If `e^(itheta)=costheta+is intheta,` find the value of `|1e^(ipi//3)e^(ipi//4)e^(-ipi//3)1e^(i2pi//3)e^(-ipi//4)e^(-i2pi//3)1|`

Explanation

To find the value of the determinant

\[ D = \begin{vmatrix} 1 & e^{i\pi/3} & e^{i\pi/4} \\ e^{-i\pi/3} & 1 & e^{i2\pi/3} \\ e^{-i\pi/4} & e^{-i2\pi/3} & 1 \end{vmatrix} \]

we will use the properties of determinants and the exponential form of complex numbers.

### Step 1: Write the Determinant
We start with the given determinant:

\[ D = \begin{vmatrix} 1 & e^{i\pi/3} & e^{i\pi/4} \\ e^{-i\pi/3} & 1 & e^{i2\pi/3} \\ e^{-i\pi/4} & e^{-i2\pi/3} & 1 \end{vmatrix} \]

### Step 2: Use the Determinant Expansion
Using the determinant expansion (Sarrus' rule for 3x3 matrices), we can calculate:

\[ D = 1 \cdot \begin{vmatrix} 1 & e^{i2\pi/3} \\ e^{-i2\pi/3} & 1 \end{vmatrix} - e^{i\pi/3} \cdot \begin{vmatrix} e^{-i\pi/3} & e^{i2\pi/3} \\ e^{-i\pi/4} & 1 \end{vmatrix} + e^{i\pi/4} \cdot \begin{vmatrix} e^{-i\pi/3} & 1 \\ e^{-i\pi/4} & e^{-i2\pi/3} \end{vmatrix} \]

### Step 3: Calculate the 2x2 Determinants
1. For the first determinant:

\[ \begin{vmatrix} 1 & e^{i2\pi/3} \\ e^{-i2\pi/3} & 1 \end{vmatrix} = 1 \cdot 1 - e^{i2\pi/3} \cdot e^{-i2\pi/3} = 1 - 1 = 0 \]

2. For the second determinant:

\[ \begin{vmatrix} e^{-i\pi/3} & e^{i2\pi/3} \\ e^{-i\pi/4} & 1 \end{vmatrix} = e^{-i\pi/3} \cdot 1 - e^{i2\pi/3} \cdot e^{-i\pi/4} = e^{-i\pi/3} - e^{i(2\pi/3 - \pi/4)} \]

3. For the third determinant:

\[ \begin{vmatrix} e^{-i\pi/3} & 1 \\ e^{-i\pi/4} & e^{-i2\pi/3} \end{vmatrix} = e^{-i\pi/3} \cdot e^{-i2\pi/3} - 1 \cdot e^{-i\pi/4} = e^{-i(\pi/3 + 2\pi/3)} - e^{-i\pi/4} = e^{-i\pi} - e^{-i\pi/4} = -1 - e^{-i\pi/4} \]

### Step 4: Substitute Back into the Determinant
Now substituting back into the expression for \(D\):

\[ D = 1 \cdot 0 - e^{i\pi/3} \cdot (e^{-i\pi/3} - e^{i(2\pi/3 - \pi/4)}) + e^{i\pi/4} \cdot (-1 - e^{-i\pi/4}) \]

### Step 5: Simplify the Expression
This simplifies to:

\[ D = -e^{i\pi/3} \cdot (e^{-i\pi/3} - e^{i(2\pi/3 - \pi/4)}) + e^{i\pi/4} \cdot (-1 - e^{-i\pi/4}) \]

### Step 6: Final Calculation
After simplifying and calculating the trigonometric values, we find:

\[ D = -2 - \sqrt{2} \]

### Final Answer
Thus, the value of the determinant is:

\[ \boxed{-2 - \sqrt{2}} \]