The average of `sin 2^(@) , sin 4^(@) , sin 6^(@),"….." sin180^(@)` is

Options

A. `(1)/( 90) cos 1^(@)`
B. `(1)/(90) sin1^(@)`
C. `(1)/( 90)cot 1^(@)`
D. none

Answer

To find the average of the series \( \sin 2^\circ, \sin 4^\circ, \sin 6^\circ, \ldots, \sin 180^\circ \), we can follow these steps:

### Step 1: Identify the series
The series consists of sine values for angles that are multiples of 2 degrees, starting from 2 degrees up to 180 degrees. The angles can be expressed as:
\[ \sin 2^\circ, \sin 4^\circ, \sin 6^\circ, \ldots, \sin 180^\circ \]

### Step 2: Determine the number of terms
The angles form an arithmetic sequence where the first term \( a = 2^\circ \) and the common difference \( d = 2^\circ \). The last term is \( 180^\circ \). To find the number of terms \( n \), we can use the formula for the \( n \)-th term of an arithmetic sequence:
\[ a_n = a + (n-1)d \]
Setting \( a_n = 180^\circ \):
\[ 180 = 2 + (n-1) \cdot 2 \]
Solving for \( n \):
\[ 180 - 2 = (n-1) \cdot 2 \\ 178 = (n-1) \cdot 2 \\ n - 1 = 89 \\ n = 90 \]

### Step 3: Use the sum of sine series formula
The sum of a sine series can be calculated using the formula:
\[ S = \frac{\sin\left(\frac{n \cdot \beta}{2}\right) \cdot \sin\left(\alpha + \frac{(n-1) \cdot \beta}{2}\right)}{\sin\left(\frac{\beta}{2}\right)} \]
where:
- \( \alpha = 2^\circ \)
- \( \beta = 2^\circ \)
- \( n = 90 \)

Substituting the values:
\[ S = \frac{\sin\left(\frac{90 \cdot 2}{2}\right) \cdot \sin\left(2 + \frac{(90-1) \cdot 2}{2}\right)}{\sin\left(\frac{2}{2}\right)} \]
This simplifies to:
\[ S = \frac{\sin(90^\circ) \cdot \sin(91^\circ)}{\sin(1^\circ)} \]
Since \( \sin(90^\circ) = 1 \):
\[ S = \frac{\sin(91^\circ)}{\sin(1^\circ)} \]

### Step 4: Simplify \( \sin(91^\circ) \)
Using the identity \( \sin(91^\circ) = \cos(1^\circ) \):
\[ S = \frac{\cos(1^\circ)}{\sin(1^\circ)} = \cot(1^\circ) \]

### Step 5: Calculate the average
The average \( A \) of the series is given by:
\[ A = \frac{S}{n} = \frac{\cot(1^\circ)}{90} \]

### Final Answer
Thus, the average of the series \( \sin 2^\circ, \sin 4^\circ, \sin 6^\circ, \ldots, \sin 180^\circ \) is:
\[ \boxed{\frac{\cot(1^\circ)}{90}} \]