Problem 14 - Calculus of a Single Variable

Question

In Exercises 7–14, verify the identity. $$ \cosh x+\cosh y=2 \cosh \frac{x+y}{2} \cosh \frac{x-y}{2} $$

Step-by-Step Solution

Answer

\cosh x + \cosh y = 2 \cosh \frac{x+y}{2} \cosh \frac{x-y}{2} has been verified.

Step 1: Write 'cosh' function in terms of 'e'

The first step is to rewrite the right-hand side of the identity in exponential form using the formula \(cosh(x) = (e^x + e^{-x})/2.\). Hence, \[ \cosh \frac{x+y}{2} \cosh \frac{x-y}{2} = \frac{(e^{(x+y)/2} + e^{-(x+y)/2}) (e^{(x-y)/2} + e^{-(x-y)/2})}{4} \]

Step 2: Simplification

Next, we simplify the above expression by expanding and rearranging it. \[ = \frac{(e^x + e^{-x} + e^y + e^{-y})}{4} = \frac{e^x}{4} + \frac{e^{-x}}{4} + \frac{e^y}{4} + \frac{e^{-y}}{4} \]

Step 3: Write the 'cosh' function back and compare

In this step, we will translate the exponential forms back to their cosh function representations. \[ = \frac{2cosh(x) + 2cosh(y)}{4} = cosh(x) + cosh(y) \] By comparing it with the left-hand side, we can verify that both sides are equal. Thus, the identity is proven true.

Key Concepts

Hyperbolic Cosine

The hyperbolic cosine function, often denoted as \(\cosh(x)\), is an essential part of hyperbolic functions, which have similarities to the regular trigonometric functions. This function is defined using exponential functions and plays a significant role in many mathematical equations related to hyperbolic identities. Unlike the regular cosine function, \(\cosh(x)\) creates a symmetric curve that opens upwards, and it is defined as:

• \(\cosh(x) = \frac{e^x + e^{-x}}{2}\)

Here, \(e\) represents the base of the natural logarithm, approximately equal to 2.718. The hyperbolic cosine function is critical in solving various problems in calculus and differential equations. Knowing this definition is vital as it helps in transitioning to different forms, like exponential, which is often required in proofs and problem-solving.

Exponential Form

Understanding the exponential form of hyperbolic functions is crucial. The expression of hyperbolic cosine in exponential terms simplifies the process of proving identities. For example, when the exercise asks us to verify a specific equation, converting hyperbolic functions using their exponential form can make calculations more straightforward.
In the identity \( \cosh(x+y) = \cosh(x) \cosh(y) + \sinh(x) \sinh(y) \), the exponential form helps break down complex expressions.

• This allows us to express \(\cosh\) and \(\sinh\) in terms of powers of \(e\):\(cosh(x) = \frac{e^x + e^{-x}}{2}\) and \(sinh(x) = \frac{e^x - e^{-x}}{2}\).

The exponential form is a powerful tool that facilitates the simplification of expressions and the verification of hyperbolic identities, making them easier to manage.

Identity Verification

Identity verification is the process of confirming that two expressions are equal. In the context of hyperbolic functions, this often involves clever manipulation of equations. Problems like the given exercise require rewriting one side of an equation in different terms to match the other side.

• This is often done using known identities and simplifying expressions.
• Knowing how to switch between forms, such as between hyperbolic and exponential, is crucial.

For instance, to verify \(\cosh x + \cosh y = 2 \cosh \frac{x+y}{2} \cosh \frac{x-y}{2}\), you need to express each part in a common form, often using exponential equivalents, to ascertain both sides are indeed equal. It's about stepping through logical transformations and thoroughly checking each manipulation, proving the equation's truth.

Function Simplification

Function simplification involves reducing complex equations into simplified, more manageable forms. This is essential in mathematics, as it helps to reveal the true characteristics of an equation, making the solution apparent. In the provided exercise, simplification helps in verifying identities.

• For example, expanding and rearranging exponential expressions can often lead to a simplified result that matches the original equation.
• Breaking down terms to inspect their components is a common simplification strategy.

For instance, if \(cosh(x) = \frac{(e^x + e^{-x})}{2}\), then expanding the expression in the problem can eventually lead to a form that clearly shows both sides are the same. Simplification is a powerful technique to navigate complex algebraic territories and confirm mathematical truths.