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2.1 Hyperbolic functions

Syllabus
9231–2028–2029
Topic
2.1
Level
A2

Learning objectives

2.1.1Hyperbolic functions• understand the definitions of the hyperbolic functions sinh x, cosh x, tanh x, sech x, cosech x, coth x in terms of the exponential function2.1.2Hyperbolic functions• sketch the graphs of hyperbolic functions2.1.3Hyperbolic functions• prove and use identities involving hyperbolic functions e.g. cosh2 x - sinh2 x ≡ 1, sinh2x ≡ 2 sinh x cosh x, and similar results corresponding to the standard trigonometric identities.2.1.4Hyperbolic functions• understand and use the definitions of the inverse hyperbolic functions and derive and use the logarithmic forms

Build all six hyperbolic functions from two exponentials

Function Exponential definition Equivalent reciprocal/quotient
sinh⁡x\sinh x (ex−e−x)/2(e^x-e^{-x})/2 odd
cosh⁡x\cosh x (ex+e−x)/2(e^x+e^{-x})/2 even
tanh⁡x\tanh x (ex−e−x)/(ex+e−x)(e^x-e^{-x})/(e^x+e^{-x}) sinh⁡x/cosh⁡x\sinh x/\cosh x
sech⁡x\operatorname{sech}x 2/(ex+e−x)2/(e^x+e^{-x}) 1/cosh⁡x1/\cosh x
cosech⁡x\operatorname{cosech}x 2/(ex−e−x)2/(e^x-e^{-x}) 1/sinh⁡x1/\sinh x
coth⁡x\coth x (ex+e−x)/(ex−e−x)(e^x+e^{-x})/(e^x-e^{-x}) cosh⁡x/sinh⁡x\cosh x/\sinh x

Because ex>0e^x>0, cosh⁡x>0\cosh x>0 for every real x, so tanh and sech are defined on all reals. Since sinh⁡x=0\sinh x=0 only at x=0x=0, cosech and coth exclude x=0x=0. The signs under x↦−xx\mapsto-x show that sinh, tanh, cosech and coth are odd, while cosh and sech are even.

At $x=0$:\sinh0=0,\quad\cosh0=1,\quad\tanh0=0,\quad\operatorname{sech}0=1,whilecosech0andcoth0areundefined.while cosech 0 and coth 0 are undefined.

sech⁡x\operatorname{sech}x is the reciprocal 1/cosh⁡x1/\cosh x; cosh⁡−1x\cosh^{-1}x denotes an inverse function. Reciprocal and inverse notation describe different operations.

Sketch each hyperbolic graph from its range and asymptotes

y Domain and range Shape/symmetry Asymptotes
sinh⁡x\sinh x R→R\mathbb R\to\mathbb R odd, strictly increasing, through (0,0) none
cosh⁡x\cosh x R→[1,∞)\mathbb R\to[1,\infty) even, minimum (0,1) none
tanh⁡x\tanh x R→(−1,1)\mathbb R\to(-1,1) odd, strictly increasing, through (0,0) y=1y=1 as x→∞x\to\infty; y=−1y=-1 as x→−∞x\to-\infty
sech⁡x\operatorname{sech}x R→(0,1]\mathbb R\to(0,1] even, maximum (0,1), positive y=0y=0
cosech⁡x\operatorname{cosech}x R∖{0}→R∖{0}\mathbb R\setminus\{0\}\to\mathbb R\setminus\{0\} odd; decreasing on each branch x=0x=0, y=0y=0
coth⁡x\coth x R∖{0}→(−∞,−1)∪(1,∞)\mathbb R\setminus\{0\}\to(-\infty,-1)\cup(1,\infty) odd; decreasing on each branch x=0x=0, y=1y=1 right, y=−1y=-1 left

Start with parity and intercepts, then use exponential dominance as x→±∞x\to\pm\infty. For reciprocal functions, zeros of the denominator become vertical asymptotes and very large denominator magnitude makes the reciprocal approach 0. Label asymptotes and open range endpoints; none of the six basic graphs is periodic.

For x>0x>0, coth x falls from +∞+\infty near x=0+x=0^+ toward the horizontal asymptote y=1y=1 from above. This is different from tanh x, which rises from 0 toward 1 from below.

Do not transfer sine/cosine ranges or periodicity. A graph sketch must show excluded x-values, correct branch quadrants, extrema/intercepts and labelled asymptotes—not only a generic curve shape.

Prove hyperbolic identities from exponentials before using their family

Fromthedefinitions,From the definitions,\cosh^2x-\sinh^2x=\frac{(e^x+e^{-x})^2-(e^x-e^{-x})^2}{4}=1.Divide by $\cosh^2x$ or $\sinh^2x$ to obtain1-\tanh^2x=\operatorname{sech}^2x,\qquad \coth^2x-\operatorname{cosech}^2x=1.

\sinh(x\pm y)=\sinh x\cosh y\pm\cosh x\sinh y,\cosh(x\pm y)=\cosh x\cosh y\pm\sinh x\sinh y,wherethepairedsignscorrespond.where the paired signs correspond.

Setting $y=x$ gives\sinh2x=2\sinh x\cosh x,\cosh2x=\cosh^2x+\sinh^2x=2\cosh^2x-1=1+2\sinh^2x,\tanh2x=\frac{2\tanh x}{1+\tanh^2x}.

Choose the identity that matches the structure. For example, cosh⁡4x−sinh⁡4x=(cosh⁡2x−sinh⁡2x)(cosh⁡2x+sinh⁡2x)=cosh⁡2x\cosh^4x-\sinh^4x=(\cosh^2x-\sinh^2x)(\cosh^2x+\sinh^2x)=\cosh2x. When asked to prove an identity, transform one side from definitions or known identities until it equals the other; do not assume the target equality mid-proof.

The fundamental sign is minus, not the trigonometric plus. Consequently the cosh addition formula and cosh⁡2x\cosh2x use plus between the squared terms.

Inverse hyperbolic logarithms come from solving for a positive exponential

Inverse Input domain and chosen output Logarithmic form
sinh⁡−1y\sinh^{-1}y y∈Ry\in\mathbb R, output real ln⁡(y+y2+1)\ln(y+\sqrt{y^2+1})
cosh⁡−1y\cosh^{-1}y y≥1y\ge1, output ≥0\ge0 ln⁡(y+y2−1)\ln(y+\sqrt{y^2-1})
tanh⁡−1y\tanh^{-1}y ∣y∣<1|y|<1, output real 12ln⁡((1+y)/(1−y))\tfrac12\ln((1+y)/(1-y))
Reciprocal inverse Domain Reduce to
sech⁡−1y\operatorname{sech}^{-1}y 0<y≤10<y\le1, output ≥0\ge0 cosh⁡−1(1/y)\cosh^{-1}(1/y)
cosech⁡−1y\operatorname{cosech}^{-1}y y≠0y\ne0 sinh⁡−1(1/y)\sinh^{-1}(1/y)
coth⁡−1y\coth^{-1}y ∣y∣>1|y|>1 tanh⁡−1(1/y)=12ln⁡((y+1)/(y−1))\tanh^{-1}(1/y)=\tfrac12\ln((y+1)/(y-1))

Let $u=\tanh^{-1}y$, soy=\frac{e^u-e^{-u}}{e^u+e^{-u}}=\frac{e^{2u}-1}{e^{2u}+1}.ThenThene^{2u}(1-y)=1+y,and because $|y|<1$, both sides of the logarithm are positive:u=\frac12\ln\left(\frac{1+y}{1-y}\right).

For sinh or cosh, set t=eu>0t=e^u>0 and solve the resulting quadratic. Positivity selects the valid t before taking u=ln⁡tu=\ln t; for cosh, restricting u≥0u\ge0 makes the even function one-to-one. Substitute back into the original hyperbolic function to check the branch.

The superscript −1-1 here means inverse function, not reciprocal. Every logarithmic form carries its input domain and branch restriction; algebraic roots outside them are not alternative answers.