Prove that ` cos theta + cos (120^(@) + theta) + cos ( 120^(@)-theta) =0. `
Hence deduce that `cos ^(3) theta + cos ^(3) ( 120^(@) + theta ) + cos^3( 120^(@)-theta) = 3/4 cos 3 theta.`

Answer

To prove the equation \( \cos \theta + \cos(120^\circ + \theta) + \cos(120^\circ - \theta) = 0 \), we can use the cosine addition and subtraction formulas. Let's go through the solution step by step.

### Step 1: Use the Cosine Addition Formula
We start with the expression:
\[ \cos(120^\circ + \theta) + \cos(120^\circ - \theta) \]
Using the cosine addition formula:
\[ \cos(a + b) + \cos(a - b) = 2 \cos a \cos b \]
Let \( a = 120^\circ \) and \( b = \theta \). Thus, we have:
\[ \cos(120^\circ + \theta) + \cos(120^\circ - \theta) = 2 \cos(120^\circ) \cos(\theta) \]

### Step 2: Substitute the Value of \( \cos(120^\circ) \)
We know that:
\[ \cos(120^\circ) = -\frac{1}{2} \]
Substituting this value into our equation gives:
\[ \cos(120^\circ + \theta) + \cos(120^\circ - \theta) = 2 \left(-\frac{1}{2}\right) \cos(\theta) = -\cos(\theta) \]

### Step 3: Combine the Terms
Now, we can substitute this back into our original equation:
\[ \cos \theta + \cos(120^\circ + \theta) + \cos(120^\circ - \theta) = \cos \theta - \cos \theta = 0 \]
Thus, we have proven that:
\[ \cos \theta + \cos(120^\circ + \theta) + \cos(120^\circ - \theta) = 0 \]

### Step 4: Deduce the Second Equation
Now, we will deduce that:
\[ \cos^3 \theta + \cos^3(120^\circ + \theta) + \cos^3(120^\circ - \theta) = \frac{3}{4} \cos 3\theta \]
Using the identity for the sum of cubes:
\[ a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - ac - bc) \]
Where \( a = \cos \theta \), \( b = \cos(120^\circ + \theta) \), and \( c = \cos(120^\circ - \theta) \). Since we have already established that \( a + b + c = 0 \), we can simplify:
\[ \cos^3 \theta + \cos^3(120^\circ + \theta) + \cos^3(120^\circ - \theta) = 3 \cos \theta \cos(120^\circ + \theta) \cos(120^\circ - \theta) \]

### Step 5: Calculate the Product
Using the product of cosines:
\[ \cos(120^\circ + \theta) \cos(120^\circ - \theta) = \cos^2(120^\circ) \cos^2(\theta) - \sin^2(120^\circ) \sin^2(\theta) \]
Substituting \( \cos(120^\circ) = -\frac{1}{2} \) and \( \sin(120^\circ) = \frac{\sqrt{3}}{2} \):
\[ = \left(-\frac{1}{2}\right)^2 \cos^2(\theta) - \left(\frac{\sqrt{3}}{2}\right)^2 \sin^2(\theta) \]
\[ = \frac{1}{4} \cos^2(\theta) - \frac{3}{4} \sin^2(\theta) \]
\[ = \frac{1}{4} (\cos^2(\theta) - 3\sin^2(\theta)) \]
Using \( \sin^2(\theta) = 1 - \cos^2(\theta) \):
\[ = \frac{1}{4} (4\cos^2(\theta) - 3) = \frac{1}{4} (4\cos^2(\theta) - 3) \]

### Step 6: Substitute Back
Thus, we have:
\[ \cos^3 \theta + \cos^3(120^\circ + \theta) + \cos^3(120^\circ - \theta) = 3 \cos \theta \cdot \frac{1}{4} (4 \cos^2 \theta - 3) \]
\[ = \frac{3}{4} (4 \cos^3 \theta - 3 \cos \theta) \]
This simplifies to:
\[ = \frac{3}{4} \cos 3\theta \]
Thus, we have proven that:
\[ \cos^3 \theta + \cos^3(120^\circ + \theta) + \cos^3(120^\circ - \theta) = \frac{3}{4} \cos 3\theta \]