Defining sinh(x) and cosh(x) with e and analytic Geometry
The standard definition of hyperbolic sine and cosine, written sinh and cosh, are as follows;
and
. Their graphs look like those in figure 1 and 2. We
can make sense of these graphs if we look at the functions end behavior. For example take
cosh(x), as x increases the value of
increases exponentially and the value of
decreases
exponentially. When x decreases, the value of
decreases and the value of
increases.
When x=0, the function is 1. The same can be said for sinh(x), as the value of x increases the
value of
increases. As the value of x decreases, the value of
of
increases, but the sign flips the graph over the x-axis.
decreases to 0 and the value
There is also a graphical definition for sinh and cosh just like there is for sin and cos. Remember
how sine and cosine can be used to describe a circle? Well sinh and cosh can be used to
describe a hyperbola. From now on I will refer to sine and cosine as circular trig functions. In
figure 3 you can see how any point on the unit circle
can be represented by cosine
and sine as a function of time or degree measure. In figure 4 you can see how cosh and sinh are
used to describe any point on a unit regular hyperbola, granted this only accounts for the positive
half. Note: cosh(x) is always positive in the graph in figure 1. Interesting fact: the area of the
hyperbolic sector (in red) is equal to half the measure of the angle.
Figure 4 The formulas for hyperbolic trig functions can also be derived from the graph of a hyperbola.
Let equal the hyperbolic angle between the x-axis and the point (x, y) on the unit regular
hyperbola
. Then
where A is the area of the hyperbolic sector. The area of
the hyperbolic sector can be determined via integration. Thus,
simply
, or more
.
Let
, so
. Also
t=x, t=arcsec(x). Making this substitution,
. Note: when t=1, u=0, and when
In order to solve the integral part of this equation we must use integration by parts. Our parts
will be, v=tan(u) and dw=sec(u)tan(u)du, so dv=sec2(u)du and w=sec(u). Thus our integral is
now,
Adding
to both sides of the above equation we obtain
. Substituting this in to our original equation
we get,
. Since
,
. This integral can be easily solved by a simple
substitution. Let
u=0, v=1, and when
. Note: when
. Thus,
, so
. Now we simply
solve for x in terms of the angle. So,
Dividing both sides by
we get,
, squaring both side yields,
, but
, so
. Using
.
we solve for the y
coordinate in terms of the angle alpha.
. The relation
between the hyperbolic angle and the (x, y) coordinates of a regular unit hyperbola can be
expressed as (cosh(a), sinh(a)).
Defining sinh(x) and cosh(x) with e and analytic Geometry
The standard definition of hyperbolic sine and cosine, written sinh and cosh, are as follows;
and
. Their graphs look like those in figure 1 and 2. We
can make sense of these graphs if we look at the functions end behavior. For example take
cosh(x), as x increases the value of
increases exponentially and the value of
decreases
exponentially. When x decreases, the value of
decreases and the value of
increases.
When x=0, the function is 1. The same can be said for sinh(x), as the value of x increases the
value of
increases. As the value of x decreases, the value of
of
increases, but the sign flips the graph over the x-axis.
decreases to 0 and the value
There is also a graphical definition for sinh and cosh just like there is for sin and cos. Remember
how sine and cosine can be used to describe a circle? Well sinh and cosh can be used to
describe a hyperbola. From now on I will refer to sine and cosine as circular trig functions. In
figure 3 you can see how any point on the unit circle
can be represented by cosine
and sine as a function of time or degree measure. In figure 4 you can see how cosh and sinh are
used to describe any point on a unit regular hyperbola, granted this only accounts for the positive
half. Note: cosh(x) is always positive in the graph in figure 1. Interesting fact: the area of the
hyperbolic sector (in red) is equal to half the measure of the angle.
Figure 4 The formulas for hyperbolic trig functions can also be derived from the graph of a hyperbola.
Let equal the hyperbolic angle between the x-axis and the point (x, y) on the unit regular
hyperbola
. Then
where A is the area of the hyperbolic sector. The area of
the hyperbolic sector can be determined via integration. Thus,
simply
, or more
.
Let
, so
. Also
t=x, t=arcsec(x). Making this substitution,
. Note: when t=1, u=0, and when
In order to solve the integral part of this equation we must use integration by parts. Our parts
will be, v=tan(u) and dw=sec(u)tan(u)du, so dv=sec2(u)du and w=sec(u). Thus our integral is
now,
Adding
to both sides of the above equation we obtain
. Substituting this in to our original equation
we get,
. Since
,
. This integral can be easily solved by a simple
substitution. Let
u=0, v=1, and when
. Note: when
. Thus,
, so
. Now we simply
solve for x in terms of the angle. So,
Dividing both sides by
we get,
, squaring both side yields,
, but
, so
. Using
.
we solve for the y
coordinate in terms of the angle alpha.
. The relation
between the hyperbolic angle and the (x, y) coordinates of a regular unit hyperbola can be
expressed as (cosh(a), sinh(a)).
Application for sinh(x) and cosh(x) Uses for hyperbolic trig functions can be found throughout the field of engineering. Mainly when solving second order ordinary differential equations. The cosh function can be used to trace out a particular curve known as a catenary curve. Simply hanging a string from two equally high points can form the catenary curve. The curve created by gravity pulling down on the string is described by the cosh function. This curve has only to support its own weight, thus it is in complete tension, when inverted the curve is complete compression because it supports its weight only. One of the best examples of hyperbolic trig functions being used in the real world is in architecture. A common misconception is that a parabola can be used to construct an arch. If we look at the function for a parabola we see that the slope at any point is given by 2x and is changing linearly. On the other hand if we look at the function cosh(x), we see that the slope at any point is given by . Thus the slope of a catenary curve is changing exponentially. I.e., the legs of an inverted catenary curve will be straighter at the base of the arch compared to an inverted parabola, giving the structure more horizontal support. Determining the equation of the Saint Louis Gateway Arch. Using the general form we can fit the curve to roughly the mean of the outer and inner arch. We can adjust for the height and the fact the arch is an inverted curve by multiplying by ‐1 and adding 625. , where the height of the arch, the maximum cross sectional area at the base, the minimum cross sectional area at the apex. A effects the rate of openness of the curve and the height of the apex. Let . The inverse cosh function takes the horizontal distance from the apex to points on the curve, so it makes sense to adjust the openness by this factor. The openness will also be affected by the width of the base from center to either leg . Thus our formula for the curve is .