Inverse Hyperbolic Functions Formula

Inverse hyperbolic functions are the inverse form of a hyperbolic function. They take the curvature of a circular path if plotted along the x and y axis of a graph. 

• Inverse hyperbolic sine, cosine, tangent, cosecant, secant, and cotangent are the most commonly used functions.
• The inverse hyperbolic function calculates the hyperbolic angle.
• All hyperbolic functions are one-to-one functions and so have an inverse.

Read More: Hyperbolic Functions Formula

Key Terms: Hyperbolic functions, Sine, Cosine, Tangent, Hyperbola.

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Hyperbolic Functions

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A hyperbola is a plane curve with a fast moving point and two endpoints remain constant. The foci are the two fixed locations that are the midpoints of the line segment connecting the foci, which is the center of the hyperbola.

• Hyperbolic functions are exponential functions that have features comparable to trigonometric functions.
• The curvature of hyperbolic functions are hyperbola, just like trigonometric functions are related to the circle. 
• Some forms of partial differential equations can be solved using hyperbolic functions. 

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Conic Sections Detailed Video Explanation:

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Inverse Hyperbolic Function

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Inverse hyperbolic functions are hyperbolic functions but denoted as inverse trigonometric functions with the letter 'h' added for each function. For example, the inverse of below mentioned hyperbolic functions will look like– 

• Inverse hyperbolic functions are also called hyperbolic area functions.
• Hyperbolic functions are exponential functions and their inverses are logarithmic. 
• The inverse hyperbolic functions are single-valued and continuous except for cosh^-1x, which is two-valued.
• Below mentioned formulae express the inverse hyperbolic functions in terms of the inverse trigonometric functions–

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Examples of Inverse Hyperbolic Functions

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1. Inverse hyperbolic function of sinh^-1 x, where domain is (-∞, ∞) & the range is (-∞, ∞)

Solution: Considering, sinh^-1 x = z, where z ∈ R

⇒ x = sinh z

From sine hyperbolic function we get,

⇒ x = (e^z – e^-z)/2

⇒ 2x = e^z – e^-z

⇒ e^2z – 2xe^z – 1 = 0

As per the roots of an equation ax² + bx + c = 0 are x = [-b ± √(b² – 4ac)]/2a

So, e^z = x ± √(x² + 1)

Since z is a real number, e should be a positive number.

Hence, e^z = x + √(x² + 1)

⇒ z = ln[x + √(x² + 1)]

⇒ sinh^-1 x = ln[x + √(x² + 1)]

 sinh^-1 x = ln[x + √(x² + 1)]

2. Inverse hyperbolic function of cosh^-1 x, where domain is (1, ∞) & the range is (0, ∞).

Solution: Considering, cosh^-1 x = z, where z ∈ R

⇒ x = cosh z

From the cosine hyperbolic function we get,

⇒ x = (e^z + e^-z)/2

⇒ 2x = e^z + e^-z

⇒ e^2z – 2xe^z + 1 = 0

From the roots of an equation ax² + bx + c = 0 are x = [-b ± √(b² – 4ac)]/2a

So, e^z = x ± √(x² – 1)

Since z is a real number, e must be a positive number.

Hence, e^z = x + √(x² – 1)

⇒ z = ln[x + √(x² – 1)]

⇒ cosh^-1 x = ln[x + √(x² – 1)]

3. Inverse hyperbolic function of tanh^-1 x, where domain is (-1, 1) & the range is (-∞, ∞).

Solution: Considering, tanh^-1 x = z, where z ∈ R

⇒ x = tanh z

From the tangent hyperbolic function we get,

tanh z = (e^z – e^-z)/(e^z + e^-z)

x = \([\frac{(e^z – e^{-z})}{(e^z + e^{-z})}] \times [\frac{e^z}{e^z}]\)

⇒ x = (e^2z – 1)/(e^2z + 1)

⇒ x (e^2z + 1) = (e^2z – 1)

⇒ (x – 1) e^2z + (x + 1) = 0

⇒ e^2z = -[(x +1)/(x – 1)]

⇒ e^2z = [(x + 1)/(1 – x)]

⇒ 2^z = ln [(x + 1)/(1 – x)]

⇒ z = ½ ln[(1 + x)/(1 – x)] = ½ [ln(1 + x) – ln(1 – x)]

⇒ tanh^-1 x = ½ ln[(1 + x)/(1 – x)] = ½ [ln(1 + x) – ln(1 – x)]

4. Inverse hyperbolic function of csch^-1 x, where domain is (-∞, ∞) & the range is (-∞, ∞).

Solution: ⇒ x = csch z

From the cosecant hyperbolic function we get,

csch z = 2/(e^z – e^-z)

⇒ x = 2/(e^z – e^-z)

⇒ x = \([\frac{2}{(e^z – e^{-z})}] \times [\frac{e^z}{e^z}]\)

⇒ x = 2e^z/(e^2z – 1)

⇒ x (e^2z – 1) = 2e^z

⇒ xe^2z − 2e^z – x = 0

From the roots of an equation ax² + bx + c = 0 are x = [-b ± √(b² – 4ac)]/2a

⇒ e^z = (1 + √(x² + 1)/x

⇒ z = ln \(\frac{1 + \sqrt{(1 + x)^2}}{x}\)

⇒ csch^-1x = ln \(\frac{1 + \sqrt{(1 + x)^2}}{x}\)= ln[1 + √(1 + x²)] – ln(x)

5. Inverse hyperbolic function of sech^-1 x, where domain is(0, 1] & the range is [0, ∞).

Solution: Considering, sech^-1 x = z, where z ∈ R

⇒ x = sech z

From the secant hyperbolic function we get,

sech z = 2/(e^z + e^-z)

⇒ x = 2/(e^z + e^-z)

⇒ x = \([\frac{2}{(e^z + e^{-z})}] \times [\frac{e^z}{e^z}]\)

⇒ x = 2e^z/(e^2z + 1)

⇒ x (e^2z +1) = 2e^z

⇒ xe^2z − 2e^z + x = 0

From the roots of an equation ax² + bx + c = 0 are x = [-b ± √(b² – 4ac)]/2a

So, by simplifying we get, 

e^z = \(\frac{1 + \sqrt{(1 - x)^2}}{x}\)

z = ln \(\frac{1 + \sqrt{(1 - x)^2}}{x}\)= ln[1 + √(1 – x²)] – ln(x)

⇒ sech^-1 x = ln \(\frac{1 + \sqrt{(1 - x)^2}}{x}\)= ln[1 + √(1 – x²)] – ln(x)

6. Inverse hyperbolic function of coth^-1 x, where domain is(-∞, -1) or (1, ∞) & the range is (-∞, ∞).

Solution: Considering, coth^-1 x = z, where z ∈ R

⇒ x = coth z

From the cotangent hyperbolic function we get,

coth z = (e^z + e^-z)/(e^z – e^-z)

⇒ x = (e^z + e^-z)/(e^z – e^-z)

⇒ x = \([\frac{e^z + e^{-z}}{e^z – e^{-z}}] \times [\frac{e^z}{e^z}]\)

⇒ x = (e^2z + 1)/(e^2z – 1)

⇒ x (e^2z – 1) = (e^2z + 1)

⇒ (x – 1) e^2z – (x + 1) = 0

⇒ e^2z = [(x +1)/(x – 1)]

⇒ 2^z = ln [(x + 1)/(x – 1)]

⇒ z = ½ ln[(x + 1)/(x – 1)] = ½[ln(x + 1) – ln(x – 1)]

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Plotting of Inverse Hyperbolic Functions

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1. Plotted graph for sinh^-1x

2. Plotted graph for cosh^-1x

3. Plotted graph for tanh^-1x

4. Plotted graph for csch^-1x

5. Plotted graph for sech^-1x

6. Plotted graph for coth^-1x

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Solved Examples

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Example 1: Evaluate the following tanh ^-1 \((-\frac{7}{8})\)

Solution: As per the expression known, tanh ^-1 x = \(\frac{1}{2}\) In \(\frac{1 + x}{1 - x}\)

tanh ^-1 \((-\frac{7}{8})\) = \(\frac{1}{2}\) In \(|\frac{1 - \frac{7}{8}}{1 + \frac{7}{8}}|\)

=  \(\frac{1}{2}\) In \(|\frac{\frac{1}{8}}{ \frac{15}{8}}|\)

=  \(\frac{1}{2}\) In \(\frac{1}{15}\)

≈ – 1.35

Example 2: Evaluate the following sinh ^-1 3

Solution: As per the expression known, sinh ^-1 x = In ( x + \(\sqrt{x^2 + 1}\))

sinh ^-1 3 = In ( 3 + \(\sqrt{3^2 + 1}\))

= In ( 3 + \(\sqrt{10}\))

≈ 1.82

Also Read:

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Things to Remember

• A hyperbola is a plane curve created by a point that moves so fast that the distance between two fixed points remains constant. 
• Inverse hyperbolic functions are the inverse form of a hyperbolic function.
• Inverse hyperbolic sine, inverse hyperbolic cosine, inverse hyperbolic tangent, inverse hyperbolic cosecant, inverse hyperbolic secant, and inverse hyperbolic cotangent are the most commonly used functions.
• All hyperbolic functions are one-to-one functions.
• Few forms of partial differential equations can be solved using hyperbolic functions. 
• Hyperbolic functions are exponential functions and their inverses are logarithmic.