Circular and hyperbolic functions differ by rotations

The difference between a circular function and a hyperbolic function is a rotation or two.

For example, cosh(z) = cos(iz). You can read that as saying that to find the hyperbolic cosine of z, first you rotate z a quarter turn to the left (i.e. multiply by i) and then take the cosine.

For another example, sinh(z) = −i sin(iz). This says that you can calculate the hyperbolic sine of z by rotating z to the left, taking the sine, and then rotating to the right.

You can relate each trig function “foo” with its hyperbolic counterpart “fooh” by applying one of the functions to iz then multiplying by a constant c that depends on foo: c = i for sin and tan, c = 1 for cos and sec, and c = −i for csc and cot.

Note that if the constant for foo is c, the constant for 1/foo is 1/c. So, for example, the constant for tan is i and the constant for cot is 1/i = −i.

We have four groups of equations, depending on whether the left side of the equation is foo(iz), fooh(iz), foo(z), or fooh(z).

foo(iz)

$\begin{align*} \sin(iz) & = \phantom{-}i\sinh(z) \\ \cos(iz) & = \phantom{-i}\cosh(z) \\ \tan(iz) & = \phantom{-}i\tanh(z) \\ \csc(iz) & = -i\text{csch}(z) \\ \sec(iz) & = \phantom{-i}\text{sech}(z) \\ \cot(iz) & = -i\coth(z) \\ \end{align*}$

fooh(iz)

$\begin{align*} \sinh(iz) & = \phantom{-}i\sin(z) \\ \cosh(iz) & = \phantom{-i}\cos(z) \\ \tanh(iz) & = \phantom{-}i\tan(z) \\ \text{csch}(iz) & = -i\csc(z) \\ \text{sech}(iz) & = \phantom{-i}\sec(z) \\ \coth(iz) & = -i\cot(z) \\ \end{align*}$

foo(z)

$\begin{align*} \sin(z) & = -i\sinh(iz) \\ \cos(z) & = \phantom{-i}\cosh(iz) \\ \tan(z) & = -i\tanh(iz) \\ \csc(z) & = \phantom{-}i\text{csch}(iz) \\ \sec(z) & = \phantom{-i}\text{sech}(iz) \\ \cot(z) & = \phantom{-}i\coth(iz) \\ \end{align*}$

fooh(z)

$\begin{align*} \sinh(z) & = -i\sin(iz) \\ \cosh(z) & = \phantom{-i}\cos(iz) \\ \tanh(z) & = -i\tan(iz) \\ \text{csch}(z) & = \phantom{-}i\csc(iz) \\ \text{sech}(z) & = \phantom{-i}\sec(iz) \\ \coth(z) & = \phantom{-}i\cot(iz) \\ \end{align*}$

foo(iz)

fooh(iz)

foo(z)

fooh(z)

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