Prove that sin(θ+30°)+cos(θ+60°)= cos θ.

Answer

To prove that \( \sin(\theta + 30^\circ) + \cos(\theta + 60^\circ) = \cos \theta \), we will start with the left-hand side (LHS) and simplify it step by step.

### Step 1: Write down the LHS
We start with the expression:
\[ \text{LHS} = \sin(\theta + 30^\circ) + \cos(\theta + 60^\circ) \]

### Step 2: Apply the sine addition formula
Using the sine addition formula, \( \sin(a + b) = \sin a \cos b + \cos a \sin b \), we can expand \( \sin(\theta + 30^\circ) \):
\[ \sin(\theta + 30^\circ) = \sin \theta \cos 30^\circ + \cos \theta \sin 30^\circ \]
Substituting the values of \( \cos 30^\circ = \frac{\sqrt{3}}{2} \) and \( \sin 30^\circ = \frac{1}{2} \):
\[ \sin(\theta + 30^\circ) = \sin \theta \cdot \frac{\sqrt{3}}{2} + \cos \theta \cdot \frac{1}{2} \]

### Step 3: Apply the cosine addition formula
Next, we expand \( \cos(\theta + 60^\circ) \) using the cosine addition formula, \( \cos(a + b) = \cos a \cos b - \sin a \sin b \):
\[ \cos(\theta + 60^\circ) = \cos \theta \cos 60^\circ - \sin \theta \sin 60^\circ \]
Substituting the values of \( \cos 60^\circ = \frac{1}{2} \) and \( \sin 60^\circ = \frac{\sqrt{3}}{2} \):
\[ \cos(\theta + 60^\circ) = \cos \theta \cdot \frac{1}{2} - \sin \theta \cdot \frac{\sqrt{3}}{2} \]

### Step 4: Combine the results
Now we can combine the results from Steps 2 and 3:
\[ \text{LHS} = \left( \sin \theta \cdot \frac{\sqrt{3}}{2} + \cos \theta \cdot \frac{1}{2} \right) + \left( \cos \theta \cdot \frac{1}{2} - \sin \theta \cdot \frac{\sqrt{3}}{2} \right) \]
This simplifies to:
\[ \text{LHS} = \sin \theta \cdot \frac{\sqrt{3}}{2} + \cos \theta \cdot \frac{1}{2} + \cos \theta \cdot \frac{1}{2} - \sin \theta \cdot \frac{\sqrt{3}}{2} \]
The \( \sin \theta \cdot \frac{\sqrt{3}}{2} \) terms cancel out:
\[ \text{LHS} = \cos \theta \cdot \frac{1}{2} + \cos \theta \cdot \frac{1}{2} = \cos \theta \]

### Step 5: Conclusion
Thus, we have shown that:
\[ \sin(\theta + 30^\circ) + \cos(\theta + 60^\circ) = \cos \theta \]
Therefore, we can conclude:
\[ \text{LHS} = \text{RHS} \]
Hence, the statement is proved.