4.07e, 4.07f - Inverse hyperbolic functions as logarithms

4.07e, 4.07f - Inverse hyperbolic functions as logarithms

Understand the inverse hyperbolic functions, including the domain and range of each, in 4.07e. In 4.07f you will derive their logarithmic expressions from exponentials and use them for exact values and equations, keeping careful track of the branch of inverse cosine hyperbolic.

Inverses and the cosine branch

Why does cosh⁡x=54\cosh x=\frac54 have two solutions, while cosh⁡−1(54)\cosh^{-1}(\frac54) has only one value? You already know the equation has the solutions x=±ln⁡2x=\pm\ln2. An inverse function must return just one output, so it cannot simply return both.

You already know

sinh⁡\sinh and tanh⁡\tanh are strictly increasing, with ranges R\mathbb R and (−1,1)(-1,1). The even function cosh⁡\cosh has range [1,∞)[1,\infty) and is strictly increasing when its domain is restricted to [0,∞)[0,\infty). An inverse swaps a function's input and output.

A one-to-one function has an inverse because each output came from exactly one input. To invert cosh⁡\cosh, choose its non-negative-input branch. The reflection in y=xy=x then becomes a function, just as it does for the other two hyperbolic functions.

[DIAGRAM: asset_name: Hyperbolic functions and their inverses; asset_slug: h245_inverse_hyperbolic_graphs; caption: Each inverse is the reflection of its original function in y = x, using only the non-negative-input branch of cosine hyperbolic; description: Three equal-scale panels show blue original functions and clay inverse functions reflected in the dotted line y = x. The sine panel shows sinh and its inverse through the origin. The cosine panel shows cosh only for x at least zero, beginning at (0, 1), and its inverse beginning at (1, 0). The tangent panel shows horizontal asymptotes y = plus or minus 1 for tanh and vertical asymptotes x = plus or minus 1 for its inverse. Domain and range statements appear beneath every panel]
Each inverse is the reflection of its original function in y = x, using only the non-negative-input branch of cosine hyperbolic

The output restrictions are part of the definitions, not an extra choice made while solving a particular question.

Inverse hyperbolic functions

The real inverse hyperbolic functions are defined by

y=sinh⁡−1x  ⟺  x=sinh⁡yx,y∈Ry=cosh⁡−1x  ⟺  x=cosh⁡yx≥1, y≥0y=tanh⁡−1x  ⟺  x=tanh⁡y−1<x<1, y∈R\begin{aligned} y=\sinh^{-1}x&\iff x=\sinh y && x,y\in\mathbb R \\ y=\cosh^{-1}x&\iff x=\cosh y && x\ge1,\ y\ge0 \\ y=\tanh^{-1}x&\iff x=\tanh y && -1<x<1,\ y\in\mathbb R \end{aligned}

The alternative notations are arsinh x\mathrm{arsinh}\,x, arcosh x\mathrm{arcosh}\,x and artanh x\mathrm{artanh}\,x, respectively.

For example, sinh⁡−10=0\sinh^{-1}0=0, cosh⁡−11=0\cosh^{-1}1=0 and tanh⁡−10=0\tanh^{-1}0=0. There is no real value of cosh⁡−10\cosh^{-1}0 or tanh⁡−11\tanh^{-1}1. An inverse function is not a reciprocal: sinh⁡−1x\sinh^{-1}x means the input whose sine hyperbolic is xx, while 1/sinh⁡x1/\sinh x means a division.

FunctionDomainRange
sinh⁡−1x\sinh^{-1}xR\mathbb RR\mathbb R
cosh⁡−1x\cosh^{-1}x[1,∞)[1,\infty)[0,∞)[0,\infty)
tanh⁡−1x\tanh^{-1}x(−1,1)(-1,1)R\mathbb R

These domains and ranges follow by swapping those of the original functions, with the stated restriction for cosh⁡\cosh. The domain/range table is not in the formula booklet; learn it. It also explains a useful boundary case: cosh⁡−1(cosh⁡x)=∣x∣\cosh^{-1}(\cosh x)=|x|, not xx, because the inverse output must be non-negative. This composition rule is not in the formula booklet; learn it.

Key idea

cosh⁡−1x\cosh^{-1}x returns the non-negative input; solving a cosine hyperbolic equation may require its negative partner as well.

Use the definition rather than interpreting the superscript as an ordinary power.

The logarithmic form of inverse sine hyperbolic

To turn an inverse into a logarithm, start with its defining equation and solve for an exponential.

Let y=sinh⁡−1xy=\sinh^{-1}x. Then x=sinh⁡yx=\sinh y, so with t=ey>0t=e^y>0,

2x=t−1texponential definitiont2−2xt−1=0multiply by tt=x±x2+1quadratic formula\begin{aligned} 2x&=t-\frac1t && \text{exponential definition} \\ t^2-2xt-1&=0 && \text{multiply by }t \\ t&=x\pm\sqrt{x^2+1} && \text{quadratic formula} \end{aligned}

Because x2+1>∣x∣\sqrt{x^2+1}>|x|, the plus choice is positive and the minus choice is negative for every real xx. Hence ey=x+x2+1e^y=x+\sqrt{x^2+1}, and taking logarithms gives the inverse. Even when xx is a large negative number, the argument of this logarithm stays positive.

Inverse sine hyperbolic as a logarithm

For every real xx,

sinh⁡−1x=ln⁡(x+x2+1)\sinh^{-1}x=\ln\left(x+\sqrt{x^2+1}\right)

The logarithm's argument is positive for every real xx.
In the formula booklet.

The quadratic equation belongs in a derivation: quoting the printed formula is not a derivation of it. When merely finding a value, however, you can use the booklet result directly.

Worked example 1: a positive exact value

Question. Find the exact value of sinh⁡−1(34)\sinh^{-1}(\frac{3}{4}), giving your answer as a logarithm.

Plan. Substitute the entire fraction into the logarithmic expression and simplify its square root.

sinh⁡−134=ln⁡(34+916+1)booklet formula=ln⁡(34+54)positive square root=ln⁡2combine fractions\begin{aligned} \sinh^{-1}\frac34 &=\ln\left(\frac34+\sqrt{\frac9{16}+1}\right) && \text{booklet formula} \\ &=\ln\left(\frac34+\frac54\right) && \text{positive square root} \\ &=\ln2 && \text{combine fractions} \end{aligned}

Why this step? The square root contains the square of the entire input, so (3/4)2=9/16(3/4)^2=9/16, not 9/49/4.

Answer. ln⁡2\ln2.

Check. sinh⁡(ln⁡2)=12(2−12)=34\sinh(\ln2)=\frac12(2-\frac12)=\frac34.

The next example changes only the sign of the input. The same positive square root now combines with a negative first term, producing a logarithm of a number below 11.

Worked example 2: a negative exact value

Question. Find the exact value of sinh⁡−1(−34)\sinh^{-1}(-\frac{3}{4}), giving your answer as a logarithm.

Plan. Keep the negative sign outside the square root and simplify the resulting positive argument.

sinh⁡−1(−34)=ln⁡(−34+916+1)booklet formula=ln⁡(−34+54)square unchanged=ln⁡12=−ln⁡2logarithm law\begin{aligned} \sinh^{-1}\left(-\frac34\right) &=\ln\left(-\frac34+\sqrt{\frac9{16}+1}\right) && \text{booklet formula} \\ &=\ln\left(-\frac34+\frac54\right) && \text{square unchanged} \\ &=\ln\frac12=-\ln2 && \text{logarithm law} \end{aligned}

Why this step? A negative inverse value does not require a negative logarithm argument. The logarithm is negative when its positive argument lies between 00 and 11.

Answer. −ln⁡2-\ln2.

Check. sinh⁡(−ln⁡2)=−34\sinh(-\ln2)=-\frac34, agreeing with the odd sine hyperbolic function.

A sign mistake in the quadratic root can make a perfectly valid real inverse appear undefined.

Watch out

Wrong: ey=x−x2+1e^y=x-\sqrt{x^2+1} is another possible real branch. That expression is always negative, whereas ey>0e^y>0. Only x+x2+1x+\sqrt{x^2+1} survives.

Apply the same calculation to a new negative input.

The logarithmic form of inverse cosine hyperbolic

The cosine calculation produces two positive exponential roots, so the branch restriction must do more work than it did for sine.

Continue this lesson.

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