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Explain Inverse Hyperbolic Functions Formula

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  • Last Updated : 01 Jul, 2022
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In mathematics, the inverse functions of hyperbolic functions are referred to as inverse hyperbolic functions or area hyperbolic functions. There are six inverse hyperbolic functions, namely, inverse hyperbolic sine, inverse hyperbolic cosine, inverse hyperbolic tangent, inverse hyperbolic cosecant, inverse hyperbolic secant, and inverse hyperbolic cotangent functions. These functions are depicted as sinh-1 x, cosh-1 x, tanh-1 x, csch-1 x, sech-1 x, and coth-1 x. With the help of an inverse hyperbolic function, we can find the hyperbolic angle of the corresponding hyperbolic function.

Function name

Function

Formula

Domain

Range
Inverse hyperbolic sine

sinh-1 x

ln[x + √(x2 + 1)]

(-∞, ∞)

(-∞, ∞)

Inverse hyperbolic cosine

cosh-1x

ln[x + √(x2 – 1)

[1, ∞)

[0, ∞)

Inverse hyperbolic tangent

tanh-1 x

½ ln[(1 + x)/(1 – x)]

(-1,1)

(-∞, ∞)

Inverse hyperbolic cosecant

csch-1 x

ln[(1 + √(x2 + 1)/x]

(-∞, ∞)

(-∞, ∞)

Inverse hyperbolic secant

sech-1 x

ln[(1 + √(1 – x2)/x]

(0, 1]

[0, ∞)

Inverse hyperbolic cotangent

coth-1 x

½ ln[(x + 1)/(x – 1)]

(-∞, -1) or (1, ∞)

(-∞, ∞)

Inverse hyperbolic sine Function

sinh-1 x = ln[x + √(x2 + 1)]

Proof:

Let sinh-1 x = z, where z ∈ R

⇒ x = sinh z

Using the sine hyperbolic function we get,

⇒ x = (ez – e-z)/2

⇒ 2x = ez – e-z

⇒ e2z – 2xez – 1 = 0

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

So, ez = x ± √(x2 + 1)

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

Hence, ez = x + √(x2 + 1)

⇒ z = ln[x + √(x2 + 1)]

⇒ sinh-1 x =  ln[x + √(x2 + 1)]

 sinh-1 x =  ln[x + √(x2 + 1)]

Inverse hyperbolic cosine Function

cosh-1 x = ln[x + √(x2 – 1)]

Proof:

Let cosh-1 x = z, where z ∈ R

⇒ x = cosh z

Using the cosine hyperbolic function we get,

⇒ x = (ez + e-z)/2

⇒ 2x = ez + e-z

⇒ e2z – 2xez + 1 = 0

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

So, ez = x ± √(x2 – 1)

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

Hence, ez = x + √(x2 – 1)

⇒ z = ln[x + √(x2 – 1)]

⇒ cosh-1 x =  ln[x + √(x2 – 1)]

cosh-1 x =  ln[x + √(x2 – 1)]

Inverse hyperbolic tangent function

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

Proof:

 Let tanh-1 x = z, where z ∈ R

⇒ x = tanh z

Using the tangent hyperbolic function we get,

tanh z = (ez – e-z)/(ez + e-z)

x = \left[\frac{e^{z}-e^{-z}}{e^{z}+e^{-z}}\right]\times\left[\frac{e^{z}}{e^{z}}\right]   

⇒ x = (e2z – 1)/(e2z + 1)

⇒ x (e2z + 1) = (e2z – 1)

⇒ (x – 1) e2z + (x + 1) = 0

⇒ e2z = -[(x +1)/(x – 1)]

⇒ e2z = [(x + 1)/(1 – x)]

⇒ 2z = 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)]

tanh-1 x = ½ ln[\frac{1+x}{1-x}] = ½ [ln(1 + x) – ln(1 – x)]

Inverse hyperbolic cosecant function

csch-1 x = ln[(1 + √(x2 + 1)/x]

Proof:

Let csch-1 x = z, where z ∈ R

⇒ x = csch z

Using the cosecant hyperbolic function we get,

csch z = 2/(ez – e-z)

⇒ x =  2/(ez – e-z)

⇒ x = \left[\frac{2}{e^{z}-e^{-z}}\right]\times\left[\frac{e^{z}}{e^{z}}\right]

⇒ x = 2ez/(e2z – 1)

⇒ x (e2z – 1) = 2ez

⇒  xe2z − 2ez – x = 0

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

⇒ ez = (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)

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

Inverse hyperbolic secant function

sech-1 x = ln[(1 + √(1 – x2)/x]

Proof:

Let sech-1 x = z, where z ∈ R

⇒ x = sech z

Using the secant hyperbolic function we get,

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

⇒ x =  2/(ez + e-z)

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

⇒ x = 2ez/(e2z + 1)

⇒ x (e2z +1) = 2ez

⇒ xe2z − 2ez + x = 0

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

So, by simplifying we get, 

ez =\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) 

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

Inverse hyperbolic cotangent function

coth-1 x = ½ ln[(x + 1)/(x – 1)]

Proof:

Let coth-1 x = z, where z ∈ R

⇒ x = coth z

Using the cotangent hyperbolic function we get,

coth z = (ez + e-z)/(ez – e-z)

⇒ x = (ez + e-z)/(ez – e-z)

⇒ x = \left[\frac{e^{z}+e^{-z}}{e^{z}-e^{-z}}\right]\times\left[\frac{e^{z}}{e^{z}}\right]

⇒ x = (e2z + 1)/(e2z – 1)

⇒ x (e2z – 1) = (e2z + 1)

⇒ (x – 1) e2z – (x + 1) = 0

⇒ e2z = [(x +1)/(x – 1)]

⇒ 2z = ln [(x + 1)/(x – 1)]

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

coth-1 x = ½ ln[(x + 1)/(x – 1)] = ½ [ln(x + 1) – ln(x – 1)]

Derivates of inverse hyperbolic functions

Inverse hyperbolic function

Derivative

sinh-1x

1/√(x2 + 1)

cosh-1 x

1/√(x2 – 1), x>1

tanh-1x

1/(1 – x2), |x| < 1

csch-1 x

1/{|x|√(1 + x2)}, x ≠ 0

sech-1 x

-1/[x√(1 – x2)], 0 < x < 1

coth-1 x

1/(1 – x2), |x| > 1

Sample Problems

Problem 1: If sinh x = 4, then prove that x = loge(4 + √17).

Solution:

Given, sinh x = 4

⇒ x = sinh-1 (4)

We know that,

sinh-1 (x) = loge [x + √(x2 + 1)]

⇒ x = loge[4 + √(42 + 1)] = loge(4 + √17)

Hence, x = loge(4 + √17)

Problem 2: Prove that tanh-1 (sin x) = cosh-1 (sec x).

Solution:

We know that,

tanh-1 x = 1/2 ln[(1+x)/(1-x)]

Now, tanh-1 (sin x) = 1/2 log[(1 + sin x)/(1 – sin x)]

We have,

cosh-1 x = ln(x + √[x2-1])

Now, cosh-1 (sec x) = ln[sec x + √(sec2 x – 1)]

= ln[sec x + √tan2 x] {Since, sec2 x – 1 = tan2 x}

= ln[sec x + tan x]

= ln[1/cos x + sin x/cos x]

= ln[(1 + sin x)/cos x]

Now, multiply and divide the term with 2

= 1/2 × 2 ln[(1 + sin x)/cos x]

= 1/2 ln[(1 + sin x)/cos x]2  {since, 2 ln x = ln x2}

= 1/2 ln[(1 + sin x)2/cos2 x]

We know, cos2 x = 1 – sin2x = (1 + sin x)(1 – sin x)

Hence, (1 + sin x)2/cos2 x = [(1 + sin x)(1+ sin x)]/[(1 + sin x)(1 – sin x)] = (1 + sin x)/(1 – sin x)

= 1/2 ln[(1 + sin x)/(1 – sin x)]

= tanh-1 (sin x)

Hence, tanh-1 (sin x) = cosh-1 (sec x)

Problem 3: Find the value of tanh-1 (1/5).

Solution:

We know,

tanh-1 x = 1/2 ln[(1+x)/(1-x)]

⇒ tanh-1 (1/5) = 1/2 ln[(1+(1/5))/(1 – (1/5)]

= 1/2 ln[(6/5)/(4/5)]

=1/2 ln(3/2)

Hence, tanh-1 (1/5) = 1/2 ln(3/2)

Problem 4: Find the value of sech-1 (3/8).

Solution:

We know,

sech-1 x = ln[(1 + √(1 – x2)/x]

So,  sech-1 (3/8) = ln\left[\frac{1+\sqrt{(1-(\frac{3}{8}})^{2}}{\frac{3}{8}}\right]

= ln[(8 + √(64 – 9))/3]

ln\left[\frac{8+\sqrt{55}}{3}\right]

Hence, sech-1(3/8) = ln\left[\frac{8+\sqrt{55}}{3}\right]

Problem 5: Find the derivative of [sinh-1 (5x + 1)]2.

Solution: 

Let y = [sinh-1 (5x + 1)]2

Now derivative of the given function is,

dy/dx = d([sinh-1 (5x + 1)]2)/dx

= 2[sinh-1 (5x + 1)] d/dx [sinh-1 (5x + 1) 

We know d(sinh-1 x)/dx = 1/√(x2 + 1)

= 2 [sinh-1 (5x + 1)] {1/√[(5x+1)2 + 1]} d(5x+1)/dx

= 2 [sinh-1 (5x + 1)] × {1/√(25×2 + 10x + 2)} × 5

= 10 sinh-1(5x+1)/[√(25x2+10x+2)]

Hence, the derivative of [sinh-1(5x+1)] = \frac{10sinh^{-1}(5x+1)}{\sqrt{(25x^{2}+10x+2)}} .


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