If `f(s) =(b^(s) +b^(-s))//2`, where `b gt 0`. Find `f(s+t) +f(s-t)`.

Options

A. `f(s) -f(t)`
B. `2f(s).f(t)`
C. `4f(s).f(t)`
D. `f(s)+f(t)`

Answer

To solve the problem, we need to find \( f(s+t) + f(s-t) \) given the function \( f(s) = \frac{b^s + b^{-s}}{2} \).

### Step-by-Step Solution:

1. **Calculate \( f(s+t) \)**:
\[ f(s+t) = \frac{b^{s+t} + b^{-(s+t)}}{2} \]
This can be rewritten using the properties of exponents:
\[ f(s+t) = \frac{b^s \cdot b^t + \frac{1}{b^s \cdot b^t}}{2} = \frac{b^s b^t + \frac{1}{b^s b^t}}{2} \]

2. **Calculate \( f(s-t) \)**:
\[ f(s-t) = \frac{b^{s-t} + b^{-(s-t)}}{2} \]
Similarly, this can be rewritten:
\[ f(s-t) = \frac{b^s \cdot b^{-t} + \frac{1}{b^s \cdot b^{-t}}}{2} = \frac{b^s b^{-t} + \frac{1}{b^s b^{-t}}}{2} \]

3. **Add \( f(s+t) \) and \( f(s-t) \)**:
Now we need to add both results:
\[ f(s+t) + f(s-t) = \frac{b^s b^t + \frac{1}{b^s b^t}}{2} + \frac{b^s b^{-t} + \frac{1}{b^s b^{-t}}}{2} \]
Combine the fractions:
\[ = \frac{1}{2} \left( b^s b^t + \frac{1}{b^s b^t} + b^s b^{-t} + \frac{1}{b^s b^{-t}} \right) \]

4. **Simplify the expression**:
Notice that \( b^s b^t = b^{s+t} \) and \( b^s b^{-t} = b^{s-t} \):
\[ = \frac{1}{2} \left( b^{s+t} + b^{s-t} + \frac{1}{b^{s+t}} + \frac{1}{b^{s-t}} \right) \]

5. **Final Result**:
Thus, we have:
\[ f(s+t) + f(s-t) = \frac{1}{2} \left( b^{s+t} + b^{s-t} + \frac{1}{b^{s+t}} + \frac{1}{b^{s-t}} \right) \]