The arithmetic mean of the numbers `2sin2^(@), 4sin4^(@), 6 sin 6^(@), …………. 178 sin 178^(@), 180 sin 180^(@)`

Options

A. `sin1^(@)`
B. `cot1^(@)`
C. `tan1^(@)`
D. `cos 1^(@)`

Answer

To find the arithmetic mean of the numbers \(2 \sin(2^\circ), 4 \sin(4^\circ), 6 \sin(6^\circ), \ldots, 178 \sin(178^\circ), 180 \sin(180^\circ)\), we can follow these steps:

### Step 1: Identify the series
The series consists of terms of the form \(2n \sin(n^\circ)\) where \(n\) takes values from 1 to 90. Thus, we can express our series as:
\[ S = 2 \sin(2^\circ) + 4 \sin(4^\circ) + 6 \sin(6^\circ) + \ldots + 180 \sin(180^\circ) \]

### Step 2: Rewrite the series
We can factor out 2 from each term:
\[ S = 2 \left( \sin(2^\circ) + 2 \sin(4^\circ) + 3 \sin(6^\circ) + \ldots + 90 \sin(90^\circ) \right) \]

### Step 3: Count the number of terms
The series has \(90\) terms since \(n\) ranges from \(1\) to \(90\).

### Step 4: Calculate the sum of the series
We can denote the sum of the series inside the parentheses as \(T\):
\[ T = \sin(2^\circ) + 2 \sin(4^\circ) + 3 \sin(6^\circ) + \ldots + 90 \sin(90^\circ) \]

### Step 5: Use the sine addition formula
To simplify \(T\), we can use the identity for the sum of sines:
\[ \sum_{k=1}^{n} k \sin(k \theta) = \frac{n \sin\left(\frac{n \theta}{2}\right) \sin\left(\frac{(n+1) \theta}{2}\right)}{\sin\left(\frac{\theta}{2}\right)} \]
In our case, \(\theta = 2^\circ\) and \(n = 90\).

### Step 6: Substitute values into the formula
Substituting \(n = 90\) and \(\theta = 2^\circ\):
\[ T = \frac{90 \sin(90^\circ) \sin(91^\circ)}{\sin(1^\circ)} \]
Since \(\sin(90^\circ) = 1\), we have:
\[ T = \frac{90 \sin(91^\circ)}{\sin(1^\circ)} \]

### Step 7: Calculate the arithmetic mean
The arithmetic mean \(A\) is given by:
\[ A = \frac{S}{n} = \frac{2T}{90} = \frac{2}{90} \cdot \frac{90 \sin(91^\circ)}{\sin(1^\circ)} = \frac{2 \sin(91^\circ)}{\sin(1^\circ)} \]
Using the fact that \(\sin(91^\circ) = \sin(89^\circ)\):
\[ A = \frac{2 \sin(89^\circ)}{\sin(1^\circ)} \]

### Step 8: Final simplification
Since \(\sin(89^\circ) \approx \sin(90^\circ) = 1\):
\[ A \approx \frac{2}{\sin(1^\circ)} \]

### Conclusion
Thus, the arithmetic mean of the given series is approximately:
\[ A \approx 2 \cdot \frac{1}{\sin(1^\circ)} \]