The value of `"cos"^(4)(pi)/(8)+"cos"^(4)(3pi)/(8)+"cos"^(4)(5pi)/(8)+"cos"^(4)(7pi)/(8)` is

Options

A. `3//2`
B. `1//2`
C. `3//4`
D. `1//4`

Answer

To solve the problem of finding the value of

\[ \cos^4\left(\frac{\pi}{8}\right) + \cos^4\left(\frac{3\pi}{8}\right) + \cos^4\left(\frac{5\pi}{8}\right) + \cos^4\left(\frac{7\pi}{8}\right), \]

we can follow these steps:

### Step 1: Simplify the terms
Notice that:
- \(\cos\left(\frac{5\pi}{8}\right) = \cos\left(\pi - \frac{3\pi}{8}\right) = -\cos\left(\frac{3\pi}{8}\right)\)
- \(\cos\left(\frac{7\pi}{8}\right) = \cos\left(\pi - \frac{\pi}{8}\right) = -\cos\left(\frac{\pi}{8}\right)\)

Thus, we can rewrite the expression as:
\[ \cos^4\left(\frac{\pi}{8}\right) + \cos^4\left(\frac{3\pi}{8}\right) + \left(-\cos\left(\frac{3\pi}{8}\right)\right)^4 + \left(-\cos\left(\frac{\pi}{8}\right)\right)^4 \]
This simplifies to:
\[ \cos^4\left(\frac{\pi}{8}\right) + \cos^4\left(\frac{3\pi}{8}\right) + \cos^4\left(\frac{3\pi}{8}\right) + \cos^4\left(\frac{\pi}{8}\right) \]
or
\[ 2\cos^4\left(\frac{\pi}{8}\right) + 2\cos^4\left(\frac{3\pi}{8}\right) \]

### Step 2: Factor out the common terms
We can factor out the 2:
\[ 2\left(\cos^4\left(\frac{\pi}{8}\right) + \cos^4\left(\frac{3\pi}{8}\right)\right) \]

### Step 3: Use the identity for \(A^4 + B^4\)
We can use the identity:
\[ A^4 + B^4 = (A^2 + B^2)^2 - 2A^2B^2 \]
Let \(A = \cos^2\left(\frac{\pi}{8}\right)\) and \(B = \cos^2\left(\frac{3\pi}{8}\right)\):
\[ \cos^4\left(\frac{\pi}{8}\right) + \cos^4\left(\frac{3\pi}{8}\right) = \left(\cos^2\left(\frac{\pi}{8}\right) + \cos^2\left(\frac{3\pi}{8}\right)\right)^2 - 2\cos^2\left(\frac{\pi}{8}\right)\cos^2\left(\frac{3\pi}{8}\right) \]

### Step 4: Calculate \(\cos^2\left(\frac{\pi}{8}\right)\) and \(\cos^2\left(\frac{3\pi}{8}\right)\)
Using the identity \(\cos^2\theta = \frac{1 + \cos(2\theta)}{2}\):
- \(\cos^2\left(\frac{\pi}{8}\right) = \frac{1 + \cos\left(\frac{\pi}{4}\right)}{2} = \frac{1 + \frac{1}{\sqrt{2}}}{2} = \frac{2 + \sqrt{2}}{4}\)
- \(\cos^2\left(\frac{3\pi}{8}\right) = \frac{1 + \cos\left(\frac{3\pi}{4}\right)}{2} = \frac{1 - \frac{1}{\sqrt{2}}}{2} = \frac{2 - \sqrt{2}}{4}\)

### Step 5: Substitute back into the equation
Now we can substitute these values back into our equation:
\[ \cos^2\left(\frac{\pi}{8}\right) + \cos^2\left(\frac{3\pi}{8}\right) = \frac{2 + \sqrt{2}}{4} + \frac{2 - \sqrt{2}}{4} = \frac{4}{4} = 1 \]
Thus,
\[ \left(\cos^2\left(\frac{\pi}{8}\right) + \cos^2\left(\frac{3\pi}{8}\right)\right)^2 = 1^2 = 1 \]

### Step 6: Calculate \(2\cos^2\left(\frac{\pi}{8}\right)\cos^2\left(\frac{3\pi}{8}\right)\)
Now we need to calculate \(2\cos^2\left(\frac{\pi}{8}\right)\cos^2\left(\frac{3\pi}{8}\right)\):
\[ \cos^2\left(\frac{\pi}{8}\right)\cos^2\left(\frac{3\pi}{8}\right) = \frac{(2 + \sqrt{2})(2 - \sqrt{2})}{16} = \frac{4 - 2}{16} = \frac{2}{16} = \frac{1}{8} \]
Thus,
\[ 2\cos^2\left(\frac{\pi}{8}\right)\cos^2\left(\frac{3\pi}{8}\right) = 2 \cdot \frac{1}{8} = \frac{1}{4} \]

### Step 7: Final Calculation
Putting everything together:
\[ 2\left(1 - \frac{1}{4}\right) = 2 \cdot \frac{3}{4} = \frac{3}{2} \]

Thus, the final value is:
\[ \boxed{\frac{3}{2}} \]