f(x)=1-2sin^2(x+兀/8)+2sin(x+兀/8)cos(x+兀/8) 求f(x)的最小正周期第2问:f(x)的单调增区间
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/30 04:22:31
![f(x)=1-2sin^2(x+兀/8)+2sin(x+兀/8)cos(x+兀/8) 求f(x)的最小正周期第2问:f(x)的单调增区间](/uploads/image/z/15115493-29-3.jpg?t=f%28x%29%3D1-2sin%5E2%28x%2B%E5%85%80%2F8%29%2B2sin%28x%2B%E5%85%80%2F8%29cos%28x%2B%E5%85%80%2F8%29+%E6%B1%82f%28x%29%E7%9A%84%E6%9C%80%E5%B0%8F%E6%AD%A3%E5%91%A8%E6%9C%9F%E7%AC%AC2%E9%97%AE%3Af%28x%29%E7%9A%84%E5%8D%95%E8%B0%83%E5%A2%9E%E5%8C%BA%E9%97%B4)
f(x)=1-2sin^2(x+兀/8)+2sin(x+兀/8)cos(x+兀/8) 求f(x)的最小正周期第2问:f(x)的单调增区间
f(x)=1-2sin^2(x+兀/8)+2sin(x+兀/8)cos(x+兀/8) 求f(x)的最小正周期
第2问:f(x)的单调增区间
f(x)=1-2sin^2(x+兀/8)+2sin(x+兀/8)cos(x+兀/8) 求f(x)的最小正周期第2问:f(x)的单调增区间
1-2sin^2(x+兀/8)=cos(2x+兀/4)
2sin(x+兀/8)cos(x+兀/8)=sin(2x+兀/4)
f(x)=cos(2x+兀/4)+sin(2x+兀/4)
=√2*[sin兀/4*cos(2x+兀/4)+cos兀/4*sin(2x+兀/4)]
=√2*sin(2x+兀/2)
T=2兀/2=兀
f(x)的单调增区间:
-兀/2/+2k兀≤2x+兀/2≤兀/2+2k兀
-兀/2/8+K兀≤x≤0+k兀
k属于Z
f(sin 2/x )=1+cosx 求f(x)
傅里叶级数作图f(x)=2sin[x] - sin[2x] + 2/3sin[3x] - 1/2sin[4x]我用mathematica输入程序Plot[{2sin[x],-2sin[x],2sin[x] - sin[2x],-2sin[x] + sin[2x],2sin[x] - sin[2x] + 2/3sin[3x],-2sin[x] + sin[2x] - 2/3sin[3x],2sin[x] - sin[2x] + 2/3si
f(x)=1-2sin^2(x+π/8)+2sin(x+π/8)cos(x+π/8)
f(sin^2x)=x/sinx 求f(x)
f(x)={(1-e^x)/sin(x/2) ,x>0 ;ae^2x,x
f(x)=1-2sin^2(x+兀/8)+2sin(x+兀/8)cos(x+兀/8) 求f(x)的最小正周期第2问:f(x)的单调增区间
设函数f(x)=sin(2x+∮)(-兀
f(x)=(1+cotx)sin^2x-2sin(x+π/4)sin(x-π/4)还有一题,
已知函数f(x)=(1+1 anx)sin^2x+m sin(x+π/4)sin(x-π/4)
f(x)=sin^2x+sinxcosx-1/2,化简
f(x)=(1+1/tanx)sin(x)^2化简
化简f(x)=(1+cos2x)sin^2x
f(x)=sin(πx^2)(-1
已知函数f(x)=cos(x-3/ 兀)-sin(2/兀-x).(1)求函数f(x)的最小值.
已知函数f(x)=-1/2+sin(5/2x)/2sin(x/2)(0
若f(x)=sin πx/6,则f(1)+f(2)+...+f(102)=?
若f(x)=sin(π/4)x,求f(1)+f(2)+.+f(2010)
已知f(sin-1)=cos2x+2,求f(x)