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Topos Uranos

Summary

This lesson studies the hyperbolic cosine, sine and tangent, three functions obtained by combining the natural exponential exe^x with its reciprocal e−xe^{-x}. It determines their parity, their sign, their monotonicity and their range, and proves the identity cosh⁡2x−sinh⁡2x=1\cosh^2 x - \sinh^2 x = 1, according to which the point (cosh⁡t,sinh⁡t)\left( \cosh t, \sinh t \right) traverses the right branch of the hyperbola x2−y2=1x^2 - y^2 = 1 and passes exactly once through each of its points, which is where their name comes from. It then derives the addition and double-argument formulas and obtains the inverse functions, which are expressed by means of logarithms. Finally, it describes the catenary, which is the shape of a hanging chain, and the area of the hyperbolic sector, which announces the analogy with the circular functions.

Learning objectives

  1. Define the hyperbolic cosine, sine and tangent from the natural exponential, and prove their parity, their sign, their monotonicity and their bounds.
  2. Prove the fundamental identity and use it to parametrize the branches of hyperbolas, recognizing that each point of a branch corresponds to a unique parameter.
  3. Derive the addition and double-argument formulas, and apply them to proving identities and inequalities.
  4. Obtain the inverse hyperbolic functions with their domains, by solving quadratic equations in eye^y, and use them to solve hyperbolic equations.

Definition and first properties

In the lesson on the number e and the natural exponential, the function exe^x was defined, positive and strictly increasing on the whole real line. Its reciprocal e−x=1exe^{-x} = \frac{1}{e^x} is also positive, but strictly decreasing, and its graph is the reflection of that of exe^x in the yy-axis, according to the lesson on function operations. Half the sum and half the difference of these two functions, and their quotient, appear so often in geometry, in physics and in calculus that they receive names of their own.

For every real number xx, the hyperbolic cosine, the hyperbolic sine and the hyperbolic tangent of xx are defined as half the sum of exe^x and e−xe^{-x}, half their difference, and the quotient of the second by the first, respectively.
cosh⁡x=ex+e−x2,sinh⁡x=ex−e−x2,tanh⁡x=sinh⁡xcosh⁡x=ex−e−xex+e−x\cosh x = \frac{e^x + e^{-x}}{2}, \qquad \sinh x = \frac{e^x - e^{-x}}{2}, \qquad \tanh x = \frac{\sinh x}{\cosh x} = \frac{e^x - e^{-x}}{e^x + e^{-x}}

The three functions are defined on all of R\mathbb{R}: the denominator of the tangent is a sum of two positive numbers and never vanishes. At x=0x = 0 we have e0=1e^0 = 1, so that cosh⁡0=1\cosh 0 = 1, sinh⁡0=0\sinh 0 = 0 and tanh⁡0=0\tanh 0 = 0. The adjective “hyperbolic” will be justified in the third section, where it will be seen that these functions describe a hyperbola.

LemmaParity and decomposition of the exponential

The hyperbolic cosine is an even function, and the hyperbolic sine and tangent are odd functions. Moreover, for every real xx we have ex=cosh⁡x+sinh⁡xe^x = \cosh x + \sinh x and e−x=cosh⁡x−sinh⁡xe^{-x} = \cosh x - \sinh x.

When xx is replaced by −x-x, the functions exe^x and e−xe^{-x} exchange their roles. The figure shows the graphs of the hyperbolic cosine and sine with two points of opposite abscissas on each.

Proof

  1. cosh⁡(−x)=e−x+ex2=cosh⁡x\cosh(-x) = \frac{\resaltar{e^{-x} + e^{x}}}{2} = \cosh x

    When xx is replaced by −x-x, half the sum does not change, because its two terms are swapped: the hyperbolic cosine is even, and its graph is symmetric about the yy-axis.

  2. sinh⁡(−x)=e−x−ex2=−sinh⁡x\sinh(-x) = \frac{\resaltar{e^{-x} - e^{x}}}{2} = -\sinh x

    Half the difference, on the other hand, changes sign: the hyperbolic sine is odd, and its graph is symmetric about the origin.

  3. tanh⁡(−x)=−sinh⁡xcosh⁡x=−tanh⁡x\tanh(-x) = \frac{\resaltar{-\sinh x}}{\resaltar{\cosh x}} = -\tanh x

    The tangent, the quotient of an odd function by an even one, is odd.

  4. cosh⁡x+sinh⁡x=ex,cosh⁡x−sinh⁡x=e−x\cosh x + \sinh x = \resaltar{e^x}, \qquad \cosh x - \sinh x = \resaltar{e^{-x}}

    Adding and subtracting the definitions of the hyperbolic cosine and sine recovers the two exponentials: cosh⁡x\cosh x and sinh⁡x\sinh x are, respectively, the even part and the odd part of the exponential.

ExampleThe graphs of the hyperbolic cosine and sine

The graphs of cosh⁡\cosh and of sinh⁡\sinh are built by adding and subtracting, abscissa by abscissa, the ordinates of ex2\frac{e^x}{2} and of e−x2\frac{e^{-x}}{2}.

Proof

  1. y=ex2,e02=12y = \frac{e^x}{2}, \quad \frac{e^0}{2} = \frac{1}{2}

    The graph of ex2\frac{e^x}{2} is obtained from that of exe^x by compressing it vertically by half: it crosses the yy-axis at 12\frac{1}{2}, grows without bound to the right and approaches the xx-axis to the left.

  2. y=e−x2y = \frac{e^{-x}}{2}

    The graph of e−x2\frac{e^{-x}}{2} is the reflection of the first one in the yy-axis.

  3. cosh⁡0=12+12=1\cosh 0 = \resaltar{\frac{1}{2} + \frac{1}{2}} = \resaltar{1}

    At each abscissa, cosh⁡x\cosh x is the sum of the two ordinates. At 00 both equal 12\frac{1}{2}: the graph passes through (0,1)(0, 1), and it is symmetric about the yy-axis, because the hyperbolic cosine is even.

  4. cosh⁡x−ex2=e−x2>0\cosh x - \frac{e^x}{2} = \frac{e^{-x}}{2} > 0

    The hyperbolic cosine exceeds ex2\frac{e^x}{2} by exactly e−x2\frac{e^{-x}}{2}, a positive quantity that becomes as small as desired as xx grows: to the right, the graph of cosh⁡\cosh approaches that of ex2\frac{e^x}{2} from above, and to the left, by symmetry, that of e−x2\frac{e^{-x}}{2}.

  5. sinh⁡0=12−12=0\sinh 0 = \resaltar{\frac{1}{2} - \frac{1}{2}} = \resaltar{0}

    The hyperbolic sine is the difference of the ordinates. It passes through the origin and its graph is symmetric about the origin, because the hyperbolic sine is odd.

  6. cosh⁡x−sinh⁡x=e−x>0\cosh x - \sinh x = e^{-x} > 0

    To the right, the graph of sinh⁡\sinh approaches that of ex2\frac{e^x}{2} from below, while that of cosh⁡\cosh does so from above: the two get closer to each other without ever touching, because their difference is positive.

The sign of the hyperbolic sine can be read off its definition: sinh⁡x>0\sinh x > 0 is equivalent to ex>e−xe^x > e^{-x}, and since the exponential is strictly increasing, this happens exactly when x>−xx > -x, that is, when x>0x > 0. Therefore, the hyperbolic sine is negative if x<0x < 0, zero at 00 and positive if x>0x > 0; the hyperbolic tangent, the quotient of the sine by a positive number, has the same sign. The hyperbolic cosine, by contrast, is always positive, and something more is true.

LemmaLower bound of the hyperbolic cosine

For every real xx we have cosh⁡x−1=2sinh⁡2x2\cosh x - 1 = 2 \sinh^2 \frac{x}{2}. Consequently, cosh⁡x≥1\cosh x \geq 1, and equality holds only when x=0x = 0.

The idea is to recognize in the numerator of cosh⁡x−1\cosh x - 1 the square of a difference. The figure shows the graph of the hyperbolic cosine above the line y=1y = 1, with the difference between the two as a vertical line segment.

Proof

  1. ex=(ex2)2,e−x=(e−x2)2,ex2e−x2=1e^x = \left( e^{\frac{x}{2}} \right)^2, \quad e^{-x} = \left( e^{-\frac{x}{2}} \right)^2, \quad \dato{p}{e^{\frac{x}{2}} e^{-\frac{x}{2}} = 1}

    By the laws of exponents, exe^x is the square of ex2e^{\frac{x}{2}}, e−xe^{-x} is that of e−x2e^{-\frac{x}{2}}, and the product of these two powers is e0=1e^0 = 1.

  2. cosh⁡x−1=ex−2+e−x2=(ex2−e−x2)22=2sinh⁡2x2\cosh x - 1 = \frac{\resaltar{e^x - 2 + e^{-x}}}{2} = \frac{\resaltar{\left( e^{\frac{x}{2}} - e^{-\frac{x}{2}} \right)^2}}{2} = \dato{c}{2 \sinh^2 \frac{x}{2}}

    Therefore, the numerator of cosh⁡x−1\cosh x - 1 is the square of a difference, and that difference is twice sinh⁡x2\sinh \frac{x}{2}.

  3. cosh⁡x−1=2sinh⁡2x2≥0\cosh x - 1 = \resaltar{2 \sinh^2 \frac{x}{2}} \geq 0

    A square is not negative: the hyperbolic cosine is at least 11.

  4. cosh⁡x=1↔ex2=e−x2↔x=0\cosh x = 1 \leftrightarrow e^{\frac{x}{2}} = e^{-\frac{x}{2}} \leftrightarrow \resaltar{x = 0}

    Equality requires the base of the square to be zero, that is, ex2=e−x2e^{\frac{x}{2}} = e^{-\frac{x}{2}}; since the exponential is injective, this requires x2=−x2\frac{x}{2} = -\frac{x}{2}, that is, x=0x = 0.

The inequality cosh⁡x≥1\cosh x \geq 1 is the inequality between the arithmetic mean and the geometric mean of the positive numbers exe^x and e−xe^{-x}, whose geometric mean is 11.