Question

If \( \sinh x = \frac{12}{5} \), then \( \sinh 3x + \cosh 3x \) = ?

• A. \( 125 \)
• B. \( 144 \)
• C. \( 169 \)
• D. \( 216 \)

Hint:

When solving hyperbolic equations, use the standard identities for \( \sinh x \) and \( \cosh x \) to simplify expressions. Recognizing the triple-angle formulas can make calculations easier.

Answer 1

The Correct Option is A

We are given: \[ \sinh x = \frac{12}{5} \] 

Step 1: Compute \( \cosh x \) using the identity 
Using the hyperbolic identity: \[ \cosh^2 x - \sinh^2 x = 1 \] Substituting \( \sinh x = \frac{12}{5} \): \[ \cosh^2 x - \left( \frac{12}{5} \right)^2 = 1 \] \[ \cosh^2 x - \frac{144}{25} = 1 \] \[ \cosh^2 x = \frac{144}{25} + \frac{25}{25} = \frac{169}{25} \] \[ \cosh x = \frac{13}{5} \] 

Step 2: Compute \( \sinh 3x \) and \( \cosh 3x \) using triple angle formulas 
The standard identities for triple angles are: \[ \sinh 3x = 3 \sinh x \cosh^2 x + \sinh^3 x \] \[ \cosh 3x = 4\cosh^3 x - 3\cosh x \] Substituting the values: \[ \sinh 3x = 3 \times \frac{12}{5} \times \left(\frac{13}{5}\right)^2 + \left(\frac{12}{5}\right)^3 \] \[ = 3 \times \frac{12}{5} \times \frac{169}{25} + \frac{1728}{125} \] \[ = \frac{3 \times 12 \times 169}{125} + \frac{1728}{125} \] \[ = \frac{6084}{125} + \frac{1728}{125} \] \[ = \frac{7812}{125} \] \[ = 62.5 \] Similarly, \[ \cosh 3x = 4 \times \left(\frac{13}{5}\right)^3 - 3 \times \frac{13}{5} \] \[ = 4 \times \frac{2197}{125} - \frac{39}{5} \] \[ = \frac{8788}{125} - \frac{975}{125} \] \[ = \frac{7813}{125} \] 

Step 3: Compute \( \sinh 3x + \cosh 3x \) 
\[ \sinh 3x + \cosh 3x = 62.5 + 62.5 = 125 \] Thus, the correct answer is: \[ \mathbf{125} \]