coth( z )
The hyperbolic cotangent of z in SageMath. Defined by
\[ \coth z = \frac{ \cosh z }{ \sinh z } \]Plot on the real axis:
Series expansion about the origin:
Special values:
Related functions: tanh arccoth acoth
Function category: sagemath-docs
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
RangeInverse hyperbolic sinesinh-1 x
ln[x + √(x2 + 1)]
(-∞, ∞)
(-∞, ∞)
Inverse hyperbolic cosinecosh-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 =
⇒ 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[
] = 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 =
⇒ 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[
]⇒ csch-1x = ln[
] = ln[1 + √(1 + x2)] – ln(x)csch-1 x = ln[
] = 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 =
⇒ 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 =
z = ln[
] = ln[1 + √(1 – x2)] – ln(x)⇒ sech-1 x = ln[
] = ln[1 + √(1 – x2)] – ln(x)sech-1 x = ln[
] = 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 =
⇒ 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)