91Ó°ÊÓ

The hyperbolic sine function, denoted by \(\sinh\ x\), is similar but intimately different from the sinusoidal function we commonly know. Its formula is expressed as \(\sinh\ x = \frac{e^x - e^{-x}}{2}\).
Hyperbolic functions are important because they help in modeling the shapes and curves that occur in both mathematics and physical sciences, like those you see in a hanging cable or chain, known as a catenary.
Notably, the inverse hyperbolic sine function, \(\sinh^{-1} x\), provides a method for finding the value of \(x\) given that the hyperbolic sine of \(y\) equals \(x\).
In the provided problem, we are tasked to prove a particular property involving this function, specifically that \(\sinh^{-1} t = \ln\left(t + \sqrt{t^2 + 1}\right)\). Understanding these relationships is vital because they simplify complex expressions and solve hyperbolic equations efficiently.

A quadratic equation is a second-order polynomial equation in a single variable \(x\) with a nonzero coefficient for \(x^2\). Its general form is \(ax^2 + bx + c = 0\).
Part of solving our exercise involves translating the equation \(e^{2y} = 2e^y t + 1\) into a quadratic form by letting \(e^y = x\). Doing so results in the quadratic equation: \(x^2 - 2tx - 1 = 0\).
The quadratic formula, \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), is then applied to solve for \(x\).
In our context, \(a = 1\), \(b = -2t\), and \(c = -1\). Solving this gives the roots \(x = t \pm \sqrt{t^2 + 1}\).
Since \(e^y\) (or \(x\) in our substitution) must always be positive, we select \(x = t + \sqrt{t^2 + 1}\) as our viable solution.

The natural logarithm, denoted as \(\ln\), is the logarithm to the base \(e\) (where \(e\) is approximately 2.71828). It is the inverse function of the exponential function. This means that if \(y = \ln(x)\), then \(e^y = x\).
Natural logarithms are common in natural processes and exponential growth or decay situations commonly found in science and engineering.
In our exercise, after solving the quadratic equation, we find\(e^y = t + \sqrt{t^2 + 1}\).
To find the value of \(y\), we simply apply the natural logarithm to both sides: \(y = \ln(t + \sqrt{t^2 + 1})\).
This shows that the statement \(\sinh^{-1} t = \ln(t + \sqrt{t^2 + 1})\) holds true, enclosing our proof with a natural logarithm, cementing the result of inverse hyperbolic trigonometric functions.