What is the argument of `( 1- sin theta ) + icos theta ` ? `( i=sqrt(-1))`

Options

A. `( pi )/(2) - ( theta )/( 2)`
B. `(pi )/( 2) + ( theta )/( 2)`
C. `( pi )/( 4) - ( theta )/( 2)`
D. `( pi )/( 4) + ( theta)/( 2)`

Answer

To find the argument of the complex number \( z = (1 - \sin \theta) + i \cos \theta \), we will follow these steps:

### Step 1: Identify the real and imaginary parts
The given complex number can be expressed as:
- Real part \( a = 1 - \sin \theta \)
- Imaginary part \( b = \cos \theta \)

### Step 2: Use the formula for the argument
The argument \( \arg(z) \) of a complex number \( z = a + bi \) is given by:
\[ \arg(z) = \tan^{-1}\left(\frac{b}{a}\right) \]
Substituting the values of \( a \) and \( b \):
\[ \arg(z) = \tan^{-1}\left(\frac{\cos \theta}{1 - \sin \theta}\right) \]

### Step 3: Simplify the expression
To simplify \( \frac{\cos \theta}{1 - \sin \theta} \), we can multiply the numerator and denominator by \( 1 + \sin \theta \):
\[ \frac{\cos \theta}{1 - \sin \theta} \cdot \frac{1 + \sin \theta}{1 + \sin \theta} = \frac{\cos \theta (1 + \sin \theta)}{(1 - \sin \theta)(1 + \sin \theta)} \]
The denominator simplifies to:
\[ (1 - \sin \theta)(1 + \sin \theta) = 1 - \sin^2 \theta = \cos^2 \theta \]
Thus, we have:
\[ \frac{\cos \theta (1 + \sin \theta)}{\cos^2 \theta} = \frac{1 + \sin \theta}{\cos \theta} \]

### Step 4: Find the argument
Now we can rewrite the argument:
\[ \arg(z) = \tan^{-1}\left(\frac{1 + \sin \theta}{\cos \theta}\right) \]
Using the identity \( \tan^{-1}\left(\frac{y}{x}\right) = \tan^{-1}(y) - \tan^{-1}(x) \):
\[ \arg(z) = \tan^{-1}(1 + \sin \theta) - \tan^{-1}(\cos \theta) \]

### Step 5: Final expression for the argument
Thus, the argument of the complex number \( (1 - \sin \theta) + i \cos \theta \) is:
\[ \arg(z) = \tan^{-1}\left(\frac{1 + \sin \theta}{\cos \theta}\right) \]