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Why are sinh and cosh named after sinusoidal functions?

  1. Jan 7, 2013 #1
    We learned about those functions last semester but they seemed to have to do nothing with sine and cosine? They were defined using the exponential function.
  2. jcsd
  3. Jan 7, 2013 #2


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    Actually, I would object to the phrase "sinusoidal functions"! The sine is just one of those functions. The correct term would be either "trigonometric functions" or "circular functions". In any case, there are a number of similarities between the circular functions (as I am going to call them) and hyperbolic functions. Here are three of them:

    1) The graph of [itex]x^2+ y^2= 1[/itex] is a circle with center at (0, 0) passing through (1, 0). A standard way of defining circular functions is: starting at (1, 0), measure a distance t counter-clockwise around the circumference of the circle (clock-wise if t is negative). Then, whatever the coordinates of the ending point, we define cos(t) to be the x-coordinate, sin(t) to be the y-coordinate.

    Similarly, we can graph the hyperbola [itex]x^2- y^2= 1[/itex] (which is where "hyperbolic" comes from), start at (1, 0) and measure a distance t on the graph. We define cosh(t) to be the x-coordinate of the endpoint and sinh(t) to be the y-coordinate.

    2) We often define the "hyperbolic functions" by [itex]cosh(x)= (e^x+ e^{-x})/2[/itex], [itex]sin(x)= (e^x- e^{-x})/2[/itex]. And, we can use, say Taylor's series for the exponential and sine and cosine functions to show that [itex]cos(x)= (e^{ix}+ e^{-ix})/2[/itex] and [itex]sin(x)= (e^{ix}- e^{-ix})/2i[/itex].

    3) We can show that y= sin(x) satisfies the differential equation y''= -y with initial conditions y(0)= 0, y'(0)= 1 and that y= cos(x) satisfies the same differential equation with y(0)= 1, y'(0)= 0.

    And we can show that y= sinh(x) satisfies the diferential equation y''= y with initial conditions y(0)= 0, y'(0)= 1 and that y= cosh(x) satisfies the same differential equation with y(0)= 0, y'(0)= 1.
    Last edited by a moderator: Jan 8, 2013
  4. Jan 8, 2013 #3
    Just to elaborate on Halls' comment,

    I've always looked at sinhx and coshx as similar to cosx and sinx in the way in which they are derived.

    (ex−e−x)/2 = sinhx

    (ex+e−x)/2 = coshx

    When we derive (ex−e−x)/2 we get d/dx(ex)/2 -d/dx(e−x)/2

    = 1/2(ex) - (-1/2(e−x) = (ex+e−x)/2

    Which is just simple differentiation and d/dx(Coshx) = -Sinhx, in this way Coshx and Sinhx follow the same cycling pattern of differentiation as coax and sins.
    Last edited by a moderator: Jan 8, 2013
  5. Jan 8, 2013 #4
    If you think of trig functions in terms of what they mean in the geometry of the unit circle and what the hyperbolic trig functions mean in terms of the geometry of the unit hyperbola you can see that they are exactly the same thing.
    The sinh(x) function gives the y-axis projection of a line segment from the origin to some point on the hyperbola. tanh(x) gives the length of the line segment from the x axis to the point where some line intersects the tangent to the tip of the hyperbola and so on.. The geometrical interpretation is the same for the circle.
    Last edited: Jan 8, 2013
  6. Jan 8, 2013 #5


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    cos(x) = cosh(ix)

    i sin(x) = sinh(ix) :wink:
  7. Jan 8, 2013 #6


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  8. Jan 8, 2013 #7


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    The derivative of cosh(x) is positive sinh(x).
  9. Jan 8, 2013 #8
    I find it amazing that there is even such a connection at all. In my mind, there's no reason that this should be so but it just is. I love math.
  10. Jan 8, 2013 #9
    I have a poor understanding of hyperbolic functions to begin with. It wasn't until I realized how similar a circle is to a hyperbola that I understood the answer to my question. I had to rearrange for y in x2+y2=1 and x2-y2=1 and do some graphing on wolfram-alpha to understand why the functions look as they do when they are graphed. In both equations, y is squared which means after rearranging for y, the right side would be under a radical with plus or minus sign.

    e.g.[tex]{ x }^{ 2 }-{ y }^{ 2 }=1\\ { y }^{ 2 }={ x }^{ 2 }-1\\ |y|=\sqrt { { x }^{ 2 }-1 } \\ y=\pm \sqrt { { x }^{ 2 }-1 }[/tex]

    Knowing that the minus square root of x2-1 is just the positive square root of x2-1 mirrored on the y-axis, I began to look at why there was nothing in between -1 and 1.


    Now if I were to plot what is inside the radical, you would see that it's a simple parabola which is under the y-axis in between -1 and 1. The square root essentially erases anything below the y-axis (because the square root of a negative number is imaginary).


    Do the same to x^2+y^2=1 and you will see that the circle and hyperbola are very similar.
  11. Jan 10, 2013 #10
    solve: z=Asin(2πft+α) where A=0.06,α=58degree
  12. Jan 10, 2013 #11
    Isn't this a physics problem for simple harmonic waves...?
  13. Jan 10, 2013 #12


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    welcome to pf!

    hi bluesky20! welcome to pf! :smile:

    for each new question, please start a new thread by going to one of the homework forums (index page), and clicking on the "NEW TOPIC" button near the top :wink:

    (also make sure that you've ticked the box that stops you getting logged-out and losing your work :redface:)
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