`cos ^(2) alpha + cos ^(2) (alpha + 120^(@)) + cos ^(2) (alpha - 120^(@))` is equal to

Options

A. `3/2`
B. `1`
C. `1/2`
D. `0`

Answer

To solve the expression \( \cos^2 \alpha + \cos^2 (\alpha + 120^\circ) + \cos^2 (\alpha - 120^\circ) \), we can follow these steps:

### Step 1: Use the Cosine Addition Formula
We know that:
\[ \cos(\alpha + 120^\circ) = \cos \alpha \cos 120^\circ - \sin \alpha \sin 120^\circ \]
\[ \cos(\alpha - 120^\circ) = \cos \alpha \cos (-120^\circ) - \sin \alpha \sin (-120^\circ) \]
Using the values \( \cos 120^\circ = -\frac{1}{2} \) and \( \sin 120^\circ = \frac{\sqrt{3}}{2} \), we can write:
\[ \cos(\alpha + 120^\circ) = \cos \alpha \left(-\frac{1}{2}\right) - \sin \alpha \left(\frac{\sqrt{3}}{2}\right) \]
\[ \cos(\alpha - 120^\circ) = \cos \alpha \left(-\frac{1}{2}\right) + \sin \alpha \left(\frac{\sqrt{3}}{2}\right) \]

### Step 2: Substitute and Simplify
Now substituting these values into the original expression:
\[ \cos^2 \alpha + \left(-\frac{1}{2} \cos \alpha - \frac{\sqrt{3}}{2} \sin \alpha\right)^2 + \left(-\frac{1}{2} \cos \alpha + \frac{\sqrt{3}}{2} \sin \alpha\right)^2 \]

### Step 3: Expand the Squares
Expanding the squares:
\[ \left(-\frac{1}{2} \cos \alpha - \frac{\sqrt{3}}{2} \sin \alpha\right)^2 = \frac{1}{4} \cos^2 \alpha + \frac{3}{4} \sin^2 \alpha + \frac{\sqrt{3}}{2} \cos \alpha \sin \alpha \]
\[ \left(-\frac{1}{2} \cos \alpha + \frac{\sqrt{3}}{2} \sin \alpha\right)^2 = \frac{1}{4} \cos^2 \alpha + \frac{3}{4} \sin^2 \alpha - \frac{\sqrt{3}}{2} \cos \alpha \sin \alpha \]

### Step 4: Combine All Terms
Now, adding all the terms together:
\[ \cos^2 \alpha + \left(\frac{1}{4} \cos^2 \alpha + \frac{3}{4} \sin^2 \alpha + \frac{\sqrt{3}}{2} \cos \alpha \sin \alpha\right) + \left(\frac{1}{4} \cos^2 \alpha + \frac{3}{4} \sin^2 \alpha - \frac{\sqrt{3}}{2} \cos \alpha \sin \alpha\right) \]
Combining like terms:
\[ \cos^2 \alpha + \frac{1}{2} \cos^2 \alpha + \frac{3}{2} \sin^2 \alpha = \frac{3}{2} \cos^2 \alpha + \frac{3}{2} \sin^2 \alpha \]

### Step 5: Use Pythagorean Identity
Using the identity \( \cos^2 \alpha + \sin^2 \alpha = 1 \):
\[ \frac{3}{2} (\cos^2 \alpha + \sin^2 \alpha) = \frac{3}{2} \cdot 1 = \frac{3}{2} \]

### Final Result
Thus, the expression simplifies to:
\[ \cos^2 \alpha + \cos^2 (\alpha + 120^\circ) + \cos^2 (\alpha - 120^\circ) = \frac{3}{2} \]