cosx+cos2x周期

宓府巧4472函数y=1+cos2x/sin2x的周期是多少 -
后底玲15317732165 ______ y=(1+cos2x)/sin2x=(2cosx^2)/(2sinxcosx)=cotx 周期是2π

宓府巧4472求数学大神Y=cosX/cos2X 的周期怎么算 -
后底玲15317732165 ______[答案] 分子周期是2π 分母周期是π 所以y周期是他们的最小公倍数 所以是2π

宓府巧4472对于函数y=sinxcosx+cos2x,求最小正周期和函数的值域 -
后底玲15317732165 ______[答案] y=sinxcosx+cos2x =1/2sin2x+cos2x =√5/2sin(2x+a) T=2∏/2=∏ 值域[-√5/2,√5/2]

宓府巧4472cosx(x - 1)周期 -
后底玲15317732165 ______ f(x)=(1+cos2x)/2-1/2=(cos2x)/2 周期t=2π/2=π

宓府巧4472为何y=sin2x+cosx是周期函数 -
后底玲15317732165 ______[答案] sin2x 周期是T=π cosx周期是T=2π 得到函数周期是T=2π f(x)=sin2x+cosx f(x+2π)=sin(2x+4π)+cos(x+2π)=sin2x+cosx=f(x)

宓府巧4472求下列函数的周期:y=cos2x+sin2x. -
后底玲15317732165 ______[答案] y=cos2x+sin2x= 2( 2 2cos2x+ 2 2sin2x)= 2sin(2x+ π 4), 则三角函数的周期T= 2π 2=π.

宓府巧4472(cos2x+sin2x)/(cos2x - sin2x)的周期是什么?特别写出过程, -
后底玲15317732165 ______[答案] (cos2x+sin2x)/(cos2x-sin2x)=cos2x²-sin2x²=cos2x*cos2x-sin2x*sin2x=cos(2x+2x)=cos4x 故周期是2π/4=π/2

宓府巧4472求函数y=│cosx│+cosx的最小正周期 -
后底玲15317732165 ______[答案] cosx≥0,y=│cosx│+cosx=2cosx; cosx最小正周期2π

宓府巧4472求函数y=sin(60 - 2x)+cos2x的周期 -
后底玲15317732165 ______[答案] y=sin(60-2x)+cos2x =sin60cos2x-cos60sin2x+cos2x =(1+根号3)/2*cos2x-1/2*sin2x =根号下[((1+根号3)/2)^2+(-1/2)^2]*sin(2x+w) 所以周期为2π/2=π

宓府巧4472求函数y=(2根号3)sinxcosx+2cos2x+1的最小正周期和单调区间 -
后底玲15317732165 ______[答案] 答: y=2√3sinxcosx+2cos²x+1 =√3sin2x+cos2x+1+1 =2*[(√3/2)sin2x+(1/2)cos2x]+2 =2sin(2x+π/6)+2 所以: 最小正周期T=2π/2=π 单调递增区间满足:2kπ-π/2单调递减区间满足:2kπ+π/2k∈Z