1. Key Concepts and Formulas

• Arithmetic Mean - Geometric Mean (AM-GM) Inequality: For any non-negative real numbers $a$ and $b$, $\frac{a+b}{2} \ge \sqrt{ab}$. Equality holds if and only if $a=b$.
• Range of $a \sin x + b \cos x$: The expression $a \sin x + b \cos x$ has a range of $[-\sqrt{a^2+b^2}, \sqrt{a^2+b^2}]$.
• Properties of Exponents: $a^m \cdot a^n = a^{m+n}$, $\sqrt{a^m} = a^{m/2}$, and $a^m \cdot a^n = a^{m+n}$.
• Monotonicity of Exponential Functions: For a base $b > 1$, the function $f(t) = b^t$ is strictly increasing. This means that to minimize $b^t$, we need to minimize $t$.

2. Step-by-Step Solution

We are asked to find the minimum value of the expression $E = 2^{\sin x} + 2^{\cos x}$.

Step 1: Identify applicable inequalities and properties. The terms $2^{\sin x}$ and $2^{\cos x}$ are always positive since the base $2$ is positive. This allows us to use the AM-GM inequality. We also need to find the range of $\sin x + \cos x$ to determine the minimum value of the exponent.

Step 2: Apply the AM-GM Inequality. Let $a = 2^{\sin x}$ and $b = 2^{\cos x}$. Since both are positive, we can apply the AM-GM inequality: $\frac{2^{\sin x} + 2^{\cos x}}{2} \ge \sqrt{2^{\sin x} \cdot 2^{\cos x}}$

Step 3: Simplify the inequality using exponent rules. Simplify the right-hand side (RHS) of the inequality: $\sqrt{2^{\sin x} \cdot 2^{\cos x}} = \sqrt{2^{\sin x + \cos x}} = 2^{\frac{\sin x + \cos x}{2}}$ So, the inequality becomes: $\frac{2^{\sin x} + 2^{\cos x}}{2} \ge 2^{\frac{\sin x + \cos x}{2}}$ Multiply both sides by 2: $2^{\sin x} + 2^{\cos x} \ge 2 \cdot 2^{\frac{\sin x + \cos x}{2}}$ Using the property $a^m \cdot a^n = a^{m+n}$, where $2 = 2^1$: $2^{\sin x} + 2^{\cos x} \ge 2^{1 + \frac{\sin x + \cos x}{2}}$ Let the lower bound be $L(x) = 2^{1 + \frac{\sin x + \cos x}{2}}$.

Step 4: Determine the range of the exponent's trigonometric part. To find the minimum value of $E$, we need to find the minimum value of the lower bound $L(x)$. Since the base of the exponential function is $2 > 1$, $L(x)$ is minimized when its exponent is minimized. The exponent is $1 + \frac{\sin x + \cos x}{2}$. Therefore, we need to find the minimum value of $\sin x + \cos x$.

For an expression of the form $a \sin x + b \cos x$, its range is $[-\sqrt{a^2+b^2}, \sqrt{a^2+b^2}]$. Here, $a=1$ and $b=1$. So, the range of $\sin x + \cos x$ is $[-\sqrt{1^2+1^2}, \sqrt{1^2+1^2}] = [-\sqrt{2}, \sqrt{2}]$. The minimum value of $\sin x + \cos x$ is $-\sqrt{2}$.

Step 5: Substitute the minimum value of the trigonometric expression into the lower bound. The minimum value of $\sin x + \cos x$ is $-\sqrt{2}$. Substitute this into the exponent of $L(x)$: Minimum value of exponent $= 1 + \frac{-\sqrt{2}}{2} = 1 - \frac{\sqrt{2}}{2} = 1 - \frac{1}{\sqrt{2}}$. Therefore, the minimum value of $E$ is: $E_{min} \ge 2^{1 - \frac{1}{\sqrt{2}}}$

Step 6: Verify the equality condition for AM-GM. The minimum value derived from an inequality is achievable if the equality condition for that inequality can be met. For the AM-GM inequality, equality holds when $2^{\sin x} = 2^{\cos x}$, which implies $\sin x = \cos x$.

We found that the minimum value of $\sin x + \cos x$ is $-\sqrt{2}$, which occurs when $\sin x = -\frac{1}{\sqrt{2}}$ and $\cos x = -\frac{1}{\sqrt{2}}$. This happens, for example, at $x = \frac{5\pi}{4}$. At this value of $x$, $\sin x = \cos x = -\frac{1}{\sqrt{2}}$. Therefore, the condition for equality in AM-GM ($\sin x = \cos x$) is satisfied when the exponent is minimized. Hence, the minimum value $2^{1 - \frac{1}{\sqrt{2}}}$ is indeed achievable.

3. Common Mistakes & Tips

• Forgetting the Equality Condition: It is crucial to verify that the minimum value obtained from the inequality can actually be attained. If the equality condition cannot be met for the values of $x$ that minimize the exponent, then the derived minimum might not be the true minimum.
• Confusing Maximum and Minimum: When dealing with exponential functions with a base greater than 1, minimizing the function requires minimizing its exponent. Conversely, maximizing the function requires maximizing the exponent.
• Incorrect Range of Trigonometric Sums: Ensure you correctly recall or derive the range of expressions like $a \sin x + b \cos x$.

4. Summary

The problem involves finding the minimum value of a sum of exponential terms with trigonometric arguments. We successfully applied the AM-GM inequality to establish a lower bound for the expression. By determining the range of the trigonometric part of the exponent and finding its minimum value, we found the minimum possible value of the lower bound. Crucially, we verified that the conditions for equality in the AM-GM inequality coincide with the conditions that yield the minimum value of the exponent, confirming that the derived lower bound is indeed the minimum value of the expression.

The minimum value of $2^{\sin x} + 2^{\cos x}$ is $2^{1 - \frac{1}{\sqrt{2}}}$.

This corresponds to option (B).

The final answer is \boxed{\text{2^{1 - {1 \over {\sqrt 2 }}}}}