The value of `{sin(logi^i)}^3+{cos(logi^i)}^3,` is

Options

A. 1
B. `-1`
C. 2
D. 2i

Explanation

To solve the expression \({\sin(\log(i^i))}^3 + {\cos(\log(i^i))}^3\), we will follow these steps:

### Step 1: Simplify \(\log(i^i)\)
Using the property of logarithms, we can express \(\log(i^i)\) as:
\[ \log(i^i) = i \cdot \log(i) \]

### Step 2: Find \(\log(i)\)
The complex logarithm of \(i\) can be expressed as:
\[ \log(i) = \log(1) + i \cdot \frac{\pi}{2} = 0 + i \cdot \frac{\pi}{2} = i \frac{\pi}{2} \]

### Step 3: Substitute \(\log(i)\) back into the expression
Now substituting \(\log(i)\) back into our expression for \(\log(i^i)\):
\[ \log(i^i) = i \cdot \left(i \frac{\pi}{2}\right) = -\frac{\pi}{2} \]

### Step 4: Substitute \(\log(i^i)\) into the sine and cosine functions
Now we substitute \(-\frac{\pi}{2}\) into the sine and cosine functions:
\[ \sin(\log(i^i)) = \sin\left(-\frac{\pi}{2}\right) = -1 \]
\[ \cos(\log(i^i)) = \cos\left(-\frac{\pi}{2}\right) = 0 \]

### Step 5: Compute the cubes
Now we compute the cubes:
\[ \sin^3\left(-\frac{\pi}{2}\right) = (-1)^3 = -1 \]
\[ \cos^3\left(-\frac{\pi}{2}\right) = 0^3 = 0 \]

### Step 6: Add the results
Finally, we add the results:
\[ \sin^3(\log(i^i)) + \cos^3(\log(i^i)) = -1 + 0 = -1 \]

### Final Answer
Thus, the value of \({\sin(\log(i^i))}^3 + {\cos(\log(i^i))}^3\) is:
\[ \boxed{-1} \]