if cos (1-i) = a+ib, where a , b `in ` R and `i = sqrt(-1)` , then

Options

A. `a = (1)/(2)(e-(1)/(e))cos 1, b = (1)/(2)(e+(1)/(e))sin 1 `
B. `a=(1)/(2)(e+(1)/(e))cos 1,b=(1)/(2)(e-(1)/(e))sin 1`
C. `a=(1)/(2)(e+(1)/(e))cos 1,b=(1)/(2)(e+(1)/(e))sin 1`
D. `a=(1)/(2)(e-(1)/(e))cos 1,b=(1)/(2)(e-(1)/(e))sin 1`

Answer

To solve the problem \( \cos(1 - i) = a + ib \), where \( a, b \in \mathbb{R} \) and \( i = \sqrt{-1} \), we can follow these steps:

### Step 1: Use the definition of cosine in terms of exponential functions
The cosine function can be expressed as:
\[ \cos(z) = \frac{e^{iz} + e^{-iz}}{2} \]
In our case, we need to find \( \cos(1 - i) \).

### Step 2: Substitute \( z = 1 - i \)
Using the definition, we have:
\[ \cos(1 - i) = \frac{e^{i(1 - i)} + e^{-i(1 - i)}}{2} \]

### Step 3: Simplify the exponentials
Calculating the exponentials:
\[ e^{i(1 - i)} = e^{i} \cdot e^{1} = e \cdot (\cos(1) + i \sin(1)) \]
\[ e^{-i(1 - i)} = e^{-i} \cdot e^{1} = e \cdot (\cos(-1) + i \sin(-1)) = e \cdot (\cos(1) - i \sin(1)) \]

### Step 4: Combine the results
Now substituting back into the cosine formula:
\[ \cos(1 - i) = \frac{e(\cos(1) + i \sin(1)) + e(\cos(1) - i \sin(1))}{2} \]
\[ = \frac{e \cos(1) + ie \sin(1) + e \cos(1) - ie \sin(1)}{2} \]
\[ = \frac{2e \cos(1)}{2} = e \cos(1) \]
The imaginary parts cancel out, so:
\[ \cos(1 - i) = e \cos(1) \]

### Step 5: Separate real and imaginary parts
Now we can express this in the form \( a + ib \):
\[ \cos(1 - i) = e \cos(1) + 0i \]
Thus, we have:
\[ a = e \cos(1), \quad b = 0 \]

### Step 6: Compare with the given options
Now we can check the options given in the problem:
- \( a = \frac{1}{2} e^{-1} \cos(1) \)
- \( b = \frac{1}{2} e (1 + e) \sin(1) \)

From our calculation:
- \( a = e \cos(1) \)
- \( b = 0 \)

### Conclusion
The correct values for \( a \) and \( b \) do not match the provided options, indicating that the options might have been misinterpreted or incorrectly stated.