If `((4i^(3)-i)/(2i+1))^(2)=r(cos theta+isin theta)`, then `cos theta+sin theta` is equal to (where, `i^(2)=-1`)

Answer

To solve the problem, we need to simplify the expression given and find the value of \( \cos \theta + \sin \theta \).

### Step-by-Step Solution:

1. **Identify the expression**:
We start with the expression:
\[ \left( \frac{4i^3 - i}{2i + 1} \right)^2 = r(\cos \theta + i \sin \theta) \]

2. **Substitute \( i^2 = -1 \)**:
Recall that \( i^3 = i^2 \cdot i = -i \). Therefore, we can rewrite \( 4i^3 \):
\[ 4i^3 = 4(-i) = -4i \]
Thus, the numerator becomes:
\[ 4i^3 - i = -4i - i = -5i \]

3. **Rewrite the expression**:
Now the expression simplifies to:
\[ \left( \frac{-5i}{2i + 1} \right)^2 \]

4. **Rationalize the denominator**:
To simplify \( \frac{-5i}{2i + 1} \), we multiply the numerator and denominator by the conjugate of the denominator:
\[ \frac{-5i(2i - 1)}{(2i + 1)(2i - 1)} = \frac{-10i^2 + 5i}{4i^2 - 1} \]
Since \( i^2 = -1 \):
\[ = \frac{-10(-1) + 5i}{4(-1) - 1} = \frac{10 + 5i}{-4 - 1} = \frac{10 + 5i}{-5} = -2 - i \]

5. **Square the result**:
Now we square the result:
\[ (-2 - i)^2 = (-2)^2 + 2(-2)(-i) + (-i)^2 = 4 + 4i - 1 = 3 + 4i \]

6. **Express in polar form**:
We can express \( 3 + 4i \) in polar form:
\[ r = |3 + 4i| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \]
The angle \( \theta \) can be found using:
\[ \tan \theta = \frac{4}{3} \]

7. **Find \( \cos \theta \) and \( \sin \theta \)**:
From the polar form, we have:
\[ r \cos \theta = 3 \quad \text{and} \quad r \sin \theta = 4 \]
Thus:
\[ \cos \theta = \frac{3}{5} \quad \text{and} \quad \sin \theta = \frac{4}{5} \]

8. **Calculate \( \cos \theta + \sin \theta \)**:
Now we can find:
\[ \cos \theta + \sin \theta = \frac{3}{5} + \frac{4}{5} = \frac{7}{5} \]

### Final Answer:
\[ \cos \theta + \sin \theta = \frac{7}{5} \]