If \( \cosh x = \frac{5}{4} \), then \( \tanh 3x = \)
Show Hint
- \( \frac{25}{26} \)
- \( \frac{63}{65} \)
- \( \frac{65}{67} \)
- \( \frac{252}{265} \)
The Correct Option is B
Approach Solution - 1
Given: \( \cosh x = \frac{5}{4} \). We want to find \( \tanh 3x \).
First, recall that \(\cosh^2 x - \sinh^2 x = 1\). Thus, \(\sinh x = \sqrt{\cosh^2 x - 1}\).
Substitute \(\cosh x = \frac{5}{4}\):
\(\sinh x = \sqrt{\left(\frac{5}{4}\right)^2 - 1} = \sqrt{\frac{25}{16} - \frac{16}{16}} = \sqrt{\frac{9}{16}} = \frac{3}{4}\).
Now, compute \(\tanh x = \frac{\sinh x}{\cosh x} = \frac{\frac{3}{4}}{\frac{5}{4}} = \frac{3}{5}\).
We utilize the triple angle formula for \(\tanh 3x\):
\(\tanh 3x = \frac{3\tanh x + \tanh^3 x}{1 + 3\tanh^2 x}\).
Substitute \(\tanh x = \frac{3}{5}\):
\(\tanh^3 x = \left(\frac{3}{5}\right)^3 = \frac{27}{125}\),
\(3\tanh x = 3 \times \frac{3}{5} = \frac{9}{5}\),
\(3\tanh^2 x = 3 \left(\frac{3}{5}\right)^2 = 3 \times \frac{9}{25} = \frac{27}{25}\).
Calculate:
\(\tanh 3x = \frac{\frac{9}{5} + \frac{27}{125}}{1 + \frac{27}{25}}\).
Simplify the terms:
\(\frac{9}{5} = \frac{225}{125}\),\(\frac{9}{5} + \frac{27}{125} = \frac{225}{125} + \frac{27}{125} = \frac{252}{125}\),
\(1 + \frac{27}{25} = \frac{25}{25} + \frac{27}{25} = \frac{52}{25}\).
Thus, \(\tanh 3x = \frac{\frac{252}{125}}{\frac{52}{25}} = \frac{252}{125} \times \frac{25}{52} = \frac{252 \times 25}{125 \times 52} = \frac{6300}{6500} = \frac{63}{65}\).
However, this is incorrect. The correct simplification reveals:
\(\tanh 3x = \frac{252 \times 25}{6500} = \frac{6300}{6500} = \frac{63}{65}\).
The correct answer is \(\frac{63}{65}\).
Approach Solution -2
\[ \cosh x = \frac{5}{4} \]
We are required to find \( \tanh 3x \).
Step 1: Use identity \( \cosh^2 x - \sinh^2 x = 1 \)
Given \( \cosh x = \frac{5}{4} \), use the identity:
\[ \sinh^2 x = \cosh^2 x - 1 = \left( \frac{5}{4} \right)^2 - 1 = \frac{25}{16} - 1 = \frac{9}{16} \Rightarrow \sinh x = \frac{3}{4} \]
Step 2: Compute \( \tanh x \)
\[ \tanh x = \frac{\sinh x}{\cosh x} = \frac{3/4}{5/4} = \frac{3}{5} \]
Step 3: Use triple angle identity for hyperbolic tangent
\[ \tanh 3x = \frac{3 \tanh x + \tanh^3 x}{1 + 3 \tanh^2 x} \]
Substitute \( \tanh x = \frac{3}{5} \):
- \( \tanh^2 x = \left( \frac{3}{5} \right)^2 = \frac{9}{25} \)
- \( \tanh^3 x = \left( \frac{3}{5} \right)^3 = \frac{27}{125} \)
Numerator:
\[ 3 \cdot \frac{3}{5} + \frac{27}{125} = \frac{9}{5} + \frac{27}{125} = \frac{225 + 27}{125} = \frac{252}{125} \]
Denominator:
\[ 1 + 3 \cdot \frac{9}{25} = 1 + \frac{27}{25} = \frac{52}{25} \]
Step 4: Divide numerator by denominator
\[ \tanh 3x = \frac{252}{125} \div \frac{52}{25} = \frac{252}{125} \cdot \frac{25}{52} = \frac{252 \cdot 25}{125 \cdot 52} = \frac{6300}{6500} = \frac{63}{65} \]
Final Answer:
\[ \boxed{\frac{63}{65}} \]
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