高一数学y=tan(x+π/4)+1/tan(x+π/4)y=tan(x+π/4)+1/tan(x+π/4)求该函数的值域及单调递增区间
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![高一数学y=tan(x+π/4)+1/tan(x+π/4)y=tan(x+π/4)+1/tan(x+π/4)求该函数的值域及单调递增区间](/uploads/image/z/3841116-60-6.jpg?t=%E9%AB%98%E4%B8%80%E6%95%B0%E5%AD%A6y%3Dtan%28x%2B%CF%80%2F4%29%2B1%2Ftan%28x%2B%CF%80%2F4%29y%3Dtan%28x%2B%CF%80%2F4%29%2B1%2Ftan%28x%2B%CF%80%2F4%29%E6%B1%82%E8%AF%A5%E5%87%BD%E6%95%B0%E7%9A%84%E5%80%BC%E5%9F%9F%E5%8F%8A%E5%8D%95%E8%B0%83%E9%80%92%E5%A2%9E%E5%8C%BA%E9%97%B4)
高一数学y=tan(x+π/4)+1/tan(x+π/4)y=tan(x+π/4)+1/tan(x+π/4)求该函数的值域及单调递增区间
高一数学y=tan(x+π/4)+1/tan(x+π/4)
y=tan(x+π/4)+1/tan(x+π/4)求该函数的值域及单调递增区间
高一数学y=tan(x+π/4)+1/tan(x+π/4)y=tan(x+π/4)+1/tan(x+π/4)求该函数的值域及单调递增区间
设tan(x+π/4)=t
则t属于(-∞,+∞)
当t=2
值域是(-∞,-2]并[2,+∞)
因为y=t+1/t在(-∞,-1)并(1,+∞)上是单调递增的
而tan(-π/4+kπ)=-1 tan(π/4+kπ)=1
所以单调递增区间是(kπ-3π/4,kπ-π/2)并(kπ,kπ+π/4),其中k属于Z(整数)
设tan(x+π/4)=t
y=t+1/t (t属于R
值域 负无穷——负2, & 2——正无穷
增区间 负无穷——负1 & 1——正无穷
直接化简.
y=(1+tanx)/(1-tanx)+(1-tanx)/(1+tanx)
=(2+2tan^2x)/(1-tan^2x)
..............
1+(tanα)^2=(secα)^2.....接着化简,时间关系,我得回去了,明天接着补充.
tan(x+π/4)+1/tan(x+π/4)
=sin(x+π/4)/cos(x+π/4)+cos(x+π/4)/sin(x+π/4)
=[sin²(x+π/4)+cos²(x+π/4)]/sin(x+π/4)cos(x+π/4)
=1/sin(x+π/4)cos(x+π/4)
=2/2sin(x+π/4)cos(x+π/4)
=2/s...
全部展开
tan(x+π/4)+1/tan(x+π/4)
=sin(x+π/4)/cos(x+π/4)+cos(x+π/4)/sin(x+π/4)
=[sin²(x+π/4)+cos²(x+π/4)]/sin(x+π/4)cos(x+π/4)
=1/sin(x+π/4)cos(x+π/4)
=2/2sin(x+π/4)cos(x+π/4)
=2/sin(2x+π/2)
=2/cos2x
令m=cos2x
则y=2/m
cos2x的值域为-1≤cos2x≤1
所以y=2/m 定义域为-1≤m<0或0<m≤1
对于反比例函数
当m∈(0,1]为减函数
当m∈[-1,0)为增函数
所以
y=2/m的值域为
y≥2或y≤-2
当m∈[-1,0) 函数y=tan(x+π/4)+1/tan(x+π/4)单调递增
即-1≤cos2x<0
解得kπ+π/4<x<kπ+3π/4 K为整数
收起
用WORD打出来截的图,不然太难打了。