Let `P(k)=(1+cospi/(4k))` `(1+cos((2k-1)pi)/(4k))` `(1+cos((2k+1)pi)/(4k))(1+cos((4k-1)pi)/(4k))dotT h e n` `P(3)=1/(16)` (b) `P(4)=(2-sqrt(2))/(16)` `P(5)=(3-sqrt(5))/(32)` (d) `P(6)(2-sqrt(3))/(16)`

Options

A. `P(3)=(1)/(16)`
B. `P(4)=(2-sqrt(2))/(16)`
C. `P(5)=(3-sqrt(5))/(32)`
D. `P(6)=(2-sqrt(3))/(16)`

Explanation

To solve the problem step by step, we will analyze the expression for \( P(k) \) and evaluate it for the given values of \( k \).

### Step 1: Write down the expression for \( P(k) \)
The expression for \( P(k) \) is given as:
\[ P(k) = (1 + \cos(\frac{\pi}{4k})) \cdot (1 + \cos(\frac{(2k-1)\pi}{4k})) \cdot (1 + \cos(\frac{(2k+1)\pi}{4k})) \cdot (1 + \cos(\frac{(4k-1)\pi}{4k})) \]

### Step 2: Simplify the cosine terms
Using the cosine angle subtraction and addition identities:
- \( \cos(\frac{\pi}{2} - x) = \sin(x) \)
- \( \cos(\frac{\pi}{2} + x) = -\sin(x) \)
- \( \cos(\pi - x) = -\cos(x) \)

We can rewrite the cosine terms:
- \( \cos(\frac{(2k-1)\pi}{4k}) = \cos(\frac{\pi}{2} - \frac{\pi}{4k}) = \sin(\frac{\pi}{4k}) \)
- \( \cos(\frac{(2k+1)\pi}{4k}) = \cos(\frac{\pi}{2} + \frac{\pi}{4k}) = -\sin(\frac{\pi}{4k}) \)
- \( \cos(\frac{(4k-1)\pi}{4k}) = \cos(\pi - \frac{\pi}{4k}) = -\cos(\frac{\pi}{4k}) \)

Thus, we can rewrite \( P(k) \) as:
\[ P(k) = (1 + \cos(\frac{\pi}{4k})) \cdot (1 + \sin(\frac{\pi}{4k})) \cdot (1 - \sin(\frac{\pi}{4k})) \cdot (1 - \cos(\frac{\pi}{4k})) \]

### Step 3: Use the identity \( a^2 - b^2 = (a-b)(a+b) \)
We can simplify the product:
\[ (1 + \sin(\frac{\pi}{4k}))(1 - \sin(\frac{\pi}{4k})) = 1 - \sin^2(\frac{\pi}{4k}) = \cos^2(\frac{\pi}{4k}) \]
\[ (1 + \cos(\frac{\pi}{4k}))(1 - \cos(\frac{\pi}{4k})) = 1 - \cos^2(\frac{\pi}{4k}) = \sin^2(\frac{\pi}{4k}) \]

Thus, we have:
\[ P(k) = \cos^2(\frac{\pi}{4k}) \cdot \sin^2(\frac{\pi}{4k}) \]

### Step 4: Rewrite using double angle identity
Using the double angle identity \( \sin(2x) = 2 \sin(x) \cos(x) \):
\[ P(k) = \frac{1}{4} \sin^2(\frac{\pi}{2k}) \]

### Step 5: Evaluate \( P(k) \) for specific values of \( k \)
Now we will evaluate \( P(k) \) for \( k = 3, 4, 5, 6 \).

1. **For \( k = 3 \)**:
\[ P(3) = \frac{1}{4} \sin^2(\frac{\pi}{6}) = \frac{1}{4} \left(\frac{1}{2}\right)^2 = \frac{1}{4} \cdot \frac{1}{4} = \frac{1}{16} \]

2. **For \( k = 4 \)**:
\[ P(4) = \frac{1}{4} \sin^2(\frac{\pi}{8}) = \frac{1}{4} \cdot \left(\frac{\sqrt{2 - \sqrt{2}}}{2}\right)^2 = \frac{1}{4} \cdot \frac{2 - \sqrt{2}}{4} = \frac{2 - \sqrt{2}}{16} \]

3. **For \( k = 5 \)**:
\[ P(5) = \frac{1}{4} \sin^2(\frac{\pi}{10}) = \frac{1}{4} \cdot \left(\frac{\sqrt{5} - 1}{4}\right)^2 = \frac{1}{4} \cdot \frac{(5 - 2\sqrt{5} + 1)}{16} = \frac{6 - 2\sqrt{5}}{64} = \frac{3 - \sqrt{5}}{32} \]

4. **For \( k = 6 \)**:
\[ P(6) = \frac{1}{4} \sin^2(\frac{\pi}{12}) = \frac{1}{4} \cdot \left(\frac{\sqrt{6} - \sqrt{2}}{4}\right)^2 = \frac{1}{4} \cdot \frac{(6 - 2\sqrt{12})}{16} = \frac{2 - \sqrt{3}}{16} \]

### Final Results:
- \( P(3) = \frac{1}{16} \)
- \( P(4) = \frac{2 - \sqrt{2}}{16} \)
- \( P(5) = \frac{3 - \sqrt{5}}{32} \)
- \( P(6) = \frac{2 - \sqrt{3}}{16} \)