Problem 4
Question
Show that for any real numbers \(x\) and \(y\), $$ \sinh (x+y)=\sinh (x) \cosh (y)+\sinh (y) \cosh (x) $$ and $$ \cosh (x+y)=\cosh (x) \cosh (y)+\sinh (x) \sinh (y) $$
Step-by-Step Solution
Verified Answer
The identities for \(\sinh(x+y)\) and \(\cosh(x+y)\) are verified using definitions and properties of exponents.
1Step 1: Recall Definitions of Hyperbolic Functions
First, recall the definitions of the hyperbolic sine and cosine functions: \[ \sinh(x) = \frac{e^x - e^{-x}}{2}, \quad \cosh(x) = \frac{e^x + e^{-x}}{2}. \] These are analogous to the definitions of the sine and cosine for circular functions, but use exponential functions instead.
2Step 2: Expand Hyperbolic Sine of Sum
Use the definition of \(\sinh(x+y)\): \[ \sinh(x+y) = \frac{e^{x+y} - e^{-(x+y)}}{2}. \] Expand and simplify this using properties of exponents:\[ \sinh(x+y) = \frac{e^x e^y - e^{-x} e^{-y}}{2}. \]
3Step 3: Expand Expressions for Hyperbolic Product
Express the terms \(e^x e^y\) and \(e^{-x} e^{-y}\) using the identities:\[ e^x e^y = (e^x + e^{-x})(e^y + e^{-y})/4 - (e^x - e^{-x})(e^y - e^{-y})/4, \]\[ e^{-x} e^{-y} = (e^x + e^{-x})(e^y + e^{-y})/4 + (e^x - e^{-x})(e^y - e^{-y})/4. \]
4Step 4: Simplify Products to Hyperbolic Functions
Substitute the above expressions and simplify:\[ \sinh(x+y) = \frac{1}{2} \left( (\frac{e^x - e^{-x}}{2})(\frac{e^y + e^{-y}}{2}) + (\frac{e^y - e^{-y}}{2})(\frac{e^x + e^{-x}}{2}) \right). \]This simplifies to:\[ \sinh(x+y) = \sinh(x) \cosh(y) + \sinh(y) \cosh(x). \]
5Step 5: Prove Hyperbolic Cosine of Sum
Now recall and expand: \[ \cosh(x+y) = \frac{e^{x+y} + e^{-(x+y)}}{2} = \frac{e^x e^y + e^{-x} e^{-y}}{2}. \]Simplify using the same identities from hyperbolic sine:\[ \cosh(x+y) = \frac{1}{2}(\frac{e^x + e^{-x}}{2})(\frac{e^y + e^{-y}}{2}) + \frac{1}{2}(\frac{e^x - e^{-x}}{2})(\frac{e^y - e^{-y}}{2}). \]This simplifies to: \[ \cosh(x+y) = \cosh(x) \cosh(y) + \sinh(x) \sinh(y). \]
6Step 6: Conclude Proof
Thus, we have shown that both hyperbolic trigonometric identities:\[ \sinh(x+y) = \sinh(x) \cosh(y) + \sinh(y) \cosh(x) \]and\[ \cosh(x+y) = \cosh(x) \cosh(y) + \sinh(x) \sinh(y) \]are true for any real numbers \(x\) and \(y\). This completes the proof.
Key Concepts
Hyperbolic IdentitiesReal NumbersExponential Functions
Hyperbolic Identities
Hyperbolic identities are similar to trigonometric identities but involve hyperbolic functions such as sinh (sine hyperbolic) and cosh (cosine hyperbolic). These identities are useful in various areas of mathematics and physics, particularly in solving differential equations and in the field of hyperbolic geometry.
The hyperbolic identities covered in this exercise show how the hyperbolic sine and cosine of a sum can be expressed in terms of the hyperbolic sine and cosine of the individual terms. Specifically:
The hyperbolic identities covered in this exercise show how the hyperbolic sine and cosine of a sum can be expressed in terms of the hyperbolic sine and cosine of the individual terms. Specifically:
- For sinh: \( \sinh(x+y) = \sinh(x) \cosh(y) + \sinh(y) \cosh(x) \)
- For cosh: \( \cosh(x+y) = \cosh(x) \cosh(y) + \sinh(x) \sinh(y) \)
Real Numbers
Real numbers are the set of numbers that include all the rational and irrational numbers. This means they encompass integers, fractions, and non-repeating, non-terminating decimals. Each real number corresponds to a point on the number line.
In the context of hyperbolic functions, real numbers serve as inputs or arguments for these functions. When the exercise says "for any real numbers \(x\) and \(y\)", it means that these hyperbolic identities are universally applicable for any possible pair of real number inputs.
The concept of real numbers is central because:
In the context of hyperbolic functions, real numbers serve as inputs or arguments for these functions. When the exercise says "for any real numbers \(x\) and \(y\)", it means that these hyperbolic identities are universally applicable for any possible pair of real number inputs.
The concept of real numbers is central because:
- They provide a comprehensive framework for most of algebraic and calculus operations.
- They allow us to apply hyperbolic and other mathematical functions in practical and theoretical problems.
Exponential Functions
Exponential functions are mathematical functions of the form \(f(x) = a^{x}\), where \(a\) is a constant and \(e\) is the base of the natural logarithm when specifically dealing with personal exponential growth, represented as \(e^x\).
In the realm of hyperbolic functions, exponential functions play a critical role because hyperbolic sine and cosine are defined using them:
Exponential functions' properties, such as their unique characteristic of continuous growth, enable applications ranging from simple calculations to complex differential equations and hyperbolic identity proofs. Recognizing exponential functions within hyperbolic identities can simplify and provide insight into solving broader mathematical problems.
In the realm of hyperbolic functions, exponential functions play a critical role because hyperbolic sine and cosine are defined using them:
- \(\sinh(x) = \frac{e^x - e^{-x}}{2}\)
- \(\cosh(x) = \frac{e^x + e^{-x}}{2}\)
Exponential functions' properties, such as their unique characteristic of continuous growth, enable applications ranging from simple calculations to complex differential equations and hyperbolic identity proofs. Recognizing exponential functions within hyperbolic identities can simplify and provide insight into solving broader mathematical problems.
Other exercises in this chapter
Problem 3
Suppose \(a\) is a positive real number and \(f: \mathbb{R} \rightarrow \mathbb{R}\) is defined by \(f(x)=a^{x}\). Show that \(f^{\prime}(x)=a^{x} \log (a)\).
View solution Problem 3
Show that for any \(x \in \mathbb{R}\) $$ \sin (2 x)=2 \sin (x) \cos (x) $$ and $$ \cos (2 x)=\cos ^{2}(x)-\sin ^{2}(x) $$
View solution Problem 4
Show that for any \(x \in \mathbb{R}\) $$ \sin ^{2}(x)=\frac{1-\cos (2 x)}{2} $$ and $$ \cos ^{2}(x)=\frac{1+\cos (2 x)}{2} $$
View solution Problem 5
Show that for any real number \(x,\) $$ \cosh ^{2}(x)-\sinh ^{2}(x)=1 $$
View solution