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Two Inverse Trig Graphs

  1. Jan 31, 2012 #1
    Hey everyone. Sorry to post another topic, but i thought this would be easy to find on google, to help me with , but i cant find it.

    All im really lookin for is someone to let me know if my answers are right, and if not, then how i can fix them :)

    Anyways, i was asked to graph y=sin(arsinc)
    So i graphed the line y=x with the doman as -1 to 1 and the range as -pi/2 to pi/2

    Then i neded to graph y=arsin(sinx)
    So i graphed the line y=x with the domain as -pi/2 to pi/2 and the range as -1 to 1

    This is just wat i came up with based on my profs notes, but im not sure if they are right, and are just kinda guesses i came up with on what he wrote. I just tried a few things and thought this made sense. haha

    if someone could let me know if this is right, i'd greatly appreciate it !
     
  2. jcsd
  3. Feb 1, 2012 #2

    Char. Limit

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    I think you might need to switch them. I know that sin(f(x)) doesn't have a range from -pi/2 to pi/2 no matter WHAT f(x) is. But I think your domain and range in 1 is correct for 2, and your domain and range in 2 is correct for 1.
     
  4. Feb 1, 2012 #3

    Simon Bridge

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    Hang on, you have to graph:[tex]y=\sin ( \text{sinc}^{-1}x ) \qquad : \qquad \text{sinc}(x)=\frac{\sin(x)}{x}[/tex]... have I got that right?

    https://www.physicsforums.com/showthread.php?t=282941
    It is not clear what your strategy entails besides plotting the line y=x.

    restricting your domain etc, wouldn't you want to plot y=sinc(x) then reflect it in y=x to get the inverse function - then take the sine of that? You are doing this numerically I hope?
     
  5. Feb 1, 2012 #4

    Char. Limit

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    ... have I got that right?

    https://www.physicsforums.com/showthread.php?t=282941
    It is not clear what your strategy entails besides plotting the line y=x.

    restricting your domain as you did, wouldn't you want to plot y=sinc(x) then reflect it in y=x to get the inverse function - then take the sine of that? You are doing this numerically I hope?[/QUOTE]

    I think he meant arsinx, or arcsin(x). But I'm not entirely certain myself.
     
  6. Feb 1, 2012 #5
    oh holy eff, i clearly cant spell. sorry about that!
    yeah i was asked to graph

    sin(arcsinx)
    and
    arcsin(sinx)

    I just have no ideas what to label the x and y-axis for each with :(
     
  7. Feb 1, 2012 #6

    SammyS

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    I think you have a typo in your first function.

    Should it be y = sin(arcsin(x)) ?

    The domain & range you give for this are actually only for the arcsine function, although this is also the correct domain for this composite function. As for the range: What is the result of putting all possible values from -π/2 to π/2 into the sine function?

    As for the second function, arcsin(sin(x)) :
    What is the domain of sin(x) ?

    Are any of the values from the range of sin(x) which are not in the domain of arcsin(x)?​

    Answering those questions should get you started.
     
  8. Feb 1, 2012 #7

    Simon Bridge

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    Oh gotcha... that's easier.
     
  9. Feb 1, 2012 #8
    Thanks Sammy! So for the first function y=sin(arcsin(x)) by putting any value from -pi/2 to pi/2 into sinx, you get values from -1 to 1. So does this mean that the range of this function is also -1 to 1, just like the domain?

    For the second one im a little omfused. So the domain of sinx when you restrict it so that you can graph just arcsinx, is from -pi/2 to pi/2 correct?
    And for your second tip, you mean the range of sin being -1 to 1, are any of those values not in the domain of arcsin? So if the domain of just arcsin is -1 to 1, then wouldnt all of those values be in it??

    AGGHHH i hattteee trig... hahah and i've never understood how to work with graphs of composite functions :/
     
    Last edited: Feb 1, 2012
  10. Feb 1, 2012 #9

    SammyS

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    But even if you don't restrict the domain of sin(x), (and in this case there is no reason to restrict it) the range of sin(x) is [-1, 1].

    Think about x going from -π/2 to π/2. Then sin(x) goes from -1 to 1. So arcsin(sin(x)) goes from -π/2 to π/2. Correct?

    Now the strange part begins...

    Let x go from π/2 to 3π/2. What does sin(x) do? It goes from 1 to -1. But, all that the arcsin function "knows" is that it's being fed values from 1 to -1. It doesn't know where these values came from to start with ... So arcsin(sin(x)) goes from  ?  to  ?  .
     
  11. Feb 1, 2012 #10
    Ugh i dont know why im finding this so hard.. :/

    so for y=sin(arcsinx) the domain and range are both -1 to -1.
    is there a general rule to work with when dealing with composite trigonometric functions and their Domains and Ranges?

    K so for y=arcsin(sinx) where you gave me the fill in the blanks. arcsin-1 = -pi/2 and arcsin 1 = pi/2 so therefore the range of this function would go from -pi/2 to pi/2? and the domain would then be -1 to 1? Sorry if i completely misread what you were trying to explain, im just having the hardest time grasping this :/
     
  12. Feb 1, 2012 #11
    oh wait wait.
    i think i got it. after reading your posts again>
    so the D and R of sin(arcsinx) are both -1 to 1
    and the D and R of arcsin(sinx) are both -pi/2 to pi/2
    is this correct?
     
  13. Feb 1, 2012 #12

    SammyS

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    Yes to this
    No to the domain of arcsin(sin(x)), but yes to the range.

    What is sin(11π/4), for instance? This is defined, correct?

    It's (√2)/2.

    What is arcsin((√2)/2) ? It's π/4 .

    sin(x) is defined for all real numbers.
     
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