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Trigonometric and inverse trigonometric equations

  1. May 14, 2014 #1
    Given a trig equation, like: sin(x)² + cos(x)² = 1² or sin(x) = 1/csc(x), exist a correspondent inverse: arcsin(x) + arccos(x) = π/2 and arcsin(x) = arccsc(1/x), respectively. Thus, given an any trigonometric equation, how find its correspondent inverse?
     
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  3. May 14, 2014 #2

    Simon Bridge

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    Use geometry.

    i.e. in a right angle triangle with unit hypotenuse and x=adjacent side, then
    ... ##x=\sin\theta=\cos(\frac{\pi}{2}-\theta)##

    Thus ##\arcsin(x)=\theta## and ##\arccos(x)=\frac{\pi}{2}-\theta## so ##\arcsin(x)+\arccos(x)=\frac{\pi}{2}##

    But notice that this has nothing much to do with the identity: ##\sin^2x + \cos^2x =1## i.e. the "x" in that relation refers to something different... the only "correspondence" is that the expression have a "cos" and a "sin" and a "+" in them. Do not confuse correspondences in labels for correspondences in maths.
     
  4. May 15, 2014 #3
    Would you find the analogous inverse for all equations below using geometry?
    http://s8.postimg.org/82n2lmlg5/imagem.png [Broken]
     
    Last edited by a moderator: May 6, 2017
  5. May 15, 2014 #4

    Simon Bridge

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    Define analogous in this context.
     
  6. May 15, 2014 #5
    Look how a trigonometric equation have an analogous inverse trigonometric equation:
    img.png
     
  7. May 15, 2014 #6

    Simon Bridge

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    I don't see any rule that those have in common - except the general stuff that they have some elements in common. The notation has them make pretty patterns on the page but that is just numerology at best. The only way to identify a pleasing pattern is to try lots of things and look.

    Please provide the definition that you want to work to... how would you describe the analogous nature is such a way that another person can go look for it rather than rely on recognizing it after the fact?
     
  8. May 15, 2014 #7
    I think there is a genuine, though not precisely defined, duality here. For example, the dual of the identity ##2\sin^2 x = 1 - \cos 2x## is ##\arccos(1-2y^2) = 2\arcsin(\left|y\right|).##

    (Note: the absolute value comes into the picture because ##\arccos(\cos(x)) = \left|x\right|## -- at least in the range that we care about.)

    You can obtain these by simple algebra.

    EDIT: I should point out that not all of the identities have nice "inverses". For example, there isn't a nice inverse of ##\sin 2x = 2\sin x \cos x## as far as I can tell.
     
    Last edited: May 15, 2014
  9. May 15, 2014 #8

    Simon Bridge

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    Does that same duality apply to all the examples given?
     
  10. May 15, 2014 #9
    Well, I didn't check the ones involving the hyperbolic functions, but it is the same thing that is going on in the first and last sets in the above post.

    EDIT: Obviously it has nothing to do with the identity in the original post, though -- I don't think there's any clear relationship between ##\sin^2 x + \cos^2 x = 1## and ##\arcsin x + \arccos x = \pi/2##. I think the "dual" of the Pythagorean identity is more like ##\arccos(\sqrt{1-y^2}) = \arcsin(\sqrt{y^2})##.
     
    Last edited: May 15, 2014
  11. May 15, 2014 #10
    Actually, the correspondent of ##\arcsin(x)+\arccos(x)= \frac{\pi}{2}## is ##\sin(\theta) = \cos(\phi)## where ##\theta + \phi = \frac{\pi}{2}##
     
  12. May 15, 2014 #11

    Simon Bridge

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    Well if you two can thrash out a definition you can agree on, we may have something to talk about ;)

    My contention is that unless the "correspondence" or "dual" can be articulated from the start, then it will not be possible to find the relations asked for.

    The short answer to the question posed in post #1 is: you cannot, because the problem is too vaguely phrased.
     
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