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Normal and inverse trig functions cancel each other out

  1. Nov 9, 2004 #1
    I'm a bit confused at the moment. All my books say that normal and inverse trig.functions cancel eachother out like this

    [tex]\sin(\arcsin(x))=x[/tex] and [tex]\arcsin(sin(x))=x[/tex]

    But when I try this out on my calculator - TI-89 - it only wants to recognize the first equation as being equal to x. Is that true or...? :confused:
     
    Last edited: Nov 9, 2004
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  3. Nov 9, 2004 #2

    Galileo

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    For a function to have an inverse, it needs to be injective and surjective.
    The problem is that sin(x) is generally not injective, however it is injective on (eg) the interval [itex][-\pi/2,\pi/2][/itex]. This interval can be chosen as the domain of sin(x) and becomes the range of arcsin(x), so for x in this interval the second equation will hold as well.
     
    Last edited: Nov 10, 2004
  4. Nov 9, 2004 #3
    But since my calculator always accepts sin(arcsin(x)) as being equal to x, does that mean that this equation is always true no matter what? :confused:
     
  5. Nov 10, 2004 #4

    Galileo

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    That's funky. You cannot take the arcsin of x if [itex]|x|>1[/itex], since such an x wouldn't be in the range of sin(x).
    My calculator gives an error, as it should.
    (Unless you use complex numbers).
     
  6. Nov 10, 2004 #5

    krab

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    Think it through. Try x=360 degrees. sin(360 deg.)=0, but arcsin(0)=0. So
    arcsin(sin(x)) is not equal to x. OTOH, sin(arcsin(x))=x
     
  7. Nov 10, 2004 #6

    HallsofIvy

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    Another function that does not have a "true" inverse is f(x)= x2 and you might find it easier to look at that function (you can do the calculations yourself rather than use a calculator).

    If we restrict the domain to non-negative numbers, THEN we can use f-1(x)= √(x) as the inverse.

    √(x) only makes sense when x is positive. For any positive number, if you first take the square root and then square, you get x back again: f(f-1(x))= x.
    In particular, if x= 4, f-1(4)= √(4)= 2 and f(2)= 22= 4 again.

    However, f(x) is defined for x positive or negative. If x= 2, f(2)= 22= 4 and then f-1(4)= √(4)= 2. In that case, f-1[\sup](f(x))= x.

    But if x= -2, f(-2)= (-2)2= 4 and then f-1(4)= √(4)= 2.
    In that case, f-1(f(x)) is NOT equal to x.
     
  8. Nov 10, 2004 #7
    sin(arcsin(x))=x true while arcsin(sin(x))=x is not true.
     
  9. Nov 10, 2004 #8

    Galileo

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    sin(arcsin(x))=x does not make sense if |x|>1, since it is undefined.
    It's only true if |x|<=1.
     
  10. Nov 10, 2004 #9
    So how would I go about it to find the solutions for let's say tan(x/2)=0 ?
    I thought I could make it easier by using arctan(tan(x/2))=x/2, but you're telling me that this wouldn't work? What if I decided to look only at a part of tan(x/2) that is increasing or decreasing? Let's say from -pi to pi. Would this work? Because that's the requirement for a trig.function to have an inverse isn't it? Would arctan(tan(x/2))=x/2 be true then?
     
  11. Nov 10, 2004 #10

    Galileo

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    tan(x) has the same problem: It isn't injective.
    However, it IS injective if it is restricted to the interval [itex](-\frac{\pi}{2},\frac{\pi}{2})[/itex]
    Then for x in this interval [itex]\arctan(\tan(x))=x[/itex] holds.
    (Notice that range of arctan(x) is [itex](-\frac{\pi}{2},\frac{\pi}{2})[/itex])

    If you wish to solve [itex]\tan(x/2)=0[/itex], then using arctan will give you an answer that lies in the interval [itex](-\frac{\pi}{2},\frac{\pi}{2})[/itex]. (In this case x=0).
    For a complete answer, you need additional info.
    In this case you can use that the tangent is periodic (period [itex]\pi[/itex])
    so that [itex]x/2=k\pi[/itex] where k is an integer will give the general solution).
     
    Last edited: Nov 10, 2004
  12. Nov 10, 2004 #11
    Wouldn't period of tan(x/2) be 2pi? When the interval of tan(x) is restricted to <-pi/2 , pi/2>, the range of arctan(x) is the same. But when the interval of tan(x/2) is <-pi , pi>, but now the range of arctan(x/2) is not the same. Why is this? Shouldn't the range of arctan(x/2) the same as the interval of tan(x/2)?
     
  13. Nov 10, 2004 #12

    Galileo

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    Yes, the period of tan(x/2) is [itex]2\pi[/itex] (I meant the period of tan(x) in my post).

    If [itex]y=\tan(x/2)[/itex]. Then for [itex]x \in (-\pi,\pi)[/itex]:
    [tex]\arctan(y)=x/2[/tex]
    [tex]x=2\arctan(y)[/tex]
    So the inverse function of tan(x/2) is 2arctan(x) and has range [itex](-\pi,\pi)[/itex]
     
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