Prove that the average of the numbers `n sin n^@`, `n = 2,4,6...180` is `cot 1^@`

Answer

To prove that the average of the numbers \( n \sin n^\circ \) for \( n = 2, 4, 6, \ldots, 180 \) is \( \cot 1^\circ \), we will follow these steps:

### Step 1: Identify the terms and their sum
The terms we are considering are:
\[ 2 \sin 2^\circ, 4 \sin 4^\circ, 6 \sin 6^\circ, \ldots, 180 \sin 180^\circ \]
The average \( A \) of these terms can be expressed as:
\[ A = \frac{2 \sin 2^\circ + 4 \sin 4^\circ + 6 \sin 6^\circ + \ldots + 180 \sin 180^\circ}{\text{Total number of terms}} \]

### Step 2: Determine the total number of terms
The sequence \( n = 2, 4, 6, \ldots, 180 \) consists of even numbers from 2 to 180. The total number of terms can be calculated as:
\[ \text{Total number of terms} = \frac{180 - 2}{2} + 1 = 90 \]

### Step 3: Simplify the sum
Since \( \sin 180^\circ = 0 \), we only need to sum up to \( 178 \):
\[ A = \frac{2 \sin 2^\circ + 4 \sin 4^\circ + \ldots + 178 \sin 178^\circ}{90} \]

### Step 4: Use the sine identity
Using the identity \( \sin(180^\circ - x) = \sin x \), we can pair the terms:
\[ \sin 178^\circ = \sin 2^\circ, \quad \sin 176^\circ = \sin 4^\circ, \quad \ldots, \quad \sin 92^\circ = \sin 88^\circ \]
This allows us to rewrite the average as:
\[ A = \frac{(2 + 178) \sin 2^\circ + (4 + 176) \sin 4^\circ + \ldots + (88 + 92) \sin 88^\circ + 90 \sin 90^\circ}{90} \]

### Step 5: Combine terms
Each pair sums to 180:
\[ A = \frac{180 (\sin 2^\circ + \sin 4^\circ + \ldots + \sin 88^\circ) + 90}{90} \]
This simplifies to:
\[ A = 2(\sin 2^\circ + \sin 4^\circ + \ldots + \sin 88^\circ) + 1 \]

### Step 6: Use the sine summation formula
The formula for the sum of sines is:
\[ \sum_{k=1}^{n} \sin(k \theta) = \frac{\sin\left(\frac{n \theta}{2}\right) \sin\left(\frac{(n+1) \theta}{2}\right)}{\sin\left(\frac{\theta}{2}\right)} \]
For \( \theta = 2^\circ \) and \( n = 44 \):
\[ \sum_{k=1}^{44} \sin(2k^\circ) = \frac{\sin(89^\circ) \sin(90^\circ)}{\sin(1^\circ)} = \frac{\sin(89^\circ)}{\sin(1^\circ)} = \frac{\cos(1^\circ)}{\sin(1^\circ)} = \cot(1^\circ) \]

### Step 7: Substitute back into the average
Now substituting this back into our average:
\[ A = 2 \cdot \cot(1^\circ) + 1 \]

### Step 8: Final simplification
Thus, we find:
\[ A = \cot(1^\circ) \]

### Conclusion
Hence, we have proved that the average of the numbers \( n \sin n^\circ \) for \( n = 2, 4, 6, \ldots, 180 \) is indeed \( \cot(1^\circ) \).

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