The value of `cos(-89^(@))+cos(-87^(@))+cos(-85^(@))+…+cos(85^(@))+cos(87^(@))+cos(89^(@))` is equal to

Options

A. `cosec 1^(@)`
B. `sec 1^(@)`
C. `2 sec 1^(@)`
D. `2 cosec 1^(@)`

Explanation

To find the value of the expression \( \cos(-89^\circ) + \cos(-87^\circ) + \cos(-85^\circ) + \ldots + \cos(85^\circ) + \cos(87^\circ) + \cos(89^\circ) \), we can follow these steps:

### Step 1: Utilize the property of cosine
We know that \( \cos(-\theta) = \cos(\theta) \). Therefore, we can rewrite the expression as:
\[ \cos(89^\circ) + \cos(87^\circ) + \cos(85^\circ) + \ldots + \cos(85^\circ) + \cos(87^\circ) + \cos(89^\circ) \]
This means each negative angle will have a corresponding positive angle, and we can pair them up.

### Step 2: Recognize the symmetry
The series can be grouped into pairs:
\[ 2(\cos(89^\circ) + \cos(87^\circ) + \cos(85^\circ) + \ldots + \cos(1^\circ)) \]
This is because \( \cos(89^\circ) \) pairs with \( \cos(-89^\circ) \), \( \cos(87^\circ) \) pairs with \( \cos(-87^\circ) \), and so on.

### Step 3: Identify the series
The series now consists of cosines of odd degrees from \( 1^\circ \) to \( 89^\circ \). The number of terms in this series can be calculated as follows:
- The odd numbers from \( 1 \) to \( 89 \) can be expressed as \( 1, 3, 5, \ldots, 89 \).
- This is an arithmetic series where the first term \( a = 1 \), the last term \( l = 89 \), and the common difference \( d = 2 \).

### Step 4: Calculate the number of terms
The number of terms \( n \) in the series can be calculated using the formula for the \( n \)-th term of an arithmetic series:
\[ l = a + (n-1)d \implies 89 = 1 + (n-1) \cdot 2 \]
Solving for \( n \):
\[ 89 - 1 = (n-1) \cdot 2 \implies 88 = (n-1) \cdot 2 \implies n-1 = 44 \implies n = 45 \]

### Step 5: Apply the sum formula for cosines
The sum of cosines can be calculated using the formula:
\[ \sum_{k=0}^{n-1} \cos(a + kd) = \frac{\sin\left(\frac{nd}{2}\right) \cdot \cos\left(a + \frac{(n-1)d}{2}\right)}{\sin\left(\frac{d}{2}\right)} \]
In our case:
- \( a = 1^\circ \)
- \( d = 2^\circ \)
- \( n = 45 \)

### Step 6: Substitute into the formula
Substituting the values into the formula:
\[ \sum_{k=0}^{44} \cos(1^\circ + 2k^\circ) = \frac{\sin\left(\frac{45 \cdot 2^\circ}{2}\right) \cdot \cos\left(1^\circ + \frac{(45-1) \cdot 2^\circ}{2}\right)}{\sin\left(\frac{2^\circ}{2}\right)} \]
This simplifies to:
\[ = \frac{\sin(45^\circ) \cdot \cos(44^\circ)}{\sin(1^\circ)} \]

### Step 7: Final calculation
Since \( \sin(45^\circ) = \frac{\sqrt{2}}{2} \):
\[ = \frac{\frac{\sqrt{2}}{2} \cdot \cos(44^\circ)}{\sin(1^\circ)} \]
Thus, the total value of the original expression is:
\[ 2 \cdot \frac{\frac{\sqrt{2}}{2} \cdot \cos(44^\circ)}{\sin(1^\circ)} = \frac{\sqrt{2} \cdot \cos(44^\circ)}{\sin(1^\circ)} \]

### Final Answer
The value of \( \cos(-89^\circ) + \cos(-87^\circ) + \ldots + \cos(89^\circ) \) is:
\[ \frac{\sqrt{2} \cdot \cos(44^\circ)}{\sin(1^\circ)} \]