`(1+ "cos "(pi)/(8)) (1+ "cos"(3pi)/(8)) (1+ "cos" (5pi)/(8)) (1+ "cos " (7pi)/(8))`=

Answer

To evaluate 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: Rewrite the terms
We know that \(\cos(\frac{5\pi}{8})\) and \(\cos(\frac{7\pi}{8})\) can be rewritten using the identity \(\cos(\pi - x) = -\cos(x)\). Thus, we have:
\[ \cos(\frac{5\pi}{8}) = \cos(\pi - \frac{3\pi}{8}) = -\cos(\frac{3\pi}{8}), \]
\[ \cos(\frac{7\pi}{8}) = \cos(\pi - \frac{\pi}{8}) = -\cos(\frac{\pi}{8}). \]

### Step 2: Substitute the rewritten terms
Now we can substitute these into the expression:
\[ (1 + \cos(\frac{\pi}{8}))(1 + \cos(\frac{3\pi}{8}))(1 - \cos(\frac{3\pi}{8}))(1 - \cos(\frac{\pi}{8})). \]

### Step 3: Group the terms
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 4: Apply the difference of squares
Using the difference of squares formula \(a^2 - b^2 = (a + b)(a - b)\), we have:
\[ (1 - \cos^2(\frac{\pi}{8}))(1 - \cos^2(\frac{3\pi}{8})). \]

### Step 5: Use the Pythagorean identity
We know that \(1 - \cos^2(x) = \sin^2(x)\), so we can rewrite the expression as:
\[ \sin^2(\frac{\pi}{8}) \sin^2(\frac{3\pi}{8}). \]

### Step 6: Use the sine double angle identity
Next, we can use the identity \(\sin(\frac{3\pi}{8}) = \cos(\frac{\pi}{8})\) (since \(\frac{3\pi}{8} + \frac{\pi}{8} = \frac{\pi}{2}\)):
\[ \sin^2(\frac{\pi}{8}) \cos^2(\frac{\pi}{8}). \]

### Step 7: Apply the double angle formula
Using the identity \(2\sin(x)\cos(x) = \sin(2x)\), we can express this as:
\[ \frac{1}{4} \sin^2(\frac{\pi}{4}). \]

### Step 8: Evaluate \(\sin(\frac{\pi}{4})\)
Since \(\sin(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}\), we have:
\[ \sin^2(\frac{\pi}{4}) = \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2}. \]

### Step 9: Final calculation
Thus, we have:
\[ \frac{1}{4} \cdot \frac{1}{2} = \frac{1}{8}. \]

### Final Result
Therefore, the value of the expression is:
\[ \frac{1}{8}. \]

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