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已知函数F(x)=23sinxCosx+2Cos2x%1(x∈R),g(x...

(1)由f(x)=23sinxcosx+2cos2x-1,得f(x)=3(2sinxcosx)+(2cos2x-1)=3sin2x+cos2x=2sin(2x+π6),所以函数f(x)的最小正周期为π.(2)因为f(x)=2sin(2x+π6)在区间[0,π6]上为增函数,在区间[π6,π2]上为减函数,又f(0)=1,f(π...

解:(Ⅰ)由f(x)=2 sinxcosx+2cos 2 x﹣1,得f(x)= (2sinxcosx)+(2cos 2 x)﹣1)= sin2x+cos2x=2sin(2x+ )所以函数f(x)的最小正周期为π.因为f(x)=2sin(2x+ )在区间[0, ]上为增函数,在区间[ , ]上为减函数,又f(0)=1,f(...

f(x)=√3sinxcosx-1/2cos2x =√3/2sin2x-1/2cos2x =sin(2x-π/6). (1)最小值:f(x)|min=-1, 此时2x-π/6=2kπ-π/2, 即x=kπ-π/6 (k为整数); 最小正周期:T=2π/2=π. (2)f(C)=1,则 sin(2C-π/6)=1,即C=π/3. R=c/(2sinC)=√3/(2·√3/2)=1 (正...

(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...

(1)函数f(x)=5sinxcosx-53cos2x+523=52sin2x-53?1+cos2x2+523=5sin(2x-π3),故函数的周期为 2π2=π.(2)∵当x∈[0,π2]时,2x-π3∈[-π3,2π3],∴sin(2x-π3)∈[-32,1],∴5sin(2x-π3)∈[-

f(x)=2sinxcosx+2cos2x=sin2x+cos2x+1=2sin(2x+π4)+1,(Ⅰ)∵ω=2,∴函数f(x)的最小正周期为2π2=π,∵-1≤sin(2x+π4)≤1,∴1-2≤2sin(2x+π4)+1≤1+2,即1-2≤f(x)≤1+2,则函数f(x)的最小值为1-2;(Ⅱ)由α为锐角,f(α)=2,得:

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

(1)f(x)=sin2x+3sinxcosx-2cos2x=32sin2x-32cos2x-12=(32)2+(?32)2sin(2x+φ)-12.(其中tanφ=?3232=-3.故φ=?π3)=3sin(2x-π3)-12.故最小正周期T=2π2=π.故由2x-π3=kπ+π2,k∈Z得函数f(x)的图象的对称轴方程为:x=kπ2+5π12,k∈Z.(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...

f(x)=4(32sinx-12cosx)=4sin(x-π6),(Ⅰ)∵x∈[0,π],∴x-π6∈[-π6,5π6],∴-12≤sin(x-π6)≤1,即-2≤4sin(x-π6)≤4,则f(x)的最大值为4,最小值为-2;(Ⅱ)∵f(x)=23sinx-2cosx=0,即tanx=33,∴原式=cosx?sinxsinx+cosx=1?tanx1+tanx=1...

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