The value of `sum_(k=1)^(3) cos^(2)(2k-1)(pi)/(12),` is

Options

A. 0
B. `1//2`
C. `-1//2`
D. `3//2`

Answer

To solve the problem of finding the value of

\[ \sum_{k=1}^{3} \cos^2\left( (2k-1) \frac{\pi}{12} \right), \]

we will evaluate the sum step by step.

### Step 1: Expand the summation
We start by substituting the values of \( k \) from 1 to 3 into the expression.

- For \( k = 1 \):
\[ \cos^2\left( (2 \cdot 1 - 1) \frac{\pi}{12} \right) = \cos^2\left( \frac{\pi}{12} \right) \]

- For \( k = 2 \):
\[ \cos^2\left( (2 \cdot 2 - 1) \frac{\pi}{12} \right) = \cos^2\left( \frac{3\pi}{12} \right) = \cos^2\left( \frac{\pi}{4} \right) \]

- For \( k = 3 \):
\[ \cos^2\left( (2 \cdot 3 - 1) \frac{\pi}{12} \right) = \cos^2\left( \frac{5\pi}{12} \right) \]

Thus, we can rewrite the sum as:
\[ \sum_{k=1}^{3} \cos^2\left( (2k-1) \frac{\pi}{12} \right) = \cos^2\left( \frac{\pi}{12} \right) + \cos^2\left( \frac{\pi}{4} \right) + \cos^2\left( \frac{5\pi}{12} \right) \]

### Step 2: Calculate \( \cos^2\left( \frac{\pi}{4} \right) \)
We know that:
\[ \cos\left( \frac{\pi}{4} \right) = \frac{1}{\sqrt{2}} \implies \cos^2\left( \frac{\pi}{4} \right) = \left( \frac{1}{\sqrt{2}} \right)^2 = \frac{1}{2} \]

### Step 3: Calculate \( \cos^2\left( \frac{5\pi}{12} \right) \)
Using the identity \( \cos\left( \frac{5\pi}{12} \right) = \sin\left( \frac{\pi}{12} \right) \):
\[ \cos^2\left( \frac{5\pi}{12} \right) = \sin^2\left( \frac{\pi}{12} \right) \]

### Step 4: Use the Pythagorean identity
We know that:
\[ \sin^2\theta + \cos^2\theta = 1 \]
Thus,
\[ \sin^2\left( \frac{\pi}{12} \right) + \cos^2\left( \frac{\pi}{12} \right) = 1 \]
This implies:
\[ \sin^2\left( \frac{\pi}{12} \right) = 1 - \cos^2\left( \frac{\pi}{12} \right) \]

### Step 5: Substitute back into the sum
Now we can substitute back into our sum:
\[ \cos^2\left( \frac{\pi}{12} \right) + \frac{1}{2} + \sin^2\left( \frac{\pi}{12} \right) \]
Substituting \( \sin^2\left( \frac{\pi}{12} \right) \):
\[ \cos^2\left( \frac{\pi}{12} \right) + \frac{1}{2} + (1 - \cos^2\left( \frac{\pi}{12} \right)) = 1 + \frac{1}{2} = \frac{3}{2} \]

### Final Answer
Thus, the value of the sum is:
\[ \sum_{k=1}^{3} \cos^2\left( (2k-1) \frac{\pi}{12} \right) = \frac{3}{2} \]