Question

Verify the following identities. \(\cosh (x+y)=\cosh x \cosh y+\sinh x \sinh y\)

Short answer

Question: Verify the identity \(\cosh (x+y)=\cosh x \cosh y+\sinh x \sinh y\). Answer: The solution for verifying the given identity involves manipulating the expression \(\cosh (x+y)\) using the definitions of hyperbolic sine and cosine. After simplifying and combining terms, we arrive at the expression \(\cosh (x+y) = \cosh x \cosh y + \sinh x \sinh y\), verifying the given identity.

Step-by-step solution

Recall the definitions of hyperbolic sine and cosine

The hyperbolic sine and cosine functions are defined as follows: $$\cosh x = \frac{e^x + e^{-x}}{2}$$ $$\sinh x = \frac{e^x - e^{-x}}{2}$$

Express \(\cosh (x+y)\) using these definitions

Using the definition of hyperbolic cosine, we can write: $$\cosh (x+y) = \frac{e^{x+y} + e^{-(x+y)}}{2}$$

Simplify the exponent terms

We can simplify the exponent terms in the expression by using the exponent properties (\(a^{mn} = (a^m)^n\) and \(a^{-m} = \frac{1}{a^m}\)): $$\cosh (x+y) = \frac{e^x \cdot e^y + \frac{1}{e^x \cdot e^y}}{2}$$

Rewrite the expression using the definitions of hyperbolic sine and cosine again

Now, we rewrite the expression in terms of the hyperbolic functions of \(x\) and \(y\): $$\cosh (x+y) = \frac{(e^x + e^{-x})(e^y + e^{-y}) + (e^x - e^{-x})(e^y - e^{-y})}{4}$$

Expand and simplify the expression

Expanding the multiplied terms in the numerator, we get: $$\cosh (x+y) = \frac{e^{2x} + e^{2y} + e^x e^{-x} e^{2y} + e^x e^{-x} e^{2x}}{4} + \frac{e^{2x} - e^{2y} - e^x e^{-x} e^{2y} + e^x e^{-x} e^{2x}}{4}$$ Simplifying the terms, we obtain: $$\cosh (x+y) = \frac{2e^{2x} + 2e^{2y}}{4} + \frac{2e^{2x} - 2e^{2y}}{4}$$ Finally, combining the terms yields: $$\cosh (x+y) = \frac{e^{2x}}{2} + \frac{e^{2y}}{2} + \frac{e^{2x}}{2} - \frac{e^{2y}}{2}$$

Express the result using hyperbolic sine and cosine

Use the definitions of hyperbolic sine and cosine to rewrite the equation: $$\cosh (x+y) = \cosh x \cosh y + \sinh x \sinh y$$ We have verified the given identity, as the left-hand side of the equation now matches the right-hand side.

Key concepts

cosh function

The hyperbolic cosine function, often denoted as \( \cosh x \), is a counterpart to the trigonometric cosine function applied to hyperbolic geometry. Its definition is rooted in the exponential function:
\[ \cosh x = \frac{e^x + e^{-x}}{2} \]
This formula highlights that \( \cosh x \) can be seen as an average of two exponential functions. This unique characteristic allows \( \cosh x \) to exhibit properties suitable for describing hyperbolic curves and shapes.### Key Characteristics of \( \cosh x \):- **Even Function**: \( \cosh(-x) = \cosh(x) \), indicating symmetry about the y-axis.- **Range**: Always positive and increases as \( x \) gets larger.Functions like \( \cosh x \) are essential in many fields such as engineering and physics, particularly in scenarios involving hyperbolic growth patterns.

sinh function

The hyperbolic sine function, represented as \( \sinh x \), emerges from exploring hyperbolic geometry through the lens of exponential functions. Its definition is given by:
\[ \sinh x = \frac{e^x - e^{-x}}{2} \]
This expression defines \( \sinh x \) as half the difference of two exponential expressions, providing it with properties that mirror the standard sine function but in the hyperbolic plane.### Important Properties of \( \sinh x \):- **Odd Function**: \( \sinh(-x) = -\sinh(x) \), implying reflection symmetry about the origin.- **Range**: Includes all real numbers, allowing it to span positive and negative values depending on \( x \).\( \sinh x \) finds applications across numerous fields, especially wherever growth or decay in hyperbolic forms is modeled.

identity verification

Verifying mathematical identities is crucial to confirming mathematical truths and understanding relationships between different functions. Let's proceed to verify the hyperbolic identity:
\( \cosh(x + y) = \cosh x \cosh y + \sinh x \sinh y \)### Identity Verification Steps:- **Step 1**: Use the definitions of \( \cosh \) and \( \sinh \) to express \( \cosh(x+y) \) in exponential form: \[ \cosh(x+y) = \frac{e^{x+y} + e^{-(x+y)}}{2} \]- **Step 2**: Expand using properties of exponents like \( e^{a+b} = e^a \cdot e^b \): \[ \cosh(x+y) = \frac{(e^x + e^{-x})(e^y + e^{-y}) + (e^x - e^{-x})(e^y - e^{-y})}{4} \]- **Step 3**: Simplify the expression by expanding and combining like terms to match the identity form.The final step sees the expression for \( \cosh(x+y) \) equal to \( \cosh x \cosh y + \sinh x \sinh y \), thus verifying the identity. This showcases the elegant relationship between hyperbolic functions.