Hyperbolic sines and cosines

We all know that the equation for a circle is $x^2+y^2=1$. What about that for a hyperbola? It is $x^2-y^2=1$. Since $\cosh t, \sinh t$ satisfy this equation, that is, $\cosh^2 t - \sinh^2 t=1$, they are called hyperbolic functions.

In this post, we will prove two results: $$\cosh a=\frac{e^a+e^{-a}}{2}\\ \sinh a=\frac{e^a-e^{-a}}{2}.$$

In the right figure, the area of the sector of the rectangular hyperbola $x^2-y^2=1$ bounded by the x-axis, ray, and hyperbola is $\dfrac{a}{2}$.

Proof:

Prerequisite knowledge:
Area of the sector of a parametric curve
Area of the sector bounded by two radii and the arc $P_0P$ of a parametric curve is given by $A=\frac{1}{2}\int_{t_0}^t [x(t)y'(t)-y(t)x'(t)]dt$ where the parametric values, $t_0$ and t relate to the endpoints, $P_0$ and P of the arc, respectively.

Now, we can find the area of the sector.
Since $\cosh^2 a-\sinh^2 a=1$, we have a parametric curve $x=\cosh a, y=\sinh a$.
Then $x'(a)=\sinh a, y'(a)=\cosh a$.
$x(a)y'(a)=\cosh^2 a, y(a)x'(a)=\sinh^2 a$
$A=\frac{1}{2}\int_0^a (\cosh^2 a-\sinh^2 a)da=\frac{1}{2}\int_0^a da=\frac{a}{2} \:\Box$

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$\frac{a}{2}=\frac{1}{2}xy-\int_1^x y\:dx$
$a=x\sqrt{x^2-1}-2\int_1^x \sqrt{x^2-1}\:dx$
$2\int_1^x \sqrt{x^2-1}\:dx\\=2\int_0^{\sec^{-1} x} \sqrt{\sec^2 u-1} \tan u \sec u \:du\\=2\int_0^{\sec^{-1} x} \tan^2 u \sec u\:du\\=2\int_0^{\sec^{-1} x} \frac{\sin^2 u}{\cos^3 u}du\\=\int_0^{\sec^{-1} x} \sin u \: d(\cos^{-2} u)\\=\frac{\sin u}{\cos^2 u}|_0^{\sec^{-1} x}-\int_0^{\sec^{-1} x}\cos^{-2} u \cos u \: du \quad (*)\\=xy-\int_0^{\sec^{-1} x}\sec u\:du\\=xy-[\ln|\sec u+\tan u|]_0^{\sec^{-1} x}\\=xy-\ln|x+\sqrt{x^2-1}| \quad (**)$
$a=xy-xy+\ln|x+\sqrt{x^2-1}|=\ln|x+\sqrt{x^2-1}|$
$e^a=x+\sqrt{x^2-1}$
$e^{2a}=x^2+2x\sqrt{x^2-1}+x^2-1=2x(x+\sqrt{x^2-1})-1=2xe^a-1$
$e^{2a}+1=2xe^a$
$e^a+e^{-a}=2x \Rightarrow x \equiv \cosh a = \frac{e^a+e^{-a}}{2}$
$\sinh a \equiv y=\sqrt{(\frac{e^a+e^{-a}}{2})^2-1}=\sqrt{\frac{e^{2a}+2+e^{-2a}}{4}-1}=\sqrt{\frac{(e^a-e^{-a})^2}{4}}=\frac{e^a-e^{-a}}{2} \Box$

Explanations:
$(*) \frac{\sin u}{\cos^2 u}|_0^{\sec^{-1} x}=\sec u \tan u |_0^{\sec^{-1} x}$
$(**) \tan (\sec^{-1} x)=\:?\\ \sec^{-1} x=y\\ \sec y=x\\ \sec^2 y=x^2\\ 1+\tan^2 y=x^2\\ \tan^2 y=x^2-1\\ \
\therefore \tan (\sec^{-1} x)=\sqrt{x^2-1}$

Relationship to differential equation
In physics, one of the most important differential equation is $y^{\prime \prime}(x)+a^2y(x)=0$. The solution of this equation is $y=A \cos(ax)+B \sin(ax)$, where A and B are constants. (Verify it yourself: differentiate the expression twice.) From trying this yourself, you probably think that the solution to $y^{\prime \prime}(x) - a^2y(x)=0$ is $y=A \cosh(ax)+B \sinh(ax)$. In fact, it is the solution!

Application of hyperbolic trig functions
They are used in the field of engineering, and can be used to solve second order ordinary differential equations. Going beyond this, we can often find hyperbolic trig functions being used in architecture. In particular, the cosh function is used to trace out a curve called a catenary, which is formed from simply hanging a string from two equally high points. We will discuss more about catenary in another post.

References:
http://math.scu.edu/~dostrov/Hyperbolic_Functions.pdf
http://www.nabla.hr/CL-DefiniteIntAppl2.htm
http://www.mathed.soe.vt.edu/Undergraduates/EulersIdentity/HyperbolicTrig.pdf