`(1+cos.(pi)/(8))(1+cos.(3pi)/(8))(1+cos.(5pi)/(8))(1+cos.(7pi)/(8))` is equal to

Options

A. `(1)/(2)`
B. `cos.(pi)/(8)`
C. `(1)/(8)`
D. `(1+sqrt(2))/(2sqrt(2))`

Explanation

To solve the expression \((1 + \cos(\frac{\pi}{8}))(1 + \cos(\frac{3\pi}{8}))(1 + \cos(\frac{5\pi}{8}))(1 + \cos(\frac{7\pi}{8}))\), we can follow these steps:

### Step 1: Use the cosine identity
We know that \(\cos(\frac{5\pi}{8}) = -\cos(\frac{3\pi}{8})\) and \(\cos(\frac{7\pi}{8}) = -\cos(\frac{\pi}{8})\). Therefore, we can rewrite the expression as:
\[ (1 + \cos(\frac{\pi}{8}))(1 + \cos(\frac{3\pi}{8}))(1 - \cos(\frac{3\pi}{8}))(1 - \cos(\frac{\pi}{8})) \]

### Step 2: Group the terms
Now, we can group the terms:
\[ [(1 + \cos(\frac{\pi}{8}))(1 - \cos(\frac{\pi}{8}))][(1 + \cos(\frac{3\pi}{8}))(1 - \cos(\frac{3\pi}{8}))] \]

### Step 3: Apply the difference of squares
Using the identity \(a^2 - b^2 = (a + b)(a - b)\):
\[ = \sin^2(\frac{\pi}{8}) \cdot \sin^2(\frac{3\pi}{8}) \]

### Step 4: Use the double angle formula
We know that \(\sin(\frac{3\pi}{8}) = \cos(\frac{\pi}{8})\) (since \(\frac{3\pi}{8} + \frac{\pi}{8} = \frac{\pi}{2}\)). Therefore:
\[ \sin^2(\frac{3\pi}{8}) = \cos^2(\frac{\pi}{8}) \]

### Step 5: Substitute back into the expression
Now, substituting this back, we have:
\[ \sin^2(\frac{\pi}{8}) \cdot \cos^2(\frac{\pi}{8}) \]

### Step 6: Use the identity for sine and cosine
Using the identity \(\sin(2\theta) = 2\sin(\theta)\cos(\theta)\):
\[ = \frac{1}{4} \sin^2(\frac{\pi}{4}) = \frac{1}{4} \left(\frac{\sqrt{2}}{2}\right)^2 = \frac{1}{4} \cdot \frac{1}{2} = \frac{1}{8} \]

### Final Answer
Thus, the value of the expression is:
\[ \frac{1}{8} \]

---