Step-by-Step Solution & Explanation
### **Detailed Step-by-Step Solution (GATE AE 2022 Q.13)**
**1. Intuitive Physical Setup & Given Parameters**:
Let $x$ be a real number and $i = \sqrt{-1}$. We are asked to determine the real part of the complex trigonometric expression $\cos(ix)$.
**2. Step-by-Step Mathematical Derivation**:
1. **Euler's Complex Definition of Cosine**:
For any complex argument $z \in \mathbb{C}$, the cosine function is defined by:
\cos(z) = \frac{e^{iz} + e^{-iz}}{2}
2. **Substitute $z = ix$**:
\cos(ix) = \frac{e^{i(ix)} + e^{-i(ix)}}{2} = \frac{e^{i^2 x} + e^{-i^2 x}}{2}
Since $i^2 = -1$:
\cos(ix) = \frac{e^{-x} + e^{-(-x)}}{2} = \frac{e^{-x} + e^x}{2} = \frac{e^x + e^{-x}}{2}
3. **Recognition of Hyperbolic Cosine**:
By definition of the hyperbolic cosine function:
\cosh(x) = \frac{e^x + e^{-x}}{2}
Since $x$ is a real number, $e^x$ and $e^{-x}$ are real, so $\cosh(x)$ is strictly real (imaginary part is zero).
Therefore, $\text{Re}(\cos(ix)) = \cosh(x)$.
**3. Comprehensive Options & Distractor Analysis**:
- **(A) cos(x)**: ❌ **INCORRECT**. Evaluates real trigonometric cosine instead of hyperbolic.
- **(B) cosh(x)**: ✅ **CORRECT**. By identity $\cos(ix) = \cosh(x)$, which is purely real for real $x$.
- **(C) sinh(x)**: ❌ **INCORRECT**. Related to $\sin(ix) = i\sinh(x)$, which is purely imaginary.
- **(D) 0**: ❌ **INCORRECT**. $\cosh(x) \ge 1$ for all real $x$, never zero.
**4. Official Master Key & Exam Takeaway**:
- **Official Answer Key**: **B**
- **Exam Takeaway**: Memorize the fundamental connections between circular and hyperbolic functions: $\cos(ix) = \cosh(x)$, $\sin(ix) = i\sinh(x)$, $\cosh(ix) = \cos(x)$, $\sinh(ix) = i\sin(x)$.