Table of Contents
Last modified on September 19th, 2024
Inverse hyperbolic functions are the inverse functions of the hyperbolic sine, cosine, tangent, and other hyperbolic functions. They are used to solve equations involving hyperbolic functions and are expressed in terms of logarithmic formulas.
The three basic inverse hyperbolic functions are:
These functions, similar to the trigonometric (or circular) functions, are widely applied in mathematics for solving integrals and differential equations.
The formula of the inverse hyperbolic sine is given by:
sinh-1 x = ${\ln \left( x+\sqrt{1+x^{2}}\right)}$
Proof
Let sinh-1 x = y, where y ∈ ℝ
⇒ x = sinh y
Using the exponential combination of the hyperbolic sine function, we get
⇒ sinh y = ${\dfrac{e^{y}-e^{-y}}{2}}$
⇒ x = ${\dfrac{e^{y}-e^{-y}}{2}}$
⇒ 2x = ey – e-y
⇒ e2y – 2xey – 1 = 0 …..(i)
Here, equation (i) is of the form ax2 + bx + c = 0 …..(ii)
Since the roots of equation (ii) are
x = ${\dfrac{-b\pm \sqrt{b^{2}-4ac}}{2a}}$
Now, from the equation (i), we get
ey = ${\dfrac{-\left( -2x\right) \pm \sqrt{\left( -2x\right) ^{2}-4\left( 1\right) \left( -1\right) }}{2\left( 1\right) }}$
= ${\dfrac{2x\pm \sqrt{4\left( x^{2}+1\right) }}{2}}$
= ${\dfrac{2x\pm 2\sqrt{\left( x^{2}+1\right) }}{2}}$
= ${x\pm \sqrt{\left( x^{2}+1\right) }}$
Since y ∈ ℝ, e must be a positive number (i.e., e > 0).
ey = ${x+\sqrt{\left( x^{2}+1\right) }}$
⇒ y = ${ln\left( x+\sqrt{\left( x^{2}+1\right) }\right)}$
⇒ sinh-1 x = ${ln\left( x+\sqrt{\left( x^{2}+1\right) }\right)}$
Thus, sinh-1 x = ${\ln \left( x+\sqrt{1+x^{2}}\right)}$
On graphing the inverse hyperbolic sine function y = sinh-1 x, we get:
The formula of the inverse hyperbolic cosine is given by:
cosh-1 x = ${\ln \left( x+\sqrt{x^{2}-1}\right)}$
Proof
Let cosh-1 x = y, where y ∈ ℝ
⇒ x = cosh y
Using the exponential combination of the hyperbolic cosine function, we get
⇒ cosh y = ${\dfrac{e^{y}+e^{-y}}{2}}$
⇒ x = ${\dfrac{e^{y}+e^{-y}}{2}}$
⇒ 2x = ey + e-y
⇒ e2y – 2xey + 1 = 0 …..(i)
Here, the equation (i) is of the form ax2 + bx + c = 0 …..(ii)
Since the roots of equation (ii) are
x = ${\dfrac{-b\pm \sqrt{b^{2}-4ac}}{2a}}$
Now, from the equation (i), we get
ey = ${\dfrac{-\left( -2x\right) \pm \sqrt{\left( -2x\right) ^{2}-4\left( 1\right) \left( 1\right) }}{2\left( 1\right) }}$
= ${\dfrac{2x\pm \sqrt{4\left( x^{2}-1\right) }}{2}}$
= ${\dfrac{2x\pm 2\sqrt{\left( x^{2}-1\right) }}{2}}$
= ${x\pm \sqrt{\left( x^{2}-1\right) }}$
Since y ∈ ℝ, e must be a positive number (i.e., e > 0).
ey = ${x+\sqrt{\left( x^{2}-1\right) }}$
⇒ y = ${ln\left( x+\sqrt{\left( x^{2}-1\right) }\right)}$
⇒ cosh-1 x = ${ln\left( x+\sqrt{\left( x^{2}-1\right) }\right)}$
Thus, cosh-1 x = ${\ln \left( x+\sqrt{x^{2}-1}\right)}$, here x ≥ 1
On graphing the inverse hyperbolic cosine function y = cosh-1 x, we get:
The formula of the inverse hyperbolic tangent is given by:
tanh-1 x = ${\dfrac{1}{2}\ln \left( \dfrac{1+x}{1-x}\right)}$
Proof
Let tanh-1 x = y, where y ∈ ℝ
⇒ x = tanh y
Using the exponential combination of the hyperbolic tangent function, we get
tanh y = ${\dfrac{e^{y}-e^{-y}}{e^{y}+e^{-y}}}$
⇒ x = ${\left( \dfrac{e^{y}-e^{-y}}{e^{y}+e^{-y}}\right) \times \left( \dfrac{e^{y}}{e^{y}}\right)}$
⇒ x = ${\dfrac{e^{2y}-1}{e^{2y}+1}}$
⇒ x(e2y + 1) = (e2y – 1)
⇒ (x – 1)e2y + (x + 1) = 0
⇒ e2y = ${-\dfrac{x+1}{x-1}}$
⇒ e2y = ${\dfrac{1+x}{1-x}}$
⇒ 2y = ${\ln \left( \dfrac{1+x}{1-x}\right)}$
⇒ y = ${\dfrac{1}{2}\ln \left( \dfrac{1+x}{1-x}\right)}$
⇒ tanh-1 x = ${\dfrac{1}{2}\ln \left( \dfrac{1+x}{1-x}\right)}$
Thus, tanh-1 x = ${\dfrac{1}{2}\ln \left( \dfrac{1+x}{1-x}\right)}$, here, |x| < 1
On graphing the inverse hyperbolic tangent function y = tanh-1 x, we get:
In addition to the three basic inverse hyperbolic functions, there are three other inverse hyperbolic functions corresponding to sine, cosine, and tangent.
The formula of the inverse hyperbolic cosecant is expressed as:
cosech-1 x = ${\ln \left( \dfrac{1+\sqrt{1+x^{2}}}{x}\right)}$
Proof
Let csch-1 x = y, where y ∈ ℝ
⇒ x = csch y
Using the exponential combination of the hyperbolic cosecant function, we get
csch y or cosech y = ${\dfrac{1}{\sinh y}}$ = ${\dfrac{2}{e^{y}-e^{-y}}}$
⇒ x = ${\dfrac{2}{e^{y}-e^{-y}}}$
⇒ x = ${\left( \dfrac{2}{e^{y}-e^{-y}}\right) \times \left( \dfrac{e^{y}}{e^{y}}\right)}$
⇒ x = ${\dfrac{2e^{y}}{e^{2y}-1}}$
⇒ x(e2y – 1) = 2ey
⇒ xe2y – 2ey – x = 0 …..(i)
Here, the equation (i) is of the form ax2 + bx + c = 0 …..(ii)
Since the roots of equation (ii) are
x = ${\dfrac{-b\pm \sqrt{b^{2}-4ac}}{2a}}$
Now, from the equation (i), we get
⇒ ey = ${\dfrac{-\left( -2\right) \pm \sqrt{\left( -2\right) ^{2}-4\left( x\right) \left( -x\right) }}{2\left( x\right) }}$
⇒ ey = ${\dfrac{2\pm \sqrt{4+4x^{2}}}{2x}}$
⇒ ey = ${\dfrac{2\pm \sqrt{4\left( 1+x^{2}\right) }}{2x}}$
⇒ ey = ${\dfrac{2\pm 2\sqrt{1+x^{2}}}{2x}}$
⇒ ey = ${\dfrac{1\pm \sqrt{1+x^{2}}}{x}}$
Since y ∈ ℝ, e must be a positive number (i.e., e > 0).
ey = ${\dfrac{1+\sqrt{1+x^{2}}}{x}}$
⇒ y = ${\ln \dfrac{1+\sqrt{1+x^{2}}}{x}}$
⇒ csch-1 x = ${\ln \dfrac{1+\sqrt{1+x^{2}}}{x}}$
⇒ csch-1 x = ${\ln \dfrac{1+\sqrt{1+x^{2}}}{x}}$
Thus, csch-1 x = ${\ln \left( \dfrac{1+\sqrt{1+x^{2}}}{x}\right)}$, here, x ≠ 0
On graphing the inverse hyperbolic cosecant function y = csch-1 x, we get:
The formula of the inverse hyperbolic secant is written as:
sech-1 x = ${\ln \left( \dfrac{1+\sqrt{1-x^{2}}}{x}\right)}$
Proof
Let sech-1 x = y, where y ∈ ℝ
⇒ x = sech y
Using the exponential combination of the hyperbolic secant function, we get
sech y = ${\dfrac{1}{\cosh y}}$ = ${\dfrac{2}{e^{y}+e^{-y}}}$
⇒ x = ${\dfrac{2}{e^{y}+e^{-y}}}$
⇒ x = ${\left( \dfrac{2}{e^{y}+e^{-y}}\right) \times \left( \dfrac{e^{y}}{e^{y}}\right)}$
⇒ x = ${\dfrac{2e^{y}}{e^{2y}+1}}$
⇒ x(e2y + 1) = 2ey
⇒ xe2y – 2ey + x = 0 …..(i)
Here, the equation (i) is of the form ax2 + bx + c = 0 …..(ii)
Since the roots of equation (ii) are
x = ${\dfrac{-b\pm \sqrt{b^{2}-4ac}}{2a}}$
Now, from the equation (i), we get
⇒ ey = ${\dfrac{-\left( -2\right) \pm \sqrt{\left( -2\right) ^{2}-4\left( x\right) \left( x\right) }}{2\left( x\right) }}$
⇒ ey = ${\dfrac{2\pm \sqrt{4-4x^{2}}}{2x}}$
⇒ ey = ${\dfrac{2\pm \sqrt{4\left( 1-x^{2}\right) }}{2x}}$
⇒ ey = ${\dfrac{2\pm 2\sqrt{1-x^{2}}}{2x}}$
⇒ ey = ${\dfrac{1\pm \sqrt{1-x^{2}}}{x}}$
Since y ∈ ℝ, e must be a positive number (i.e., e > 0).
ey = ${\dfrac{1+\sqrt{1-x^{2}}}{x}}$
⇒ y = ${\ln \dfrac{1+\sqrt{1-x^{2}}}{x}}$
⇒ sech-1 x = ${\ln \dfrac{1+\sqrt{1-x^{2}}}{x}}$
Thus, sech-1 x = ${\ln \left( \dfrac{1+\sqrt{1-x^{2}}}{x}\right)}$, here, 0 < x ≤ 1
On graphing the inverse hyperbolic secant function y = sech-1 x, we get:
The formula of the inverse hyperbolic cotangent is expressed as:
coth-1 x = ${\dfrac{1}{2}\ln \left( \dfrac{x+1}{x-1}\right)}$
Proof
Let coth1 x = y, where y ∈ ℝ
⇒ x = coth y
Using the exponential combination of the hyperbolic cotangent function, we get
coth y = ${\dfrac{1}{\tanh y}}$ = ${\dfrac{e^{y}+e^{-y}}{e^{y}-e^{-y}}}$
⇒ x = ${\dfrac{e^{y}+e^{-y}}{e^{y}-e^{-y}}}$
⇒ x = ${\left( \dfrac{e^{y}+e^{-y}}{e^{y}-e^{-y}}\right) \times \left( \dfrac{e^{y}}{e^{y}}\right)}$
⇒ x = ${\dfrac{e^{2y}+1}{e^{2y}-1}}$
⇒ x(e2y – 1) = (e2y + 1)
⇒ (x – 1)e2y – (x + 1) = 0
⇒ e2y = ${\dfrac{x+1}{x-1}}$
⇒ 2y = ${\ln \left( \dfrac{x+1}{x-1}\right)}$
⇒ y = ${\dfrac{1}{2}\ln \left( \dfrac{x+1}{x-1}\right)}$
⇒ coth1 x = ${\dfrac{1}{2}\ln \left( \dfrac{x+1}{x-1}\right)}$
Thus, coth1 x = ${\dfrac{1}{2}\ln \left( \dfrac{x+1}{x-1}\right)}$, here, |x| > 1
On graphing the inverse hyperbolic cotangent function y = coth-1 x, we get:
Here is the summary of all graphs of the inverse hyperbolic functions:
Each inverse hyperbolic function has a different domain and a range that should be considered. The domains and ranges of these functions are shown in the following table:
Inverse Hyperbolic Function | Domain | Range |
---|---|---|
sinh-1 x | (-∞, ∞) | (-∞, ∞) |
cosh-1 x | [1, ∞) | [0, ∞) |
tanh-1 x | (-1, 1) | (-∞, ∞) |
cosech-1 x | (-∞, ∞) | (-∞, ∞) |
sech-1 x | (0, 1] | [0, ∞) |
coth-1 x | (-∞, -1) ∪ (1, ∞) | (-∞, ∞) |
Here are the derivatives of the inverse hyperbolic functions:
Inverse Hyperbolic Function | Formula With Restriction |
---|---|
${\dfrac{d}{dx}\left( arcsinh \ x\right)}$ | ${\dfrac{1}{\sqrt{1+x^{2}}}}$ |
${\dfrac{d}{dx}\left( arccosh \ x\right)}$ | ${\dfrac{1}{\sqrt{x^{2}-1}}}$, here x > 1 |
${\dfrac{d}{dx}\left( arctanh \ x\right)}$ | ${\dfrac{1}{1-x^{2}}}$, here |x| < 1 |
${\dfrac{d}{dx}\left( arccosech \ x\right)}$ | ${-\dfrac{1}{\left| x\right| \sqrt{1-x^{2} }}}$, here x ≠ 0 |
${\dfrac{d}{dx}\left( arcsech \ x\right)}$ | ${-\dfrac{1}{x\sqrt{1-x^{2} }}}$, here 0 < x < 1 |
${\dfrac{d}{dx}\left( arccoth \ x\right)}$ | ${\dfrac{1}{1-x^{2}}}$, here |x| > 1 |
Prove: sinh-1 (tan z) = cosh-1 (sec z)
As we know, sinh-1 x = ${\ln \left( x+\sqrt{1+x^{2}}\right)}$
Here,
sinh-1 (tan z) = ${\ln \left( \tan z+\sqrt{\tan ^{2}z+1}\right)}$ …..(i)
From the trigonometric identities, we have: tan2 x + 1 = sec2 x …..(ii)
Substituting (ii) into the equation (i),
sinh-1 (tan z) = ${\ln \left( \tan z+\sqrt{\sec ^{2}z}\right)}$
Simplifying,
⇒ sinh-1 (tan z) = ${\ln \left( \sqrt{\sec ^{2}z-1}+\sec z\right)}$
⇒ sinh-1 (tan z) = ${\ln \left( \sec z+\sqrt{\sec ^{2}z-1}\right)}$ …..(iii)
As we know, cosh-1 x = ${\ln \left( x+\sqrt{x^{2}-1}\right)}$
Thus, from (iii),
⇒ sinh-1 (tan z) = cosh-1 (sec z)
⇒ L.H.S. = R.H.S.
Hence proved
Prove: tanh-1 (sin y) = cosh-1 (sec y)
As we know, cosh-1 x = ${\ln \left( x+\sqrt{x^{2}-1}\right)}$
Here,
cosh-1 (sec y) = ${\ln \left( \sec y+\sqrt{\sec ^{2}y-1}\right)}$
⇒ cosh-1 (sec y) = ${\ln \left( \sec y+\tan y\right)}$
⇒ cosh-1 (sec y) = ${\ln \left( \dfrac{1}{\cos y}+\dfrac{\sin y}{\cos y}\right)}$
⇒ cosh-1 (sec y) = ${\ln \left( \dfrac{1+\sin y}{\cos y}\right)}$
Multiplying and dividing by 2,
⇒ cosh-1 (sec y) = ${\dfrac{1}{2}\times 2\ln \left( \dfrac{1+\sin y}{\cos y}\right)}$
Since 2 ln(x) = ln(x2)
⇒ cosh-1 (sec y) = ${\dfrac{1}{2}\ln \left[ \left( \dfrac{1+\sin y}{\cos y}\right) ^{2}\right]}$
⇒ cosh-1 (sec y) = ${\dfrac{1}{2}\ln \left( \dfrac{\left( 1+\sin y\right) \left( 1+\sin y\right) }{\cos ^{2}y}\right)}$
⇒ cosh-1 (sec y) = ${\dfrac{1}{2}\ln \left( \dfrac{\left( 1+\sin y\right) \left( 1+\sin y\right) }{\left( 1-\sin ^{2}y\right) }\right)}$
⇒ cosh-1 (sec y) = ${\dfrac{1}{2}\ln \left( \dfrac{\left( 1+\sin y\right) \left( 1+\sin y\right) }{\left( 1+\sin y\right) \left( 1-\sin y\right) }\right)}$
⇒ cosh-1 (sec y) = ${\dfrac{1}{2}\ln \left( \dfrac{1+\sin y}{1-\sin y}\right)}$ …..(i)
As we know, tanh-1 x = ${\dfrac{1}{2}\ln \left( \dfrac{1+x}{1-x}\right)}$
Thus, from (i),
⇒ cosh-1 (sec y) = tanh-1 (sin y)
⇒ R.H.S. = L.H.S.
Hence proved
Last modified on September 19th, 2024