If `theta=(pi)/(8)`, then what is the value of `(2cos theta+1)^(10)(2cos2theta-1)^(10)(2cos theta-1)^(10)(2cos4 theta-1)^(10)` ?

Options

A. A) 0
B. B) 1
C. C) 2
D. D) 4

Answer

To solve the expression \((2\cos \theta + 1)^{10}(2\cos 2\theta - 1)^{10}(2\cos \theta - 1)^{10}(2\cos 4\theta - 1)^{10}\) given that \(\theta = \frac{\pi}{8}\), we can follow these steps:

### Step 1: Substitute \(\theta\)
We start by substituting \(\theta = \frac{\pi}{8}\) into the expression:
\[ (2\cos(\frac{\pi}{8}) + 1)^{10}(2\cos(\frac{\pi}{4}) - 1)^{10}(2\cos(\frac{\pi}{8}) - 1)^{10}(2\cos(\frac{\pi}{2}) - 1)^{10} \]

### Step 2: Calculate \(2\cos(\frac{\pi}{4}) - 1\)
We know that \(\cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}\):
\[ 2\cos(\frac{\pi}{4}) - 1 = 2 \cdot \frac{1}{\sqrt{2}} - 1 = \frac{2}{\sqrt{2}} - 1 = \sqrt{2} - 1 \]

### Step 3: Calculate \(2\cos(\frac{\pi}{2}) - 1\)
We know that \(\cos(\frac{\pi}{2}) = 0\):
\[ 2\cos(\frac{\pi}{2}) - 1 = 2 \cdot 0 - 1 = -1 \]

### Step 4: Substitute values into the expression
Now we can substitute these values back into the expression:
\[ (2\cos(\frac{\pi}{8}) + 1)^{10} \cdot (\sqrt{2} - 1)^{10} \cdot (2\cos(\frac{\pi}{8}) - 1)^{10} \cdot (-1)^{10} \]
Since \((-1)^{10} = 1\), we can simplify:
\[ (2\cos(\frac{\pi}{8}) + 1)^{10} \cdot (\sqrt{2} - 1)^{10} \cdot (2\cos(\frac{\pi}{8}) - 1)^{10} \]

### Step 5: Use the identity \(a^2 - b^2\)
Notice that we can use the identity \(a^2 - b^2 = (a + b)(a - b)\) on the first two terms:
\[ (2\cos(\frac{\pi}{8}) + 1)^{10} \cdot (2\cos(\frac{\pi}{8}) - 1)^{10} = [(2\cos(\frac{\pi}{8}))^2 - 1^2]^{10} = [4\cos^2(\frac{\pi}{8}) - 1]^{10} \]

### Step 6: Calculate \(4\cos^2(\frac{\pi}{8}) - 1\)
We need to find \(\cos^2(\frac{\pi}{8})\). Using the double angle formula:
\[ \cos(2\theta) = 2\cos^2(\theta) - 1 \implies \cos(\frac{\pi}{4}) = 2\cos^2(\frac{\pi}{8}) - 1 \]
Substituting \(\cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}\):
\[ \frac{1}{\sqrt{2}} = 2\cos^2(\frac{\pi}{8}) - 1 \implies 2\cos^2(\frac{\pi}{8}) = \frac{1}{\sqrt{2}} + 1 \]
\[ \cos^2(\frac{\pi}{8}) = \frac{1 + \sqrt{2}}{2\sqrt{2}} \]

### Step 7: Substitute back into the expression
Now substituting back:
\[ 4\cos^2(\frac{\pi}{8}) - 1 = 4 \cdot \frac{1 + \sqrt{2}}{2\sqrt{2}} - 1 = \frac{4(1 + \sqrt{2})}{2\sqrt{2}} - 1 = \frac{2(1 + \sqrt{2})}{\sqrt{2}} - 1 \]
This simplifies to:
\[ \frac{2 + 2\sqrt{2} - \sqrt{2}}{\sqrt{2}} = \frac{2 + \sqrt{2}}{\sqrt{2}} \]

### Step 8: Final Expression
Thus, we have:
\[ \left( \frac{2 + \sqrt{2}}{\sqrt{2}} \right)^{10} \cdot (\sqrt{2} - 1)^{10} \]
Using the identity \( (a + b)(a - b) = a^2 - b^2 \):
\[ \left( \sqrt{2} \right)^{20} = 2^{10} \]

### Final Answer
The final value of the expression is:
\[ \boxed{1} \]