The polar form of `sin50^(@)+icos50^(@)` is . . .

Answer

To find the polar form of the complex number \( z = \sin(50^\circ) + i \cos(50^\circ) \), we will follow these steps:

### Step 1: Identify the components of the complex number
We can express the complex number \( z \) in the standard form \( z = a + ib \), where:
- \( a = \sin(50^\circ) \)
- \( b = \cos(50^\circ) \)

### Step 2: Calculate the modulus \( r \)
The modulus \( r \) of a complex number \( z = a + ib \) is given by:
\[ r = \sqrt{a^2 + b^2} \]
Substituting the values of \( a \) and \( b \):
\[ r = \sqrt{(\sin(50^\circ))^2 + (\cos(50^\circ))^2} \]
Using the Pythagorean identity \( \sin^2(\theta) + \cos^2(\theta) = 1 \):
\[ r = \sqrt{1} = 1 \]

### Step 3: Determine the argument \( \theta \)
The argument \( \theta \) of the complex number can be found using the relationship between sine and cosine:
\[ \sin(50^\circ) = \cos(40^\circ) \]
Thus, we can express the argument as:
\[ \theta = 40^\circ \]

### Step 4: Write the polar form
The polar form of a complex number is given by:
\[ z = r(\cos(\theta) + i\sin(\theta)) \]
Substituting the values of \( r \) and \( \theta \):
\[ z = 1(\cos(40^\circ) + i\sin(40^\circ)) \]
This simplifies to:
\[ z = \cos(40^\circ) + i\sin(40^\circ) \]

### Final Answer
The polar form of \( \sin(50^\circ) + i\cos(50^\circ) \) is:
\[ z = \cos(40^\circ) + i\sin(40^\circ) \]