If `sin (7 phi+9^(@))=cos2phi`, find a value of `phi`.

Answer

To solve the equation \( \sin(7\phi + 9^\circ) = \cos(2\phi) \), we can follow these steps:

### Step 1: Use the co-function identity
We know that \( \cos(2\phi) \) can be expressed in terms of sine:
\[ \cos(2\phi) = \sin(90^\circ - 2\phi) \]
Thus, we can rewrite the equation as:
\[ \sin(7\phi + 9^\circ) = \sin(90^\circ - 2\phi) \]

### Step 2: Set the arguments equal to each other
Since the sine function is periodic, we can set the arguments equal to each other:
\[ 7\phi + 9^\circ = 90^\circ - 2\phi \]

### Step 3: Solve for \( \phi \)
Now, we can solve for \( \phi \):
1. Add \( 2\phi \) to both sides:
\[ 7\phi + 2\phi + 9^\circ = 90^\circ \]
This simplifies to:
\[ 9\phi + 9^\circ = 90^\circ \]

2. Subtract \( 9^\circ \) from both sides:
\[ 9\phi = 90^\circ - 9^\circ \]
This simplifies to:
\[ 9\phi = 81^\circ \]

3. Divide both sides by 9:
\[ \phi = \frac{81^\circ}{9} = 9^\circ \]

### Conclusion
Thus, the value of \( \phi \) is:
\[ \phi = 9^\circ \]

---