Find the value of:`(1+cos""(pi)/(8))(1+cos""(3pi)/(8))(1+cos""(5pi)/(8))(1+cos""(7pi)/(8))=?`

Options

A. (a) `(1)/(4)`
B. (b) `(1)/(2)`
C. (c) `(1)/(6)`
D. (d) `(1)/(8)`

Answer

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

### Step 1: Rewrite the cosines
We notice that \( \cos(\frac{5\pi}{8}) \) and \( \cos(\frac{7\pi}{8}) \) can be rewritten using the identity \( \cos(\pi - x) = -\cos(x) \).

- \( \cos(\frac{5\pi}{8}) = \cos(\pi - \frac{3\pi}{8}) = -\cos(\frac{3\pi}{8}) \)
- \( \cos(\frac{7\pi}{8}) = \cos(\pi - \frac{\pi}{8}) = -\cos(\frac{\pi}{8}) \)

Thus, 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 \( (a + b)(a - b) = a^2 - b^2 \), we can simplify each group:
\[ = \sin^2(\frac{\pi}{8}) \cdot \sin^2(\frac{3\pi}{8}) \]

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

### Step 5: Substitute back into the expression
Now we can substitute this back into our expression:
\[ \sin^2(\frac{\pi}{8}) \cdot \cos^2(\frac{\pi}{8}) = \sin^2(\frac{\pi}{8}) \cdot (1 - \sin^2(\frac{\pi}{8})) = \sin^2(\frac{\pi}{8}) \cdot \cos^2(\frac{\pi}{8}) \]

### Step 6: Use the double angle identity
Using the identity \( \sin(2x) = 2\sin(x)\cos(x) \), we can express this as:
\[ \frac{1}{4} \sin^2(\frac{\pi}{4}) = \frac{1}{4} \cdot \left(\frac{\sqrt{2}}{2}\right)^2 = \frac{1}{4} \cdot \frac{1}{2} = \frac{1}{8} \]

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