Find the multiplicative inverse of the following
`(sin theta,cos theta)`

Answer

To find the multiplicative inverse of the complex number represented by the coordinates \((\sin \theta, \cos \theta)\), we will follow these steps:

### Step 1: Represent the complex number
The complex number can be represented as:
\[ z = \sin \theta + i \cos \theta \]

### Step 2: Write the formula for the multiplicative inverse
The multiplicative inverse of a complex number \(z\) is given by:
\[ z^{-1} = \frac{1}{z} \]
Thus, we have:
\[ z^{-1} = \frac{1}{\sin \theta + i \cos \theta} \]

### Step 3: Rationalize the denominator
To simplify this expression, we multiply the numerator and the denominator by the conjugate of the denominator:
\[ z^{-1} = \frac{1}{\sin \theta + i \cos \theta} \cdot \frac{\sin \theta - i \cos \theta}{\sin \theta - i \cos \theta} \]
This gives us:
\[ z^{-1} = \frac{\sin \theta - i \cos \theta}{(\sin \theta + i \cos \theta)(\sin \theta - i \cos \theta)} \]

### Step 4: Simplify the denominator
The denominator can be simplified using the formula for the difference of squares:
\[ (\sin \theta + i \cos \theta)(\sin \theta - i \cos \theta) = \sin^2 \theta - (i \cos \theta)^2 \]
Since \(i^2 = -1\), we have:
\[ = \sin^2 \theta - (-\cos^2 \theta) = \sin^2 \theta + \cos^2 \theta \]
Using the Pythagorean identity:
\[ \sin^2 \theta + \cos^2 \theta = 1 \]
Thus, the denominator simplifies to \(1\).

### Step 5: Write the final result
Now, substituting back, we have:
\[ z^{-1} = \frac{\sin \theta - i \cos \theta}{1} = \sin \theta - i \cos \theta \]

### Final Answer
The multiplicative inverse of \((\sin \theta, \cos \theta)\) is:
\[ \sin \theta - i \cos \theta \]

---