Express `sin50^(@)+icos50^(@)` in the polar form. Also, find r and `theta`.

Answer

To express \( \sin 50^\circ + i \cos 50^\circ \) in polar form, we can follow these steps:

### Step 1: Rewrite the expression
We start with the expression:
\[ \sin 50^\circ + i \cos 50^\circ \]
We want to express this in the form \( r (\cos \theta + i \sin \theta) \).

### Step 2: Use trigonometric identities
We can use the identity \( \sin(90^\circ - \theta) = \cos \theta \) and \( \cos(90^\circ - \theta) = \sin \theta \). Thus, we can rewrite \( \sin 50^\circ \) and \( \cos 50^\circ \) as:
\[ \sin 50^\circ = \cos(90^\circ - 50^\circ) = \cos 40^\circ \]
\[ \cos 50^\circ = \sin(90^\circ - 50^\circ) = \sin 40^\circ \]

### Step 3: Substitute back into the expression
Now substituting these identities back into our expression, we have:
\[ \sin 50^\circ + i \cos 50^\circ = \cos 40^\circ + i \sin 40^\circ \]

### Step 4: Identify the polar form
The expression \( \cos 40^\circ + i \sin 40^\circ \) can be expressed in polar form as:
\[ r (\cos \theta + i \sin \theta) \]
where \( r = 1 \) and \( \theta = 40^\circ \).

### Step 5: Conclusion
Thus, the polar form of \( \sin 50^\circ + i \cos 50^\circ \) is:
\[ 1 (\cos 40^\circ + i \sin 40^\circ) \]
where \( r = 1 \) and \( \theta = 40^\circ \).

### Summary of Results
- Polar Form: \( 1 (\cos 40^\circ + i \sin 40^\circ) \)
- \( r = 1 \)
- \( \theta = 40^\circ \)