Problem 12
Question
Prove the identity. \(\sinh 2 x=2 \sinh x \cosh x\)
Step-by-Step Solution
Verified Answer
Using the definitions of hyperbolic sine and cosine functions, we compute the expressions for \(\sinh(2x)\) and \(2 \sinh x \cosh x\). Upon simplification, we find that both expressions are equal, thus proving the identity: \(\sinh 2x = 2 \sinh x \cosh x\).
1Step 1: Recall the definitions of \(\sinh\) and \(\cosh\)
The hyperbolic sine and cosine functions are defined as follows:
\(\sinh x = \dfrac{e^x - e^{-x}}{2}\)
\(\cosh x = \dfrac{e^x + e^{-x}}{2}\)
Note that these definitions are derived from the exponential functions, and are used to express the given identity in terms of exponential functions.
2Step 2: Compute the expression for \(\sinh(2x)\)
Using the definition of the hyperbolic sine function, we can compute the expression for \(\sinh(2x)\) as follows:
\(\sinh(2x) = \dfrac{e^{2x} - e^{-2x}}{2}\)
3Step 3: Compute the expression for \(2 \sinh x \cosh x\)
Using the definitions of the hyperbolic sine and cosine functions, we can compute the expression for \(2 \sinh x \cosh x\) as follows:
\(2 \sinh x \cosh x = 2 \left(\dfrac{e^x - e^{-x}}{2}\right) \left(\dfrac{e^x + e^{-x}}{2}\right)\)
Now, let's simplify the expression:
\(2 \sinh x \cosh x = (e^x - e^{-x})(e^x + e^{-x})\)
4Step 4: Show that both expressions are equal
Now, we will show that both expressions we derived in Step 2 and Step 3 are equal:
\(\sinh(2x) = \dfrac{e^{2x} - e^{-2x}}{2}\)
\(2 \sinh x \cosh x = (e^x - e^{-x})(e^x + e^{-x})\)
Let's expand the expression for \(2 \sinh x \cosh x\):
\(2 \sinh x \cosh x = e^{2x} + e^{-2x} - e^{2x} - e^{-2x}\)
Now, we can see that both expressions are equal:
\(\sinh(2x) = 2 \sinh x \cosh x\)
Hence, we have proved the identity:
\(\sinh 2x = 2 \sinh x \cosh x\)
Key Concepts
Identity ProofHyperbolic SineHyperbolic Cosine
Identity Proof
Proving mathematical identities can seem tricky, but they help reinforce the use of definitions and algebraic manipulation. To prove an identity, we need to show that both sides of an equation represent the same numerical value for all values of the variable involved. In the case of hyperbolic functions like \(\sinh\) and \(\cosh\), we use their exponential definitions to guide our proof.
The process typically involves
Through careful expansion and simplification, we'll see that the identity holds true for any value of \(x\). It's like solving a puzzle, where our job is to make every piece fit perfectly to complete the equation.
The process typically involves
- Expressing each function using its definition.
- Simplifying or transforming each side of the equation to reveal their equality.
Through careful expansion and simplification, we'll see that the identity holds true for any value of \(x\). It's like solving a puzzle, where our job is to make every piece fit perfectly to complete the equation.
Hyperbolic Sine
The hyperbolic sine function, represented as \(\sinh(x)\), is analogous to the sine function in trigonometry but built with exponential functions. It is defined as:
\[ \sinh x = \dfrac{e^x - e^{-x}}{2} \]
Key characteristics of \(\sinh\) include:
\[ \sinh x = \dfrac{e^x - e^{-x}}{2} \]
Key characteristics of \(\sinh\) include:
- It represents the imaginary part of a complex number when expressed in terms of \(e^x\).
- Unlike the trigonometric sine, \(\sinh(x)\) can take on any real number value.
- \sinh(x) has no limit as \(x\) becomes very large or very small.
Hyperbolic Cosine
The hyperbolic cosine function, shown as \(\cosh(x)\), shares conceptual similarities with the trigonometric cosine function, but like \(\sinh(x)\), it is rooted in exponentials. Its definition is:
\[ \cosh x = \dfrac{e^x + e^{-x}}{2} \]
Useful attributes of \(\cosh\) include:
\[ \cosh x = \dfrac{e^x + e^{-x}}{2} \]
Useful attributes of \(\cosh\) include:
- It calculates the average of two exponential functions, providing stability.
- Unlike \(\sinh(x)\), \(\cosh(x)\) embodies even function properties, meaning the graph is symmetric about the y-axis.
- The smallest value of \(\cosh(x)\) is 1, which occurs at \(x = 0\).
Other exercises in this chapter
Problem 11
Use the method of cylindrical shells to find the volume of the solid generated by revolving the region bounded by the graphs of the equations and/or inequalitie
View solution Problem 11
In Exercises \(9-40\), sketch the region bounded by the graphs of the given equations and find the area of that region. $$ y=x^{2}-2 x-1, \quad y=-e^{x}-1, \qua
View solution Problem 12
Find the centroid of the region bounded by the graphs of the given equations. $$ y=\sqrt{x}, \quad y=0, \quad x=1, \quad x=4 $$
View solution Problem 12
Find the arc length of the graph of the given equation from \(P\) to \(Q\) or on the specified interval. $$ y=\left(2-x^{2 / 3}\right)^{3 / 2} ; \quad[1,2] $$
View solution