`(sinpi//8+icospi//8)^(8)/(sinpi//8-icospi//8)^(8)=`

Options

A. `-1`
B. 0
C. 1
D. 2i

Explanation

To solve the expression \((\sin \frac{\pi}{8} + i \cos \frac{\pi}{8})^8 / (\sin \frac{\pi}{8} - i \cos \frac{\pi}{8})^8\), we can use the properties of complex numbers and Euler's formula. Here’s a step-by-step breakdown of the solution:

### Step 1: Rewrite the expression using Euler's Formula
We know that \(e^{i\theta} = \cos \theta + i \sin \theta\). We can manipulate our expression to fit this form.

Let:
\[ z_1 = \sin \frac{\pi}{8} + i \cos \frac{\pi}{8} \]
\[ z_2 = \sin \frac{\pi}{8} - i \cos \frac{\pi}{8} \]

### Step 2: Express \(z_1\) and \(z_2\) in exponential form
Using the identity \(i \cos \theta = e^{i\frac{\pi}{2}} \sin \theta\), we can express \(z_1\) and \(z_2\) as:
\[ z_1 = \sin \frac{\pi}{8} + i \cos \frac{\pi}{8} = e^{i(\frac{\pi}{2} - \frac{\pi}{8})} = e^{i\frac{3\pi}{8}} \]
\[ z_2 = \sin \frac{\pi}{8} - i \cos \frac{\pi}{8} = e^{-i(\frac{\pi}{2} - \frac{\pi}{8})} = e^{-i\frac{3\pi}{8}} \]

### Step 3: Raise both to the power of 8
Now we raise both \(z_1\) and \(z_2\) to the power of 8:
\[ z_1^8 = \left(e^{i\frac{3\pi}{8}}\right)^8 = e^{i3\pi} = -1 \]
\[ z_2^8 = \left(e^{-i\frac{3\pi}{8}}\right)^8 = e^{-i3\pi} = -1 \]

### Step 4: Form the final expression
Now we substitute back into our original expression:
\[ \frac{z_1^8}{z_2^8} = \frac{-1}{-1} = 1 \]

### Conclusion
Thus, the final result of the expression \((\sin \frac{\pi}{8} + i \cos \frac{\pi}{8})^8 / (\sin \frac{\pi}{8} - i \cos \frac{\pi}{8})^8\) is:
\[ \boxed{1} \]