4.11 Trigonometric and Hyperbolic Functions

Complex Trigonometric Functions
If \(x\) is a real valuable, then \(e^{i\theta } = \cos x + i\sin x,\quad e^{-ix} = \cos x - i\sin x\,.\,\) It follows that \[e^{ix} + e^{ix} = 2\cos x\]

\[\boxed {\cos x = \frac {e^{ix} + e^{-ix}}{2}\qquad \text {and}\qquad \sin x = \frac {e^{ix} - e^{-ix}}{2i}}\]

Definition 4.68. The complex sine and cosine functions are given by \[\cos x = \frac {e^{ix} + e^{-ix}}{2}\qquad \text {and}\qquad \sin x = \frac {e^{ix} - e^{-ix}}{2i}\]

So \(\, \tan z = \frac {\sin z}{\cos z}\quad ,\quad \cot z = \frac {\cos z}{\sin z}\quad ,\quad \csc z = \frac {1}{\sin z}\)

Example 4.69. Find the values of \(\,\text {(a)}\ \cos (i)\qquad \text {(b)}\ \sin (2 + i)\qquad \text {(c)}\ \tan (\pi - 2i)\).

Solution. Throughout we use the addition formulae, valid for complex arguments, and the bridge to the hyperbolic functions, \[\cos (iy)=\cosh y,\qquad \sin (iy)=i\sinh y .\]

(a): \(\cos (i)=\cosh 1=\dfrac {e+e^{-1}}{2}\approx 1.5431\). This is real and greater than \(1\): the bound \(\left |\cos x\right |\leq 1\) holds for real arguments only, and fails badly off the real axis.
(b): \(\sin (2+i)=\sin 2\cos i+\cos 2\sin i=\sin 2\cosh 1+i\cos 2\sinh 1 \approx 1.4031-0.4891\,i\).
(c): \(\tan \) has period \(\pi \), so \(\tan (\pi -2i)=\tan (-2i)=-\tan (2i)\), and \(\tan (iy)=i\tanh y\) gives \[\tan (\pi -2i)=-i\tanh 2\approx -0.9640\,i .\]

Definition 4.70. The complex hyperbolic sine and hyperbolic cosine functions are defined by \[\sinh z = \frac {e^z - e^{-z}}{2}\qquad \text {and}\qquad \cosh z = \frac {e^z + e^{-z}}{2}\]

Note that \begin {align*} \sinh (iz) & = \frac {e^{iz} - e^{-iz}}{2} = i\Big (\frac {e^{iz} - e^{-iz}}{2i}\Big )\\ & = i\sinh z \end {align*}

\[\implies \quad -i\sinh (iz) = \sin z\]

Hence we have the following

1.: \(\sin (z) = -i\sinh (iz)\)
2.: \(\cos (z) = \cosh (iz)\)
3.: \(\sinh (z) = -i\sin (iz)\)
4.: \(\cosh (z) = i\cos (iz)\)

\begin {align*} \tan (iz) & = \frac {\sin (iz)}{\cos (iz)} = \frac {i\sinh (z)}{\cosh z}\\ & = i\tanh (z) \end {align*}

Example 4.71. \(\cosh ^2(z) - \sinh ^2(z) = 1\)

Note that \(\cosh (z) = \cos (iz)\) and \(\sinh (z) = -i\sin (iz)\) \begin {align*} \big (\cos (iz)\big )^2 - \big (-i\sin (iz)\big )^2 & = \big (\cos (iz)\big )^2 + \big (\sin (iz)\big )^2\\ & = 1\\\\ \end {align*}

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