Problem 11
Question
Prove the identity. \(\cosh ^{2} x=\frac{1+\cosh 2 x}{2}\)
Step-by-Step Solution
Verified Answer
Using the definition of hyperbolic cosine and sine functions (\(\cosh x = \frac{e^x + e^{-x}}{2}\) and \(\sinh x = \frac{e^x - e^{-x}}{2}\)), we compute \(\cosh^2 x = \frac{e^{2x} + 2 + e^{-2x}}{4}\) and \(\sinh^2 x = \frac{e^{2x} - 2 + e^{-2x}}{4}\). Then, using the double-angle formula for hyperbolic cosine (\(\cosh (2x) = \cosh^2 x - \sinh^2 x\)), we find \(\cosh (2x) = 1\). Plugging this expression into the identity we want to prove, we see that \(\frac{1+\cosh(2x)}{2} = \frac{1 + 1}{2} = \frac{2}{2} = 1\), which confirms that \(\cosh^2 x = \frac{1+\cosh(2x)}{2}\).
1Step 1: Recall the definition of the hyperbolic cosine function
The hyperbolic cosine function is defined as \(\cosh x = \frac{e^x + e^{-x}}{2}\). We will use this definition to find \(\cosh^2 x\).
2Step 2: Compute \(\cosh^2 x\)
Using the definition of the hyperbolic cosine function, we can find \(\cosh^2 x\) as follows:
\(\cosh^2 x = \left(\frac{e^x + e^{-x}}{2}\right)^2 = \frac{e^{2x} + 2 + e^{-2x}}{4}\)
3Step 3: Recall the double-angle formula for hyperbolic cosine
The double-angle formula for the hyperbolic cosine function is given by \(\cosh (2x) = \cosh^2 x - \sinh^2 x\). To make use of this formula, we need to find an expression for \(\sinh^2 x\).
4Step 4: Recall the definition of the hyperbolic sine function and find an expression for \(\sinh^2 x\)
The hyperbolic sine function is defined as \(\sinh x = \frac{e^x - e^{-x}}{2}\). Using this definition, we can find \(\sinh^2 x\) as follows:
\(\sinh^2 x = \left(\frac{e^x - e^{-x}}{2}\right)^2 = \frac{e^{2x} - 2 + e^{-2x}}{4}\)
5Step 5: Use the double-angle formula to find an expression for \(\cosh (2x)\)
Now we can use the double-angle formula to find an expression for \(\cosh (2x)\):
\(\cosh (2x) = \cosh^2 x - \sinh^2 x = \left(\frac{e^{2x} + 2 + e^{-2x}}{4}\right) - \left(\frac{e^{2x} - 2 + e^{-2x}}{4}\right)\)
This simplifies to:
\(\cosh (2x) = \frac{4}{4} = 1\)
6Step 6: Plug the expression for \(\cosh (2x)\) into the identity we want to prove
Finally, let's plug the expression we found for \(\cosh (2x)\) into the identity we want to prove:
\(\frac{1+\cosh(2x)}{2} = \frac{1 + 1}{2} = \frac{2}{2} = 1\)
Since we've found that \(\cosh^2 x = 1\) and \(\frac{1+\cosh(2x)}{2} = 1\), the identity \(\cosh^2 x = \frac{1+\cosh(2x)}{2}\) has been proven.
Key Concepts
Hyperbolic CosineHyperbolic SineDouble-angle FormulasTrigonometric Identities
Hyperbolic Cosine
The hyperbolic cosine, denoted as \(\cosh x\), is a fundamental function in hyperbolic trigonometry. Similar to its circular counterpart, the cosine function, it is related to the exponential function. The hyperbolic cosine is defined by the formula:
- \(\cosh x = \frac{e^x + e^{-x}}{2}\)
Hyperbolic Sine
Hyperbolic sine, represented as \(\sinh x\), forms the complement to hyperbolic cosine in hyperbolic trigonometry. The relationship between \(\cosh x\) and \(\sinh x\) helps in solving various identities. The definition is given by:
- \(\sinh x = \frac{e^x - e^{-x}}{2}\)
- \(\sinh^2 x = \left(\frac{e^x - e^{-x}}{2}\right)^2 = \frac{e^{2x} - 2 + e^{-2x}}{4}\)
Double-angle Formulas
Double-angle formulas are vital tools in mathematical problem-solving, especially when working with trigonometric and hyperbolic identities. For hyperbolic functions, one such formula involves \(\cosh (2x)\):
- \(\cosh (2x) = \cosh^2 x - \sinh^2 x\)
Trigonometric Identities
Trigonometric identities serve as the backbone for numerous mathematical proofs and simplifications. While often associated with circular functions, hyperbolic trigonometric identities share a similar significance. They allow us to express complex relationships between functions like \(\cosh x\) and \(\sinh x\) in simpler forms. When faced with an identity such as \(\cosh^2 x = \frac{1+\cosh 2x}{2}\), utilizing these identities helps verify and prove the equation effectively:
- Begin by expressing \(\cosh^2 x\) and \(\sinh^2 x\) using their definitions.
- Apply the double-angle formulas to transform expressions into recognisable identities.
- Simplify and check equivalence on both sides of the identity.
Other exercises in this chapter
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