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 given problem step by step, we start with the expression:

\[ \left(\frac{4i^3 - i}{2i + 1}\right)^2 = r(\cos \theta + i \sin \theta) \]

### Step 1: Simplify \(4i^3 - i\)

We know that \(i^2 = -1\), hence:

\[ i^3 = i^2 \cdot i = -1 \cdot i = -i \]

Thus,

\[ 4i^3 = 4(-i) = -4i \]

Now substituting this back into the expression:

\[ 4i^3 - i = -4i - i = -5i \]

### Step 2: Substitute into the fraction

Now we substitute \(-5i\) into the fraction:

\[ \frac{-5i}{2i + 1} \]

### Step 3: Rationalize the denominator

To rationalize the denominator, we multiply the numerator and denominator by the conjugate of the denominator, which is \(1 - 2i\):

\[ \frac{-5i(1 - 2i)}{(2i + 1)(1 - 2i)} \]

Calculating the denominator:

\[ (2i + 1)(1 - 2i) = 2i - 4i^2 + 1 = 2i + 4 + 1 = 5 + 2i \]

Now calculating the numerator:

\[ -5i(1 - 2i) = -5i + 10i^2 = -5i - 10 = -10 - 5i \]

So we have:

\[ \frac{-10 - 5i}{5 + 2i} \]

### Step 4: Simplify the fraction

Now we can write:

\[ \frac{-10 - 5i}{5 + 2i} = \frac{-10 - 5i}{5 + 2i} \cdot \frac{5 - 2i}{5 - 2i} \]

Calculating the new denominator:

\[ (5 + 2i)(5 - 2i) = 25 + 4 = 29 \]

Calculating the new numerator:

\[ (-10 - 5i)(5 - 2i) = -50 + 20i - 25i - 10i^2 = -50 - 5i + 10 = -40 - 5i \]

So we have:

\[ \frac{-40 - 5i}{29} \]

### Step 5: Express in polar form

Now we can express this in the form \(r(\cos \theta + i \sin \theta)\):

Let:

\[ z = \frac{-40}{29} - \frac{5}{29}i \]

### Step 6: Calculate \(r\) and \(\theta\)

To find \(r\):

\[ r = \sqrt{\left(-\frac{40}{29}\right)^2 + \left(-\frac{5}{29}\right)^2} = \sqrt{\frac{1600}{841} + \frac{25}{841}} = \sqrt{\frac{1625}{841}} = \frac{\sqrt{1625}}{29} \]

Now we find \(\theta\):

\[ \tan \theta = \frac{-5/29}{-40/29} = \frac{5}{40} = \frac{1}{8} \]

### Step 7: Find \(\cos \theta\) and \(\sin \theta\)

Using the triangle formed:

\[ \text{Opposite} = 1, \quad \text{Adjacent} = 8, \quad \text{Hypotenuse} = \sqrt{1^2 + 8^2} = \sqrt{65} \]

Thus:

\[ \cos \theta = \frac{8}{\sqrt{65}}, \quad \sin \theta = \frac{1}{\sqrt{65}} \]

### Step 8: Calculate \(\cos \theta + \sin \theta\)

Now we can find:

\[ \cos \theta + \sin \theta = \frac{8}{\sqrt{65}} + \frac{1}{\sqrt{65}} = \frac{9}{\sqrt{65}} \]

### Final Answer

Thus, the final answer is:

\[ \cos \theta + \sin \theta = \frac{9}{\sqrt{65}} \]