Date: 06/27/98 at 00:28:09 From: yoshi Subject: Problem Hi, I was wondering: ex: sin(angle) = side of triangle arcsin(side of triangle) = angle but if the function is sinh or cosh, what does the inverse function give you? What does the inverse of sinh or cosh mean? ex: arcsin(side or number) = angle inverse sinh(some number)=?
Date: 06/27/98 at 14:31:12 From: Doctor Joe Subject: Re: Problem Dear Yoshi, You have asked an interesting question, to which I think I have the answer. 1. Why are sinh and cosh defined in terms of the exponential functions? 2. Why are sinh and cosh called the hyperbolic functions? Let's go back to the trigonometric functions: sin and cos. The trigonometric functions are also called the circular functions (this must be some early encounter with the mensuration problems in circular measure - radians etc.). Suppose we have a unit circle of equation: x^2 + y^2 = 1 Fix a point A(1,0) on the unit circle. Take any point P on the unit circle. Without loss of generality, the point P lies in the first quadrant. Consider the area swept out by the radius of the circle as the point moves from A to P. Let the area of the sector AOP be denoted by theta/2. (Why theta/2 will be apparent in a while.) What about the angle? Indeed, the angle AOP will be of size theta radians. The corresponding coordinates of the point P (in terms of theta) will be given by (cos theta, sin theta). Suppose we work things out in a similar fashion, but on a different conic section; this time on a "unit" hyperbola instead of the unit circle. More precisely, we start off with the curve H, given by: x^2 - y^2 = 1 Fix a point A, as before, on the x-axis. A = (1,0). Note that A lies on the curve H. Take any point P on the hyperbola, again in the first quadrant. We shall use the area swept out by the "radius" from A to P, i.e. the area bounded by the line OP, the hyperbola arc PA, and the line OA, to parameterize the coordinates of the variable point P. Let the area swept out be theta/2. We want to find the coordinates of P in terms of theta. Corresponding to the coordinate functions of a point on a unit circle (which are called the circular functions), the newfound functions will be termed the hyperbolic functions). The remaining development is a series of simple exercises for you. 1. The area swept out is theta/2. Our objective is to find the coordinates of P in terms of theta. So, for the first step, write the coordinates of P as (c(theta), s(theta)) in anticipation that the functions "resemble" those of sine and cosine in the circular case. 2. Rotate the whole picture (coordinate system) counterclockwise by 45 degrees. Note that the asymptotes of the hyperbola H now become upright and it becomes convenient to set these asypmtotes as the principal axes, called Y and X. In terms of Y and X, compute the equation of a. the old coordinate axes, i.e. the previous x and y axes under this rotation; b. the hyperbola, under this rotation. 3. Now, using integration, find the area of the bounded region. 4. Knowing that the area should remain invariant under rotation, equate the expression that you have found in (3) and deduce a relation between the two functions c(theta) and s(theta). 5. Since (c(theta),s(theta)) lies on H, write down another relation linking these two functions of theta. 6. Solve this system of simultaneous equations. You should be able to deduce the definitions of hyperbolic sines and cosines. You should now have some idea about the geometrical interpretation of the inverse hyperbolic functions sinh^(-1) and cosh^(-1): they are actually the bounded area as described. Hope this helps. - Doctor Joe, The Math Forum http://mathforum.org/dr.math/
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