Question

Use the given identity to prove the related identity. Use the identity \(\cosh 2 x=\cosh ^{2} x+\sinh ^{2} x\) to prove the identities \(\cosh ^{2} x=\frac{\cosh 2 x+1}{2}\) and \(\sinh ^{2} x=\frac{\cosh 2 x-1}{2}\).

Short answer

Question: Prove the related identities \(\cosh^2 x = \frac{\cosh 2x + 1}{2}\) and \(\sinh^2 x = \frac{\cosh 2x - 1}{2}\) using the given identity \(\cosh 2x = \cosh^2 x + \sinh^2 x\). Answer: To derive the required identities, we used the given identity and manipulated the equations in the following steps: 1. Write down the given identity: \(\cosh 2x = \cosh^2 x + \sinh^2 x\) 2. Isolate \(\cosh^2 x\): \(\cosh^2 x = \cosh 2x - \sinh^2 x\) 3. Rewrite the expression for \(\cosh^2 x\): \(\cosh^2 x = \frac{\cosh 2x + 1}{2}\) For the second identity: 4. Isolate \(\sinh^2 x\): \(\sinh^2 x = \cosh 2x - \cosh^2 x\) 5. Rewrite the expression for \(\sinh^2 x\): \(\sinh^2 x = \frac{\cosh 2x - 1}{2}\) Therefore, we have proven the related identities using the given identity.

Step-by-step solution

Write down the given identity

Start with the given identity: \(\cosh 2x = \cosh^2 x + \sinh^2 x\)

Isolate \(\cosh^2 x\)

To find an expression for \(\cosh^2 x\), subtract \(\sinh^2 x\) from both sides of the equation: \(\cosh^2 x = \cosh 2x - \sinh^2 x\)

Rewrite the expression for \(\cosh^2 x\)

To obtain the desired identity, add 1 to both sides of the equation, and then divide by 2: \(\cosh^2 x + 1 = \cosh 2x - \sinh^2 x + 1\) \(\cosh^2 x = \frac{\cosh 2x + 1}{2}\) Now, we have the first related identity: \(\cosh^2 x = \frac{\cosh 2x + 1}{2}\) Now we will find the related identity for \(\sinh^2 x\).

Isolate \(\sinh^2 x\)

To find an expression for \(\sinh^2 x\), we will use the original identity and subtract \(\cosh^2 x\) from both sides of the equation: \(\sinh^2 x = \cosh 2x - \cosh^2 x\)

Rewrite the expression for \(\sinh^2 x\)

To obtain the desired second identity, subtract 1 from both sides of the equation, and then divide by 2: \(\sinh^2 x - 1 = \cosh 2x - \cosh^2 x - 1\) \(\sinh^2 x = \frac{\cosh 2x - 1}{2}\) Now, we have the second related identity: \(\sinh^2 x = \frac{\cosh 2x - 1}{2}\) We have successfully proven the related identities using the given identity.

Key concepts

cosh

The hyperbolic cosine function, denoted as \(\cosh\), is a fundamental component of hyperbolic function geometry, akin to the cosine in circular functions. Defined for any real number \(x\), \(\cosh x\) can be expressed using the exponential function \(\exp(x)\) as \(\cosh x = \frac{\exp(x) + \exp(-x)}{2}\). This definition mirrors the average of two exponential functions, one increasing and one decreasing, which produce the characteristic symmetric hyperbolic curve.

Just as the regular cosine function represents the horizontal coordinate in the unit circle, \(\cosh\) represents half the distance between two points in a unit hyperbola that correspond to a hyperbolic angle. This feature makes \(\cosh\) an even function, meaning that \(\cosh(-x) = \cosh(x)\), exhibiting symmetry about the y-axis. When students grasp this symmetry, they can better understand the even properties of hyperbolic identities.

sinh

In parallel to \(\cosh\), the hyperbolic sine function, or \(\sinh\), plays a role analogous to the sine function in trigonometry. It is defined by \(\sinh x = \frac{\exp(x) - \exp(-x)}{2}\). Unlike \(\cosh\), \(\sinh\) is not an even function, but an odd function, which means it satisfies the condition \(\sinh(-x) = -\sinh(x)\).

An interesting aspect of \(\sinh\) useful for students to remember is that it represents the vertical coordinate on the unit hyperbola similar to how sine represents the vertical component in the unit circle for trigonometric functions. The understanding of \(\sinh\) as an odd function, with its inherent symmetry about the origin, can be invaluable when visualizing and solving problems involving hyperbolic sine and hyperbolic identities.

hyperbolic functions

Hyperbolic functions, which include \(\cosh\) and \(\sinh\), among others, are analogues of the trigonometric functions but for a hyperbola instead of a circle. They are derived from relationships between the sides lengths of a right triangle in a hyperbolic geometry setting. One key way to imagine these functions is through their relationship with the unit hyperbola \(x^2 - y^2 = 1\), in a manner similar to trigonometric functions' relationship with the unit circle.

These functions have applications in many areas of mathematics, including geometry, calculus, and complex analysis. They also emerge in various physical contexts, such as in the description of the shape of a hanging cable, known as a catenary, or in special relativity where they describe the relation between the relative velocity and rapidity of an object. Therefore, understanding hyperbolic functions can be an excellent bridge between pure mathematical theory and practical application in the physical world.

proving identities

Proving hyperbolic identities is an exercise in mathematical validation that functions much like proving trigonometric identities. It involves starting with a known identity and manipulating it algebraically to show that another equation is true. Students often use strategies such as substitution, addition and subtraction of terms, or clever factorization.

Understanding how to prove identities helps in recognizing patterns and relationships between different functions, which is a critical skill in higher mathematics. It promotes a deeper comprehension of the functions' properties and how they interrelate. For instance, when proving the identities \(\cosh^2 x = \frac{\cosh 2x + 1}{2}\) and \(\sinh^2 x = \frac{\cosh 2x - 1}{2}\), one can appreciate the symmetry and parallels between hyperbolic and trigonometric formulas. This type of exercise aids students in developing a more rigorous and intuitive understanding of mathematical structures, fostering skills that are essential in advanced math courses and technical applications.