`(a+b+c) (cos A+cos B+cos C) =`

Options

A. `Sigma a sin^2 " A/2`
B. `Sigma a cos^2 " A/2`
C. `2Sigma a sin^2 " A/2`
D. `2Sigma a cos^2 " A/2`

Answer

To solve the expression \((a+b+c)(\cos A + \cos B + \cos C)\), we will follow a systematic approach:

### Step 1: Expand the Expression
We start by distributing the terms in the expression:
\[ (a+b+c)(\cos A + \cos B + \cos C) = a\cos A + a\cos B + a\cos C + b\cos A + b\cos B + b\cos C + c\cos A + c\cos B + c\cos C \]

### Step 2: Group the Terms
Now, we can group the terms based on the coefficients:
\[ = a\cos A + b\cos A + c\cos A + a\cos B + b\cos B + c\cos B + a\cos C + b\cos C + c\cos C \]
This can be rearranged as:
\[ = (a+b+c)\cos A + (a+b+c)\cos B + (a+b+c)\cos C \]

### Step 3: Factor Out Common Terms
Notice that we can factor out \((a+b+c)\):
\[ = (a+b+c)(\cos A + \cos B + \cos C) \]

### Step 4: Use the Projection Formula
Using the projection formula, we can express the terms in a different way. The projection of side lengths onto the angles gives us:
\[ = \frac{1}{2}(a(\cos A + \cos B + \cos C) + b(\cos A + \cos B + \cos C) + c(\cos A + \cos B + \cos C)) \]
This leads us to:
\[ = \frac{1}{2}(a + b + c)(\cos A + \cos B + \cos C) \]

### Step 5: Final Result
Thus, we conclude that:
\[ (a+b+c)(\cos A + \cos B + \cos C) = \frac{1}{2}(a + b + c)(\cos A + \cos B + \cos C) \]

### Final Answer
The final value of \((a+b+c)(\cos A + \cos B + \cos C)\) is:
\[ \frac{1}{2}(a + b + c)(\cos A + \cos B + \cos C) \]