Minimum value of `2^(sinx)+2^(cosx)` is

Options

A. `2^(1-1/sqrt(2))`
B. `2^(1+1/sqrt(2))`
C. `2^(1+sqrt(2))`
D. `2^(1-sqrt(2))`

Answer

To find the minimum value of the expression \(2^{\sin x} + 2^{\cos x}\), we can use the properties of exponential functions and inequalities.

### Step-by-Step Solution:

1. **Define the Variables**:
Let \(a = 2^{\sin x}\) and \(b = 2^{\cos x}\). We need to find the minimum value of \(a + b\).

2. **Apply the AM-GM Inequality**:
By the Arithmetic Mean-Geometric Mean (AM-GM) inequality, we know that:
\[ \frac{a + b}{2} \geq \sqrt{ab} \]
Therefore, we can write:
\[ a + b \geq 2\sqrt{ab} \]

3. **Calculate \(ab\)**:
We have:
\[ ab = 2^{\sin x} \cdot 2^{\cos x} = 2^{\sin x + \cos x} \]

4. **Find the Maximum of \(\sin x + \cos x\)**:
The maximum value of \(\sin x + \cos x\) can be found using the identity:
\[ \sin x + \cos x = \sqrt{2} \sin\left(x + \frac{\pi}{4}\right) \]
Thus, the maximum value of \(\sin x + \cos x\) is \(\sqrt{2}\).

5. **Substitute Back**:
Therefore, the maximum value of \(ab\) is:
\[ ab \leq 2^{\sqrt{2}} \]

6. **Use the AM-GM Result**:
From the AM-GM inequality, we have:
\[ a + b \geq 2\sqrt{2^{\sqrt{2}}} = 2^{1 + \frac{\sqrt{2}}{2}} \]

7. **Calculate the Minimum Value**:
The minimum value of \(2^{\sin x} + 2^{\cos x}\) is thus:
\[ 2^{1 - \frac{1}{\sqrt{2}}} \]

### Conclusion:
The minimum value of \(2^{\sin x} + 2^{\cos x}\) is:
\[ \boxed{2^{1 - \frac{1}{\sqrt{2}}}} \]