We are tasked to calculate $\cosh(x)$, $\text{sech}(x)$, $\text{csch}(x)$, and $\tanh(x)$ given that $\sinh(x) = \frac{1}{3}$.

Step 1: Relation between $\sinh(x)$ and $\cosh(x)$

From the hyperbolic identity: $\cosh^2(x) - \sinh^2(x) = 1$ Substituting $\sinh(x) = \frac{1}{3}$: $\cosh^2(x) - \left(\frac{1}{3}\right)^2 = 1$ $\cosh^2(x) - \frac{1}{9} = 1$ $\cosh^2(x) = 1 + \frac{1}{9} = \frac{10}{9}$ $\cosh(x) = \sqrt{\frac{10}{9}} = \frac{\sqrt{10}}{3}$

Step 2: Calculate the other hyperbolic functions

(a) $\text{sech}(x)$:

$\text{sech}(x) = \frac{1}{\cosh(x)}$ Substituting $\cosh(x) = \frac{\sqrt{10}}{3}$: $\text{sech}(x) = \frac{1}{\frac{\sqrt{10}}{3}} = \frac{3}{\sqrt{10}} = \frac{3\sqrt{10}}{10}$

(b) $\text{csch}(x)$:

$\text{csch}(x) = \frac{1}{\sinh(x)}$ Substituting $\sinh(x) = \frac{1}{3}$: $\text{csch}(x) = \frac{1}{\frac{1}{3}} = 3$

(c) $\tanh(x)$:

$\tanh(x) = \frac{\sinh(x)}{\cosh(x)}$ Substituting $\sinh(x) = \frac{1}{3}$ and $\cosh(x) = \frac{\sqrt{10}}{3}$: $\tanh(x) = \frac{\frac{1}{3}}{\frac{\sqrt{10}}{3}} = \frac{1}{\sqrt{10}} = \frac{\sqrt{10}}{10}$

Final Answers:

1. $\cosh(x) = \frac{\sqrt{10}}{3}$
2. $\text{sech}(x) = \frac{3\sqrt{10}}{10}$
3. $\text{csch}(x) = 3$
4. $\tanh(x) = \frac{\sqrt{10}}{10}$

Would you like further details on any of these calculations? Here are related questions for practice:

1. What is the significance of hyperbolic functions in geometry?
2. Can you prove the hyperbolic identity used here?
3. How do hyperbolic functions relate to exponential functions?
4. What is the range of $\tanh(x)$?
5. How do you calculate hyperbolic functions for complex numbers?

Tip: Always verify the domain and range of hyperbolic functions when solving such problems.