
#1
Apr1312, 02:10 AM

P: 72

1. The problem statement, all variables and given/known data
i wanted to know ,why is the range of cot^{1}(x) (0,π), unlike tan^{1}(x) which has its range (π/2,π/2). because for sin and cosec inverse range is same, for cos and sec inverse range is same. 2. Relevant equations 3. The attempt at a solution 1. The problem statement, all variables and given/known data 2. Relevant equations 3. The attempt at a solution 



#2
Apr1312, 09:19 AM

HW Helper
P: 2,874

The range of a function (or more strictly, its image) is determined by the domain of the inverse function.
So think about the domain of cot(x) and tan(x). For what values of x is each of those functions "undefined"? What's tan(π/2) or tan(π/2)? Now ask yourself what's cot(0) or cot(π). You might find it helpful to look at a graph of the functions. 



#3
Apr1312, 03:17 PM

P: 72

domain of cot(x) is {real numbers}  {nπ;n is integer}.
cosec(x) has the same domain. to make these functions bijective,so that we can define inverses,we restrict the domain of cosec(x) to [π/2 , 0)[itex]\cup[/itex](0 , π/2]. we can do the same with cot(x), so that we get numerically smallest angle as the output of its inverse function. also ,by doing this many formulas will simplify (for eg. tan^{1}(x)=cot^{1}(1/x) ) with usual conventions ,the above formula has two forms ,one for positive x and one for negative x. 



#4
Apr1312, 06:56 PM

HW Helper
P: 2,874

range of cot inverse 



#5
Apr1412, 02:37 AM

P: 72

but what about cosec^{1}(x)?




#6
Apr1412, 04:23 AM

HW Helper
P: 2,874

[π/2 , 0)∪(0 , π/2]." I'll explain this very carefully, so you should have the understanding to apply it to other cases. The reciprocal of csc(x) is sin(x). Look at the graph of sin(x). Notice that it passes through the origin (0,0). Notice that a maximum is reached at π/2, and a minimum at π/2. If you consider an xvalue within (0, π/2), you will find exactly one other xvalue within (π/2, π) that has the same yvalue. This means that if you allow a domain of, say, (0, π), the function sin(x) ceases to be injective (and therefore also ceases to be bijective). For an inverse to exist, a function needs to be bijective. Hence the domain of sin(x) needs to be "clipped" on the positive side at π/2. Apply the same argument with the signs reversed for the negative side. Hence the domain of the restricted sine function becomes [π/2, π/2]. Note that I'm using square brackets here, because you *can* have values of x at the extreme ends, since the sine of those values is welldefined (1 and +1, respectively). Now consider cosec(x). It's clear that the same arguments for restricting its domain follow. With one addendum  0 must be excluded. Why? Because cosec(0) is undefined. The limit of cosec(x) as x > 0 from the righthand side (positive) is +∞, whereas the limit from the lefthand side (negative) is ∞. Hence the limit at x=0 of cosec(x) does not exist. Therefore, we must exclude 0 from the domain of cosec(x). A similar argument applies to all values of the form nπ, but since we're only considering the interval [π/2, π/2], we're only concerned with 0 here. Therefore, the final domain of csc(x) becomes [π/2, 0) U (0, π/2]. Notice the curvy brackets on one side of each disjoint subdomain. That means that 0 is excluded. The range (or image) of this restricted csc(x) function is (∞,1] U [1, ∞). The restricted cosec function is sometimes written with a capitalised first letter as Csc(x) to distinguish it (the same capitalisation applies to the restricted inverse). Finally, the domain and range of the inverse is determined by the range and domain, respectively, of the original function. Hence the domain of Csc^{1}(x) is (∞,1] U [1, ∞) and its range is [π/2, 0) U (0, π/2]. Now apply this same line of reasoning to all the other trig functions and their inverses. 



#7
Apr1412, 06:56 AM

P: 72

thanks for your help Curious3141.
cot function in (pi/2,pi/2) (0 excluded) is also bijective.so inverse could be defined with its range (pi/2,pi/2) (0 excluded). it should behave similar to cosecant. 



#8
Apr1412, 07:12 AM

P: 72

the attached graph shows
graph of cot^{1}(x) when i)domain of cot(x) is restricted to ( π/2,π/2) (the blue discontinuous curve) ii)domain of cot(x) is restricted to (0,π) (the continuous blue green curve) has discontinuity got something to do with the definition of this function? 



#9
Apr1412, 09:38 AM

HW Helper
P: 2,874

The corresponding range for the inverse cot can therefore be either: [ π/2,π/2]\{0} or (0,π) The domain for the inverse cot is (∞,∞) either way. It's just a matter of how you wish to define the range of the function prior to using it. Both are equally acceptable. But some prefer to have the function return values that are smallest numerically, so they use the first range. 



#10
Apr1412, 12:35 PM

P: 72

thanks for your help.now i got it.it is just a matter of convention.



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