2.1 Hyperbolic functions

sinh x=(e^x−e^{−x})/2 and cosh x=(e^x+e^{−x})/2. Their quotient is tanh x=sinh x/cosh x. These definitions make exponential methods available for hyperbolic equations.

The key identity is cosh²x−sinh²x=1, analogous to a trigonometric identity with a sign change. sinh is odd, cosh is even, and tanh is odd.

At x=0, sinh0=0, cosh0=1 and tanh0=0. For large positive x, tanh x approaches 1 because e^{−x} becomes negligible relative to e^x.

Hyperbolic functions are not ordinary sine and cosine with a different name; their signs, domains and identities differ.

sinh x is an odd increasing curve through the origin; cosh x is an even U-shaped curve with minimum 1; tanh x is odd, increasing and bounded between −1 and 1.

The limits as x→±∞ explain the horizontal asymptotes of tanh. Evenness gives symmetry about the y-axis, while oddness gives rotational symmetry about the origin.

cosh(−x)=cosh x, so its graph mirrors across the y-axis. tanh x approaches 1 from below as x grows and −1 from above as x decreases.

Do not transfer the sine/cosine range or periodicity to hyperbolic functions; none of these basic hyperbolic graphs is periodic.

The inverse functions are defined on restricted domains: asinh x=ln(x+√(x²+1)), acosh x=ln(x+√(x²−1)) for x≥1, and atanh x=½ln((1+x)/(1−x)) for |x|<1.

The domain restrictions make the square roots real and select a one-to-one branch. Differentiate or substitute back to verify a result, and preserve absolute-value conditions when integrating logarithms.

asinh 0=0. For x=1, acosh1=0 because ln(1+0)=0; atanh x cannot accept x=1 because its denominator in the logarithmic form vanishes.

Inverse hyperbolic notation means inverse function, not reciprocal, and the domains are not optional.

Use cosh²x−sinh²x=1, tanh x=sinh x/cosh x and the exponential definitions to rewrite an equation in a form that can be factored or solved with logarithms.

Check the domain after squaring or taking square roots, and test candidates in the original equation. For expressions such as a cosh x+b sinh x, exponential substitution turns the problem into a quadratic in e^x.

Setting t=e^x>0 converts cosh x= (t+t^{-1})/2 and sinh x=(t−t^{-1})/2; solve the resulting quadratic and retain only positive t before taking ln t.

Algebraic roots of the transformed equation are not all valid x-values; positivity of e^x and the original domain must be checked.