`sin ( 240^(@) + theta) + cos (330^(@) + theta ) = 0`

Answer

To solve the equation \( \sin(240^\circ + \theta) + \cos(330^\circ + \theta) = 0 \), we will follow these steps:

### Step 1: Write down the equation
We start with the equation:
\[ \sin(240^\circ + \theta) + \cos(330^\circ + \theta) = 0 \]

### Step 2: Use the sine and cosine addition formulas
We will use the sine and cosine addition formulas:
- \( \sin(x + y) = \sin x \cos y + \cos x \sin y \)
- \( \cos(x + y) = \cos x \cos y - \sin x \sin y \)

Applying these formulas, we have:
\[ \sin(240^\circ + \theta) = \sin 240^\circ \cos \theta + \cos 240^\circ \sin \theta \]
\[ \cos(330^\circ + \theta) = \cos 330^\circ \cos \theta - \sin 330^\circ \sin \theta \]

### Step 3: Substitute the values of sine and cosine
Now we need to find the values of \( \sin 240^\circ \), \( \cos 240^\circ \), \( \cos 330^\circ \), and \( \sin 330^\circ \):
- \( \sin 240^\circ = -\frac{\sqrt{3}}{2} \)
- \( \cos 240^\circ = -\frac{1}{2} \)
- \( \cos 330^\circ = \frac{\sqrt{3}}{2} \)
- \( \sin 330^\circ = -\frac{1}{2} \)

Substituting these values into our equations gives:
\[ \sin(240^\circ + \theta) = -\frac{\sqrt{3}}{2} \cos \theta - \frac{1}{2} \sin \theta \]
\[ \cos(330^\circ + \theta) = \frac{\sqrt{3}}{2} \cos \theta + \frac{1}{2} \sin \theta \]

### Step 4: Combine the expressions
Now we can combine these expressions into our original equation:
\[ -\frac{\sqrt{3}}{2} \cos \theta - \frac{1}{2} \sin \theta + \left( \frac{\sqrt{3}}{2} \cos \theta + \frac{1}{2} \sin \theta \right) = 0 \]

### Step 5: Simplify the equation
Combining like terms:
\[ \left(-\frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2}\right) \cos \theta + \left(-\frac{1}{2} + \frac{1}{2}\right) \sin \theta = 0 \]
This simplifies to:
\[ 0 \cdot \cos \theta + 0 \cdot \sin \theta = 0 \]

### Step 6: Conclusion
Since both terms are equal to zero, we have:
\[ 0 = 0 \]
This confirms that the original equation holds true.