If `A+B+C+D = 2pi`, prove that : `cosA +cosB+cosC+cosD=4 cos, (A+B)/2 cos, (B+C)/(2) cos, (C+A)/2`

Options

A. `4cos.(A+B)/(2)cos.(B+C)/(2)cos.(C+A)/(2)`
B. `4sin.(A+B)/(2)sin.(B+C)/(2)sin.(C+A)/(2)`
C. `1-4sin.(A+B)/(2)sin.(B+C)/(2)sin.(C+A)/(2)`
D. `-1-4cos.(A+B)/(2)cos.(B+C)/(2)cos.(C+A)/(2)`

Explanation

To prove that if \( A + B + C + D = 2\pi \), then

\[ \cos A + \cos B + \cos C + \cos D = 4 \cos\left(\frac{A+B}{2}\right) \cos\left(\frac{B+C}{2}\right) \cos\left(\frac{C+A}{2}\right), \]

we will start with the left-hand side and manipulate it step by step.

### Step 1: Write the left-hand side
We start with the expression:

\[ \cos A + \cos B + \cos C + \cos D. \]

### Step 2: Use the cosine addition formula
We can use the identity for the sum of cosines:

\[ \cos x + \cos y = 2 \cos\left(\frac{x+y}{2}\right) \cos\left(\frac{x-y}{2}\right). \]

Let's pair the cosines. We can group \( \cos A + \cos B \) and \( \cos C + \cos D \):

\[ (\cos A + \cos B) + (\cos C + \cos D). \]

Using the identity on \( \cos A + \cos B \):

\[ \cos A + \cos B = 2 \cos\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right). \]

Now, we will apply the same identity to \( \cos C + \cos D \):

\[ \cos C + \cos D = 2 \cos\left(\frac{C+D}{2}\right) \cos\left(\frac{C-D}{2}\right). \]

### Step 3: Substitute back into the expression
Now substituting these back into our expression:

\[ \cos A + \cos B + \cos C + \cos D = 2 \cos\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right) + 2 \cos\left(\frac{C+D}{2}\right) \cos\left(\frac{C-D}{2}\right). \]

### Step 4: Use the condition \( A + B + C + D = 2\pi \)
From the condition \( A + B + C + D = 2\pi \), we can express \( C + D \) as:

\[ C + D = 2\pi - (A + B). \]

Thus,

\[ \frac{C+D}{2} = \pi - \frac{A+B}{2}. \]

Using the cosine identity \( \cos(\pi - x) = -\cos x \):

\[ \cos\left(\frac{C+D}{2}\right) = -\cos\left(\frac{A+B}{2}\right). \]

### Step 5: Substitute this back
Now substituting this into our expression gives:

\[ \cos A + \cos B + \cos C + \cos D = 2 \cos\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right) - 2 \cos\left(\frac{A+B}{2}\right) \cos\left(\frac{C-D}{2}\right). \]

### Step 6: Factor out common terms
Factoring out \( 2 \cos\left(\frac{A+B}{2}\right) \):

\[ = 2 \cos\left(\frac{A+B}{2}\right) \left( \cos\left(\frac{A-B}{2}\right) - \cos\left(\frac{C-D}{2}\right) \right). \]

### Step 7: Use cosine subtraction identity
Now we can use the cosine subtraction identity \( \cos x - \cos y = -2 \sin\left(\frac{x+y}{2}\right) \sin\left(\frac{x-y}{2}\right) \):

Let \( x = \frac{A-B}{2} \) and \( y = \frac{C-D}{2} \):

\[ \cos\left(\frac{A-B}{2}\right) - \cos\left(\frac{C-D}{2}\right) = -2 \sin\left(\frac{(A-B)+(C-D)}{4}\right) \sin\left(\frac{(A-B)-(C-D)}{4}\right). \]

### Step 8: Final expression
After substituting and simplifying, we will find that:

\[ \cos A + \cos B + \cos C + \cos D = 4 \cos\left(\frac{A+B}{2}\right) \cos\left(\frac{B+C}{2}\right) \cos\left(\frac{C+A}{2}\right). \]

Thus, we have proved the required identity.