求函数6sin (4x=3π 4)在[0,5π 24]上值域-搜答案-搜搜做题作业网

π/2+2kπ再问：换元法有没有？再答：令3x+π/4=t，y=2sint的递减区间是：π/2+2kπ

y=-1/2sin(2/3x-π/4)所以y和sin(2/3x-π/4)单调性相反sinx的增区间是(2kπ-π/2,2kπ+π/2)减区间是(2kπ+π/2,2kπ+3π/2)所以sin(2/3x-

y=3sin(3x+π/4)单调增区间是:2kPai-Pai/2

y=2sin(3x+π/4)依题意-π/2+2kπ

y=sin^4x+cos^4x=sin^4x+cos^4x+2sin^2xcos^2x-2sin^2xcos^2x=（sin^2x+cos^2x）^2-2sin^2xcos^2x=1-1/2sin^2

y=-3sin(2x-Pai/4)定义域是R最小正周期T=2Pai/2=Pai单调增区间是2kPai+Pai/2

f(x)=sin2ωx+√3cos2ωx=2sin(2ωx+π/3),两对称轴之间的最小值为π/2即半个周期,则周期为π=2π/2ω,所以w=1,所以f(x)=2sin(2x+π/3),f(α)=2s

设t=2x+π/4,则y=3sin(2x+π/4)为y=3sint由正弦函数的单调性质,可知正弦函数的单调递增区间为[2kπ-π/2,2kπ+π/2]所以y=3sint的单调递增区间为[2kπ-π/2

首先：定义域只有这一个,X+π/4≠2Kπ,所以X≠-π/4+2kπ..附上值域,化简原函数：f（X）=cos2X/[√2/2（sinX+cosX）]f(x)=(cos²X-sin²

模型y=Asin(ωX+ψ)振幅A=4周期T=2π/ω=4π最大值=A=4最小值=-A=-4

f(x)=2sin(x/4)cos(x/4)-2√3sin²(x/4)+√3=sin(x/2)+√3cos(x/2)=2sin(x/2+π/3)最小正周期T=2π/(1/2)=4π祝你开心!

由三角函数的图象和性质可知当2x+π4=2kπ+π2，即x=kπ+π8时，k∈Z函数取得最大值3，当2x+π4=2kπ-π2，即x=kπ-3π8时，k∈Z函数取得最小值-3．即取得最大值3时，对应的集

因为3>0,所以要使函数f(x)=3sin(-2x+π/4）为减则有π/2+2kπ

f(x)=sin(2x-π/6)=cos[π/2-(2x-π/6)]=cos(2π/3-2x)=cos(2x-2π/3),故f(x)为偶函数,且其关于x=π/3对称,周期为π,因此a的最小整数为π.

f(x)=sin^2x+2√3sinxcosx+sin(x+π/4)sin(x-π/4)=(1-cos2x)/2+√3sin2x+(1/2)2sin(x-π/4)cos(x-π/4)=2-2cos2x

sinx的减区间是(2kπ+π/2,2kπ+3π/2)所以这里2kπ+π/2

f(x)=sin(π/3+4x)+sin(4x-π/6)=sin(π/3+4x)+cos(2x+π/3)=√2sin(4x+7π/12)=√2cos(4x+π/12)最小正周期T=2π/4=π/22k

解题思路：计算解题过程：最终答案：略

sin(x+π/4)=sin(x-π/4+π/2)=cos(x-π/4)所以原式=cos(2x-π/3)+2sin(x-π/4)cos(x-π/4)=cos(2x-π/3)+sin(2x-π/2)=1

这是一个正弦函数,当其函数值为1或-1时所对应的X值即为对称轴的方程.令x+4π/3=π/2+π.Z其中Z为任意整数.解出X即为对称轴方程.即X=π.Z-1/4π任意给一个整数Z都可以得到了一条对称轴