Find the value of the `sum_(k=0)^359 k.cos k^@`.

Options

A. `-90`
B. `-180`
C. `-270`
D. `-360`

Answer

To find the value of the summation \( \sum_{k=0}^{359} k \cos(k^\circ) \), we can follow these steps:

### Step 1: Write the summation explicitly
We start with the expression:
\[ S = \sum_{k=0}^{359} k \cos(k^\circ) = 0 \cdot \cos(0^\circ) + 1 \cdot \cos(1^\circ) + 2 \cdot \cos(2^\circ) + \ldots + 359 \cdot \cos(359^\circ) \]

### Step 2: Utilize the property of cosine
Using the property of cosine, \( \cos(k^\circ) = \cos(360^\circ - k^\circ) \), we can pair terms:
\[ S = \sum_{k=0}^{179} k \cos(k^\circ) + (360-k) \cos(k^\circ) \]
This means we can rewrite the sum as:
\[ S = \sum_{k=0}^{179} (k \cos(k^\circ) + (360 - k) \cos(k^\circ)) \]
\[ = \sum_{k=0}^{179} 360 \cos(k^\circ) \]

### Step 3: Simplify the summation
Now we can factor out the constant:
\[ S = 360 \sum_{k=0}^{179} \cos(k^\circ) \]

### Step 4: Evaluate the sum of cosines
The sum \( \sum_{k=0}^{179} \cos(k^\circ) \) can be evaluated using the formula for the sum of cosines:
\[ \sum_{k=0}^{n} \cos(k \theta) = \frac{\sin\left(\frac{(n+1)\theta}{2}\right) \cos\left(\frac{n\theta}{2}\right)}{\sin\left(\frac{\theta}{2}\right)} \]
For \( n = 179 \) and \( \theta = 1^\circ \):
\[ \sum_{k=0}^{179} \cos(k^\circ) = \frac{\sin\left(90^\circ\right) \cos\left(89.5^\circ\right)}{\sin\left(0.5^\circ\right)} \]
Since \( \sin(90^\circ) = 1 \):
\[ = \frac{\cos(89.5^\circ)}{\sin(0.5^\circ)} \]

### Step 5: Note the symmetry
Notice that \( \cos(89.5^\circ) \) is very small, and the sum from \( 0^\circ \) to \( 179^\circ \) will have terms that cancel out due to symmetry around \( 90^\circ \).

### Step 6: Final evaluation
The only non-canceling term is at \( k = 180 \):
\[ S = 360 \cdot 0 + 180 \cdot (-1) = -180 \]

### Conclusion
Thus, the value of the summation is:
\[ \boxed{-180} \]