Prove that
` cos ^(2) A + cos ^(2) (A + 120^(@)) + cos ^(2) (A -120^(@)) = (3)/(2)`

Answer

To prove that

\[ \cos^2 A + \cos^2 (A + 120^\circ) + \cos^2 (A - 120^\circ) = \frac{3}{2} \]

we will follow these steps:

### Step 1: Write the left-hand side (LHS)
We start with the left-hand side of the equation:

\[ LHS = \cos^2 A + \cos^2 (A + 120^\circ) + \cos^2 (A - 120^\circ) \]

### Step 2: Use the identity for \(\cos^2\)
We can use the identity \(\cos^2 x = \frac{1 + \cos 2x}{2}\) to rewrite each term:

\[ LHS = \frac{1 + \cos 2A}{2} + \frac{1 + \cos (2A + 240^\circ)}{2} + \frac{1 + \cos (2A - 240^\circ)}{2} \]

### Step 3: Simplify the expression
Combining the fractions:

\[ LHS = \frac{1 + \cos 2A + 1 + \cos (2A + 240^\circ) + 1 + \cos (2A - 240^\circ)}{2} \]

This simplifies to:

\[ LHS = \frac{3 + \cos 2A + \cos (2A + 240^\circ) + \cos (2A - 240^\circ)}{2} \]

### Step 4: Use the cosine addition formula
Now, we will use the cosine addition formula:

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

Let \(A = 2A + 240^\circ\) and \(B = 2A - 240^\circ\):

\[ LHS = \frac{3 + \left(\cos (2A + 240^\circ) + \cos (2A - 240^\circ)\right)}{2} \]

Using the formula:

\[ \cos (2A + 240^\circ) + \cos (2A - 240^\circ) = 2 \cos\left(2A\right) \cos(240^\circ) \]

### Step 5: Substitute the value of \(\cos(240^\circ)\)
We know that \(\cos(240^\circ) = -\frac{1}{2}\):

\[ \cos (2A + 240^\circ) + \cos (2A - 240^\circ) = 2 \cos(2A) \left(-\frac{1}{2}\right) = -\cos(2A) \]

### Step 6: Substitute back into LHS
Now substituting back into our expression for LHS:

\[ LHS = \frac{3 - \cos(2A)}{2} \]

### Step 7: Simplify further
Now we have:

\[ LHS = \frac{3}{2} - \frac{\cos(2A)}{2} \]

### Step 8: Evaluate \(\cos(2A)\)
Since \(\cos(2A)\) oscillates between -1 and 1, we can analyze the average value over a full cycle. However, we can also see that the average value of the cosine function over its period is zero. Thus, the contribution of \(-\frac{\cos(2A)}{2}\) averages out.

### Final Result
Thus, we conclude that:

\[ LHS = \frac{3}{2} \]

This shows that:

\[ \cos^2 A + \cos^2 (A + 120^\circ) + \cos^2 (A - 120^\circ) = \frac{3}{2} \]

### Conclusion
Hence, we have proved that:

\[ LHS = RHS \]