Prove that `sin ( alpha + 30^(@)) = cos alpha + sin (alpha - 30^(@))`

Answer

To prove that \( \sin(\alpha + 30^\circ) = \cos \alpha + \sin(\alpha - 30^\circ) \), we will start with the right-hand side (RHS) and manipulate it to show that it equals the left-hand side (LHS).

### Step 1: Write down the RHS
We start with the right-hand side:
\[ \text{RHS} = \cos \alpha + \sin(\alpha - 30^\circ) \]

### Step 2: Use the sine subtraction formula
The sine subtraction formula states that:
\[ \sin(a - b) = \sin a \cos b - \cos a \sin b \]
Applying this to \( \sin(\alpha - 30^\circ) \):
\[ \sin(\alpha - 30^\circ) = \sin \alpha \cos 30^\circ - \cos \alpha \sin 30^\circ \]

### Step 3: Substitute known values of sine and cosine
We know that:
\[ \cos 30^\circ = \frac{\sqrt{3}}{2} \quad \text{and} \quad \sin 30^\circ = \frac{1}{2} \]
Substituting these values into the equation:
\[ \sin(\alpha - 30^\circ) = \sin \alpha \cdot \frac{\sqrt{3}}{2} - \cos \alpha \cdot \frac{1}{2} \]

### Step 4: Substitute back into the RHS
Now, substituting this back into the RHS:
\[ \text{RHS} = \cos \alpha + \left( \sin \alpha \cdot \frac{\sqrt{3}}{2} - \cos \alpha \cdot \frac{1}{2} \right) \]
This simplifies to:
\[ \text{RHS} = \cos \alpha + \frac{\sqrt{3}}{2} \sin \alpha - \frac{1}{2} \cos \alpha \]

### Step 5: Combine like terms
Combining the terms involving \( \cos \alpha \):
\[ \text{RHS} = \left(1 - \frac{1}{2}\right) \cos \alpha + \frac{\sqrt{3}}{2} \sin \alpha \]
This simplifies to:
\[ \text{RHS} = \frac{1}{2} \cos \alpha + \frac{\sqrt{3}}{2} \sin \alpha \]

### Step 6: Now, find the LHS
Now, we compute the left-hand side:
\[ \text{LHS} = \sin(\alpha + 30^\circ) \]
Using the sine addition formula:
\[ \sin(a + b) = \sin a \cos b + \cos a \sin b \]
Applying this to \( \sin(\alpha + 30^\circ) \):
\[ \sin(\alpha + 30^\circ) = \sin \alpha \cos 30^\circ + \cos \alpha \sin 30^\circ \]
Substituting the known values:
\[ \sin(\alpha + 30^\circ) = \sin \alpha \cdot \frac{\sqrt{3}}{2} + \cos \alpha \cdot \frac{1}{2} \]

### Step 7: Combine the terms
This simplifies to:
\[ \sin(\alpha + 30^\circ) = \frac{\sqrt{3}}{2} \sin \alpha + \frac{1}{2} \cos \alpha \]

### Step 8: Compare LHS and RHS
Now we compare:
\[ \text{LHS} = \frac{\sqrt{3}}{2} \sin \alpha + \frac{1}{2} \cos \alpha \]
\[ \text{RHS} = \frac{1}{2} \cos \alpha + \frac{\sqrt{3}}{2} \sin \alpha \]
Both sides are equal, thus proving:
\[ \sin(\alpha + 30^\circ) = \cos \alpha + \sin(\alpha - 30^\circ) \]

### Conclusion
Hence, we have proven that:
\[ \sin(\alpha + 30^\circ) = \cos \alpha + \sin(\alpha - 30^\circ) \]