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已知F(x)=sin(2x+π6)+12%m2(1)求函数F(x)...

(1)∵f(x)=sin(2x+π6)+12-m2,∴函数f(x)的最小正周期T=2π2=π;(2)由2kπ-π2≤2x+π6≤2kπ+π2可得kπ-π3≤x≤kπ+π6,∴函数f(x)的单调递增区间为[kπ-π3,kπ+π6](k∈Z);(3)当x∈[-π6,π3]时,2x+π6∈[-π6,5π6],∴sin(2x+π6)∈[?12,1],∴f...

(1)当m=0时,f(x)=f(x)=sin2x+sinxcosx=12(sin2x?cos2x)+12=22sin(2x?π4)+12由于x∈[π8,3π4]所以:2x?π4∈[0,5π4]sin(2x?π4)∈[?22,1]f(x)∈[0,1+22](2)由于f(x)=sin2x+sinxcosx+m2sin(x+π4)sin(x?π4)=12[sin2x?(1+m2)cos2x]+12所以...

∵y=2sin(x+π2)cos(x-π2)=2cosxsinx=sin2x,∴由题意得:sin2x=12,∴2x=2kπ+π6或2x=2kπ+5π6,∴x=kπ+π12或x=kπ+5π12,k∈Z,∵正弦曲线y=sin2x与直线y=12在y轴右侧的交点自左向右依次记为M1,M2,M3,…,∴得M1(π12,0),M2(5π12,0),M3(π+π...

y=sin2x+m?cosx+58m-32=1-cos2x+mcosx+58m?32∵x∈[0,π2],∴0≤cosx≤1y=-(cosx-m2)2+m24+5m8?12①当m2≤0,即m≤0时,ymax=5m8?12=1,得m=125(舍);②当0<m2≤1,即0<m≤2时,ymax=m24+5m8?12=1,得a=-4(舍)或a=32;③当m2>1,即m>2时,ymax=m+...

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