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Homework Help: Inverse Trigonometric Function Problems

  1. Apr 5, 2005 #1
    The math book I have does a pretty terrible job explaining this to me, because I am absolutely stumped as to why I get every question wrong in two sections: finding values of each expression in radians (can often be given in terms of ?) and finding approximate/exact values of the expressions.

    Here is a problem where I have to find values for each expression in radians:

    Sin^-1(1/2) The answer is ?/6, but as to how they got this answer I am completely stumped. I don't think I got one problem right in the section that I had to use no calculator on.

    The next ones are worse though, because even though I can do actual calculation I still get them wrong. The problem is tan(Cos^-1(12/13)) and I am supposed to find the value of the expressions:

    I did Cos^-1 which is hypotenuse/adjacent: attachment.php?attachmentid=2995&stc=1.jpg

    Then took tan, which is x/y, so I would get 13/15, but the answer is 5/12!

    Anyways, any help would be appreciated.

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  3. Apr 5, 2005 #2


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    [tex]\cos^{-1} x \neq \frac{1}{\cos x}[/tex]

    You need to review what it means to be an inverse trig function.

    (P.S. that triangle in your image cannot possibly be right)
    Last edited: Apr 5, 2005
  4. Apr 5, 2005 #3
    When your book uses the symbol ^-1 after a function, it means the inverse function, not the reciprocal of the function.
    Cos-1(x) is a function whose argument, x, is the cosine of some angle. The inverse cosine function, also called arccosine, returns the principal angle (the angle between -[tex]\pi[/tex] and [tex]\pi[/tex]) whose cosine is x. In other words, Cos-1(cos(t)) = t. The appropriately similar equations hold for the other inverse or arctrigonometric functions.
    In response to Sin-1(1/2), you are looking for the angle t such that sin(t) = 1/2. Drawing a triangle and letting the opposing leg (from t) have length 1, you see that the hypotenuse then has length 2 and the adjacent side has length [tex]\sqrt{3}[/tex]. This is a common 30-60-90 triangle, and the angle t is then 30 degrees, or [tex]\frac{\pi}{6}[/tex] radians.
    Last edited: Apr 5, 2005
  5. Apr 5, 2005 #4


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    Read the previous post (#3) really carefully and then attempt to solve the equation

    [tex] \sin x=\frac{1}{2} [/tex]

  6. Apr 5, 2005 #5
    you should have memorized the sine and cos function for 0, pi, pi/3, pi/4, and pi/6, and the corresponding angles in the other quadrents then just work backwards. For the other ones that don't correspond to those radian messures just use your calculator, almost all calculators have an arcsine or inverse sine button. And the posts above are also quite correct.
  7. Apr 5, 2005 #6
    Alright. You obviously know about the standard trigonometric function, [itex]\sin \theta, \; \cos \theta[/itex] and [itex]\tan \theta[/itex]. The [itex]\theta[/itex] in these expressions is an angle. You are undoubtedly familiar with angles from geometry. On the ortherhand, the value of the functions is not and angle, it is a ratio of lengths.

    The inverse trigonometric functions, [itex]\sin^{-1} x, \; \cos^{-1} x,[/itex] and [itex] \tan^{-1} x[/itex] work in a somewhat opposite manner. Here [itex]x[/itex] should be a ratio of lengths, and the value of each of these functions is an angle. They are such that

    [tex]\sin^{-1} (\sin x) = \sin \left(\sin^{-1} x\right) = x[/tex]


    [tex]\cos^{-1} (\cos x) = \cos \left(\cos^{-1} x\right) = x[/tex]


    [tex]\tan^{-1} (\tan x) = \tan \left(\tan^{-1} x\right) = x.[/tex]

    Now, you have also probably learned that, for example,

    [tex]\sin 0 = 0, \; \sin \frac{\pi}{2} = 1, \; \sin \frac{\pi}{6} = \frac{1}{2}, \; . \ . \ .[/tex]

    and similar identities for [itex]\cos[/itex] and [itex]\tan[/itex].

    Let's consider your first question, to find [itex]\sin^{-1} 1/2[/itex]. From what I wrote above, though, you see that

    [tex] \sin \frac{\pi}{6} = \frac{1}{2}[/tex]

    so we can rewrite

    [tex]\sin^{-1} \left(\frac{1}{2}\right) = \sin^{-1} \left( \sin \left( \frac{\pi}{6} \right) \right)[/tex]

    but from the rule that I mentioned above,

    [tex] \sin^{-1} ( \sin x ) = x,[/tex]

    so we get

    [tex] \sin^{-1} \left(\frac{1}{2}\right) = \sin^{-1} \left( \sin \left( \frac{\pi}{6} \right) \right)[/tex]

    [tex] = \frac{\pi}{6}.[/tex]

    You should try other questions in a similar way on your own.

    Now, let's look at your second question. You want

    [tex] \tan \left( \cos^{-1} \left( \frac{12}{13}\right) \right).[/tex]

    Your original intuition, to use the geometry of the triangle, was correct, but you made some mistakes. The way to approach such a question is as follows.

    Remember that, as I said above, [itex]\cos^{-1}[/itex] returns a angle. It serves us to consider it to be an angle in a right triangle, for this question.

    You should probably draw the triangle at this point. One of its angles is [itex]\pi /2[/itex] and another is [itex]\cos^{-1} (12/13)[/itex]. What does this second angle mean, though?

    Well, if [itex] \cos \theta = x[/itex], where [itex]\theta[/itex] is the angle, then as I mentioned earlier,

    [tex] \cos^{-1} (\cos \theta) = \theta[/tex]

    so, if the angle is [itex]\theta = \cos^{-1} (12/13)[/itex], as it is in our question, then we can immediately say that

    [tex] \cos \theta = \frac{12}{13}[/tex]

    ! But thinking about this a little more solves our question: remember that [itex]\cos[/itex] is the ratio of the lengths of the adjacent side and the hypotenuse of the triangle, ie.

    [tex] \cos \theta = \frac{\mbox{adjacent}}{\mbox{hypotenuse}} = \frac{12}{13}.[/tex]

    Then in our case, we can interpret the hypotenuse length to be [itex]13[/itex], and the adjacent length to be [itex]12[/itex]. What is the length of the third side of the triangle? Well, we can use the familiar pythagorean theorem to say that it is

    [tex] \mbox{opposite} = \sqrt{ \mbox{hypotenuse}^2 - \mbox{adjacent}^2} = \sqrt{13^2 - 12^2} = \sqrt{169 - 144} = \sqrt{25} = 5.[/tex]

    What we want, then is the tangent of the angle, ie. [itex]\tan \theta[/itex]. But I'm sure you remember that

    [tex] \tan \theta = \frac{ \mbox{opposite}}{\mbox{adjacent}}.[/tex]

    In our case, we have already found [itex] \mbox{opposite} = 5[/itex] and [itex]\mbox{adjacent} = 12}[/itex], so this is easy:

    [tex] \tan \theta = \frac{5}{12}.[/tex]

    Recalling that [itex]\theta[/itex] is just [itex]\cos^{-1} (12/13)[/itex], we are finished, since we have found

    [tex] \tan \left(\cos^{-1}\left(\frac{12}{13}\right)\right) = \tan \theta = \frac{5}{12}.[/tex]
  8. Apr 5, 2005 #7
    How do you guys get those math equations to be grouped as one whole item? Do you not type them out?

    P.S. thanks for the answers, I'm trying to work it out now.
  9. Apr 5, 2005 #8
    Double-click on math notation to see the code to generate it in a popup window.
  10. Apr 5, 2005 #9


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    Data,though your post is quite big,nowhere does it include the essential fact that the functions [itex] \sin,\cos [/itex] are invertible only on a restricted domain,namely [itex] \left[-\frac{\pi}{2},+\frac{\pi}{2}\right] [/itex] for sine and [itex] [0,\pi] [/itex] for cosine...

    Without that having been said,all those [itex] \sin^{-1}\left(\sin \left(\mbox{blah,blah,blah}\right)\right) [/itex] et al.make no sense...


    P.S.We use Latex code.
  11. Apr 5, 2005 #10
    Indeed. But I get the impression that he doesn't know about the periodic natures of the functions yet, so I don't want to confuse him. It doesn't even seem like he's really figured out what invertibility is yet. If I tell him that now, it'll probably just be random babbling for him to ignore :wink:
    Last edited: Apr 5, 2005
  12. Apr 5, 2005 #11
    They are using this forum's native TeX interpreter. You can see exactly what they typed to get the graphics they did by clicking on the graphic you're interested in. Note the important "tex" tags surrounding each one.
  13. Apr 5, 2005 #12


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    Wow, that's very nice, Data. Kudos for writing all that out.
  14. Apr 5, 2005 #13
    Better to explain it once clearly than several times unclearly :cool:
  15. Apr 5, 2005 #14
    I did another problem to make sure I understood what you guys said:

    [tex] \tan \left(\cos^{-1} \left(\frac{-12}{-13}\right)\right)[/tex]

    (how does the coding work so that I can just say it's a negative fraction?)

    Anyways, I had left[tex] \cos^{-1} \left(\frac{-12}{-13}\right) [/tex]

    I got [tex] -13^2 [/tex] - [tex] -12^2 [/tex], and I got [tex] 5 [/tex]

    So [tex] \tan [/tex] of that would be [tex] \frac{-5}{-12}.[/tex]

    Meaning the answer would be [tex] \tan \left(\cos^{-1} \left(\frac{-12}{-13}\right)\right) = \frac{-5}{-12} [/tex]

    Is this correct?
  16. Apr 5, 2005 #15
    Oh, another question, if I may be so bold!

    Is there anyway to figure out [tex] \sin^{-1}\frac{1}{2}[/tex] without memorizing tables?

    I know now the answer is [tex] \frac{\pi}{6}[/tex], but what exactly makes it that number? Is it just something I have to know?

    Pre-calc is not my best subject, but I want to get better at it!

    Thanks to everyone for their help; and thanks to Data for his ├╝ber explanation!
  17. Apr 5, 2005 #16
    Do you actually mean

    [tex] \cos^{-1} \left(-\frac{12}{13}\right)[/tex]


    in that case, the opposite side length is again [itex]5[/itex] and the answer is

  18. Apr 5, 2005 #17
    Yes, that is what I meant, I just couldn't figure out how to get that negative symbol where I wanted it to be; but now I know. Hurray I did it right!

    My main problem was I really didn't know what an inverse function was. I had it all backwards, as Hurkyl explained to me.
  19. Apr 5, 2005 #18
    Your just have to "think in reverse." An inverse trig function is always an angle. For Sin^-1(1/2), figure out what angle is there such that its sine is 1/2.
  20. Apr 5, 2005 #19
    As to your question about memorizing things, you do not actually have to memorize too much. The only trigonometric ratios that you will likely be expected to know are the "nice" ones (like the ones I mentioned, and a few more, such as [itex]\cos \pi / 4 = \sin \pi / 4 = 1 / \sqrt{2}[/itex]).
  21. Apr 5, 2005 #20
    Yes, but wouldn't I have to memorize a chart in order to know at what angle sine is [tex]\frac{1}{2}[/tex]?
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