Question

For any real number \(n\in \mathbb{R}\), \[ (\cosh x+\sinh x)^n= \]

• A. \(\cosh nx-\sinh nx\)
• B. \(\cosh nx+\sinh nx\)
• C. \(\cosh^2 nx+2\sinh nx\)
• D. \(\cosh nx-\sinh nx\)

Hint:

Always remember the key identity: \[ \cosh x+\sinh x=e^x \] This identity makes powers of \(\cosh x+\sinh x\) very easy to simplify.

Answer 1

The Correct Option is B

Step 1: Recall the definitions of cosh and sinh.
The hyperbolic functions are defined as: $ \cosh x = \frac{e^x + e^{-x}}{2} $ and $ \sinh x = \frac{e^x - e^{-x}}{2} $.
Step 2: Add cosh x and sinh x.
$ \cosh x + \sinh x = \frac{e^x + e^{-x}}{2} + \frac{e^x - e^{-x}}{2} = \frac{2e^x}{2} = e^x $. This is a very clean result: the sum is simply $ e^x $.
Step 3: Raise both sides to the power n.
$ (\cosh x + \sinh x)^n = (e^x)^n = e^{nx} $.
Step 4: Convert e^{nx} back to hyperbolic form.
Just as $ \cosh x + \sinh x = e^x $, replacing $ x $ by $ nx $ gives: $ \cosh(nx) + \sinh(nx) = e^{nx} $.
Step 5: Combine the results.
Therefore $ (\cosh x + \sinh x)^n = e^{nx} = \cosh(nx) + \sinh(nx) $. This is the hyperbolic analogue of De Moivre's theorem.
Step 6: State the final answer.
\[ \boxed{\cosh nx + \sinh nx} \]