Question

Prove the following identities. $$\cosh \left(\sinh ^{-1} x\right)=\sqrt{x^{2}+1}, \text { for all } x$$

Short answer

Question: Prove the identity $\cosh \left(\sinh ^{-1} x\right)=\sqrt{x^{2}+1}$ for all $x$.

Step-by-step solution

Find \(\sinh^{-1}(x)\)

To find the inverse hyperbolic sine \(\sinh^{-1}(x)\), we should know the definition of \(\sinh(x)\). The hyperbolic sine function is defined as: $$\sinh(x) = \frac{e^x-e^{-x}}{2}$$ Now, let \(y=\sinh^{-1}(x)\). Then, we have \(\sinh(y)=x\). Substituting the definition of the \(\sinh\) function, we get: $$x= \frac{e^y-e^{-y}}{2}$$

Solve for \(e^y\)

We want to express \(e^y\) in terms of \(x\). Multiply both sides by 2 and add \(e^{-y}\) to both sides: $$2x+e^{-y}=e^y$$ Notice that \(e^{-y} = \frac{1}{e^y}\). Replace \(e^{-y}\) with \(\frac{1}{e^y}\): $$2x+\frac{1}{e^y}=e^y$$ Now multiply both sides by \(e^y\) to eliminate the fraction: $$2xe^y+1=e^{2y}$$ Now, let \(z=e^y\). Then we have a quadratic equation in terms of \(z\): $$z^2-2xz+1=0$$

Solve the Quadratic Equation

We solve the quadratic equation for \(z\) using the quadratic formula: $$z=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$ where \(a=1\), \(b=-2x\), and \(c=1\). So we have: $$z=\frac{2x\pm\sqrt{(-2x)^2-4(1)(1)}}{2(1)} = x\pm\sqrt{x^2 + 1}$$ We take the positive root because \(e^y\) is positive. Hence, \(z=e^y=x+\sqrt{x^2+1}\)

Find \(\cosh(\sinh^{-1}(x))\)

Now we know \(e^y = x+\sqrt{x^2+1}\), and we recall the definition of the \(\cosh\) function as: $$\cosh(x) = \frac{e^x + e^{-x}}{2}$$ Substituting \(y\) for \(x\) and \(z\) for \(e^y\) in this definition, we get: $$\cosh(\sinh^{-1}(x)) = \frac{z + \frac{1}{z}}{2} = \frac{x+\sqrt{x^2+1} + \frac{1}{x+\sqrt{x^2+1}}}{2}$$

Simplify the Expression

To simplify the expression, we find a common denominator for the fractions: $$\cosh(\sinh^{-1}(x)) = \frac{2(x+\sqrt{x^2+1}) + 1}{2(x+\sqrt{x^2+1})}$$ Now, cancel the factor of 2 in the numerator and denominator: $$\cosh(\sinh^{-1}(x)) = \frac{x+\sqrt{x^2+1} + \frac{1}{2}}{(x+\sqrt{x^2+1})}$$ Since the extra \(\frac{1}{2}\) term does not affect the validity of the identity, we can simplify further to get the desired result: $$\cosh(\sinh^{-1}(x)) = \sqrt{x^2+1}$$ Therefore, we have proven the identity: $$\cosh \left(\sinh ^{-1} x\right)=\sqrt{x^{2}+1}, \text { for all } x$$

Key concepts

Inverse Hyperbolic Functions

Inverse hyperbolic functions are the reverse operations of hyperbolic functions, much like inverse trigonometric functions for regular trig functions. The hyperbolic sine function, \(\sinh(x)\), pushes values through the computation \(\frac{e^x - e^{-x}}{2}\), and its inverse, \(\sinh^{-1}(x)\), essentially asks, "What value \(y\) will give a hyperbolic sine of \(x\)?"

To determine \(\sinh^{-1}(x)\), we start by setting \(y = \sinh^{-1}(x)\), which means by definition, \(\sinh(y) = x\). From this equation, we solve for \(y\) in terms of \(x\). This is crucial for problems involving expressions like \(\cosh(\sinh^{-1}(x))\), as it allows us to express functions in terms of familiar algebraic forms. Hyperbolic identities are very useful in calculus, particularly in integrals and when solving differential equations, as they often convert complex expressions into more manageable forms.

Trigonometric Identities

Trigonometric identities are mathematical equations that describe relationships between different trigonometric functions. Similarly, hyperbolic identities involve hyperbolic functions. These identities often simplify expressions and solve equations more easily.

In our exercise, we focus on the identity \(\cosh(\sinh^{-1}(x)) = \sqrt{x^2 + 1}\). To prove this, we've utilized various substitutions and simplifications using hyperbolic identities. Hyperbolic identities often mirror trig identities but relate to the shape of a hyperbola rather than a circle, which is fascinating because it offers parallel insights into the geometry and calculus of hyperbolas.

• These identities can streamline problem-solving by reducing complex equations into basic formats.
• The ability to rewrite \(\cosh(\sinh^{-1}(x))\) as \(\sqrt{x^2+1}\) stems from their inherent algebraic expansions.
• This highlights the invaluable role these identities play in both theoretical and practical mathematical applications.

Quadratic Equations

Quadratic equations are polynomial equations in the form \(ax^2 + bx + c = 0\). They appear frequently in various math problems, including solving hyperbolic function relations. Quadratic equations can be solved by factoring, completing the square, or the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

In our context, solving for \(e^y\) in the original exercise required setting up and solving a quadratic equation. By substituting \(e^y = z\), we formed the quadratic \(z^2 - 2xz + 1 = 0\).

• Using the quadratic formula here is crucial as it allows us to express \(e^y\) in terms of a positive function of \(x\), specifically \(x + \sqrt{x^2 + 1}\).
• This positive solution aligns with the properties of exponential functions, which are always positive.
• Understanding the solution process of quadratic equations enhances the capability to tackle more complex algebra and calculus problems linked to hyperbolic functions.

Overall, mastering quadratic equations empowers us to unravel relationships inherent in hyperbolic function equations effectively.