`sin (60^(@) + theta) - sin ( 60^(@) - theta ) = sin theta. `

Answer

To solve the equation \( \sin(60^\circ + \theta) - \sin(60^\circ - \theta) = \sin \theta \), we will use the sine addition and subtraction formulas.

### Step-by-Step Solution:

1. **Identify the Left-Hand Side (LHS)**:
We start with the expression on the left-hand side:
\[ \sin(60^\circ + \theta) - \sin(60^\circ - \theta) \]

2. **Apply the Sine Addition Formula**:
The sine addition formula states that:
\[ \sin(a + b) = \sin a \cos b + \cos a \sin b \]
For \( a = 60^\circ \) and \( b = \theta \):
\[ \sin(60^\circ + \theta) = \sin 60^\circ \cos \theta + \cos 60^\circ \sin \theta \]
Substituting the known values:
\[ \sin 60^\circ = \frac{\sqrt{3}}{2}, \quad \cos 60^\circ = \frac{1}{2} \]
Thus,
\[ \sin(60^\circ + \theta) = \frac{\sqrt{3}}{2} \cos \theta + \frac{1}{2} \sin \theta \]

3. **Apply the Sine Subtraction Formula**:
The sine subtraction formula states that:
\[ \sin(a - b) = \sin a \cos b - \cos a \sin b \]
For \( a = 60^\circ \) and \( b = \theta \):
\[ \sin(60^\circ - \theta) = \sin 60^\circ \cos \theta - \cos 60^\circ \sin \theta \]
Thus,
\[ \sin(60^\circ - \theta) = \frac{\sqrt{3}}{2} \cos \theta - \frac{1}{2} \sin \theta \]

4. **Substitute Back into the LHS**:
Now we substitute both results back into the LHS:
\[ \sin(60^\circ + \theta) - \sin(60^\circ - \theta) = \left( \frac{\sqrt{3}}{2} \cos \theta + \frac{1}{2} \sin \theta \right) - \left( \frac{\sqrt{3}}{2} \cos \theta - \frac{1}{2} \sin \theta \right) \]

5. **Simplify the Expression**:
Simplifying the above expression:
\[ = \frac{\sqrt{3}}{2} \cos \theta + \frac{1}{2} \sin \theta - \frac{\sqrt{3}}{2} \cos \theta + \frac{1}{2} \sin \theta \]
The \( \frac{\sqrt{3}}{2} \cos \theta \) terms cancel out:
\[ = \frac{1}{2} \sin \theta + \frac{1}{2} \sin \theta = \sin \theta \]

6. **Final Result**:
Thus, we have shown that:
\[ \sin(60^\circ + \theta) - \sin(60^\circ - \theta) = \sin \theta \]
Therefore, the equation is proved.