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: Use the standard hyperbolic identity.
We know that \[ \cosh x=\frac{e^x+e^{-x}}{2} \] and \[ \sinh x=\frac{e^x-e^{-x}}{2} \] Adding both, \[ \cosh x+\sinh x = \frac{e^x+e^{-x}}{2}+\frac{e^x-e^{-x}}{2} \] \[ \cosh x+\sinh x=e^x \]

Step 2: Raise both sides to power \(n\).
Therefore, \[ (\cosh x+\sinh x)^n=(e^x)^n \] \[ =e^{nx} \]

Step 3: Convert \(e^{nx}\) into hyperbolic form.
Again, using the identity \[ e^t=\cosh t+\sinh t \] Put \[ t=nx \] So, \[ e^{nx}=\cosh nx+\sinh nx \]

Step 4: Final conclusion.
Therefore, \[ \boxed{\cosh nx+\sinh nx} \]