(1)解法一:∵f(x)=1?cos2x2+sin2x+3(1+cos2x)2=2+sin2x+cos2x=2+2sin(2x+π4)(4分)∴当2x+π4=2kπ+π2,即x=kπ+π8(k∈Z)时,f(x)取得最大值2+2.因此,f(x)取得最大值的自变量x的集合是{x|x=kπ+π8,k∈Z}. (8分)解法二:∵f(x)=(sin2...

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