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已知函数y=2sin(2x+π/4)+1 求 1.周期 2.最值以及...

已知y=2sin(2x+π/6)求:(1)最值及相应的x 2x+π/6=2kπ+π/2 2x=2kπ+π/3 x=kπ+π/6 取最大值2 2x+π/6=2kπ+3π/2 2x=2kπ+4π/3 x=kπ+2π/3 取最小值-2 (2)单调区间 递增区间: 2kπ-π/2≤2x+π/6≤2kπ+π/2 2kπ-2π/3≤2x≤2kπ+π/3 kπ-π/3≤x≤kπ+π/6 即为...

f(x)=2sin²x+sin2x+1 =(1-cos2x)+sin2ⅹ+1 =2+√2sin(2x-π/4). 1)最小正周期T=2π/2=π; 2)x∈(0,π/2), 即2x-π/4∈(-π/4,3π/4), 故sin(2x-π/4)=0,即x=π/8时, 最小值f(x)|min=2; sin(2ⅹ-π/4)=1,即x=3π/8时, 最大值f(x)|max=2+√2。

解答:(1)f(x)=2sin²(π/4+x)-根号3cos2x=1-cos(π/2+2x)-√3cos2x=sin2x-√3cos2x+1=2sin(2x-π/3)+1∵x∈[π/4,π/2]∴2x-π/3∈[π/6,2π/3]∴sin(2x-π/3)∈[1/2,1]∴2x-π/3=π/6时,f(x)有最小值22x-π/3=π/2时,f(x)有最大值3(2)|f(x)-m|m-2且f(x)m-2...

由f (x)=4sin2(π4+x)-23cos2x-1=2[1?cos(π2+2x)]?23cos2x?1=2sin2x?23cos2x+1=4(12sin2x?32cos2x)+1=4sin(2x?π3)+1(1)T=2π2=π,f(x)max=5,f(x)min=?3(2)由π2+2kπ≤2x?π3≤3π2+2kπ,解得5π6+2kπ≤2x≤11π6+2kπ,∴5π12+kπ≤x≤11π12+kπ,k∈...

解答:(1)解:函数f(x)=sin(x+7π4)+cos(x-3π4)=sinxcos7π4+cosxsin7π4+cosxcos3π4+sinxsin3π4=22sinx-22cosx-22cosx+22sinx=2sinx-2cosx=2sin(x-π4),∴f(x)的最小正周期为π,f(x)max=2,f(x)min=-2;(2)证明:cos(β-α)=cosβ...

解①当2x+π/4=π/2+2kπ时,sin(2x+π/4)最大=1 即x∈{x│x=π/8+kπ,k∈Z}时,ymax=2*1+1=3 ②由解析式可知对称轴向左平移π/8个单位 所以对称轴为x=π/4+kπ-π/8=π/8+kπ

f(x)=1/2sin(2x+6π)+5/4. 当f(x)单调递增时, 2kπ-π/2≤2x+6π≤2kπ+π/2 →kπ-13π/4≤x≤kπ-11π/4. 即单调递增区间为: [kπ-13π/4,kπ-11π/4]. 2x+6π=kπ+π/2 →x=kπ/2-11π/4. 故对称轴为: x=kπ/2-11π/4, 对称中心为: (kπ/2-11π/4,0). 当f(x)取最小值时...

(1)T=2π/2=π -π/2+2kπ

(1)f(x)=2sin(2x+π4),∵ω=2,∴最小正周期T=2πω=π,(2分)由2kπ-π2≤2x+π4≤2kπ+π2(k∈Z),解得kπ-3π8≤x≤kπ+π8(k∈Z),故函数f(x)的单调增区间是[kπ-3π8,kπ+π8](k∈Z);(7分)(2)当x∈[-π4,π4]时,(2x+π4)∈[-π4,3π4],(9分)故...

解;(1)f(x)=4cosxsin(x+π6)+a=4cosx(32sinx+12cosx)+a=3sinx+2cos2x+a=2sinx(2x+π6)+1+a∵函数的最大值为2,∴a=-1,T=2π2=π;(2)列表画图如下:.

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