If `cos18^(@)-sin18^(@)=sqrt(n)sin27^(@)`, then `n=`

Answer

To solve the equation \( \cos 18^\circ - \sin 18^\circ = \sqrt{n} \sin 27^\circ \), we will follow these steps:

### Step 1: Rewrite \( \cos 18^\circ \)
We can express \( \cos 18^\circ \) in terms of sine:
\[ \cos 18^\circ = \sin(90^\circ - 18^\circ) = \sin 72^\circ \]
So, we rewrite the left side of the equation:
\[ \cos 18^\circ - \sin 18^\circ = \sin 72^\circ - \sin 18^\circ \]

### Step 2: Use the sine subtraction formula
We can apply the sine subtraction formula:
\[ \sin A - \sin B = 2 \cos\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right) \]
Let \( A = 72^\circ \) and \( B = 18^\circ \):
\[ \sin 72^\circ - \sin 18^\circ = 2 \cos\left(\frac{72^\circ + 18^\circ}{2}\right) \sin\left(\frac{72^\circ - 18^\circ}{2}\right) \]

### Step 3: Calculate the averages
Calculating the averages:
\[ \frac{72^\circ + 18^\circ}{2} = \frac{90^\circ}{2} = 45^\circ \]
\[ \frac{72^\circ - 18^\circ}{2} = \frac{54^\circ}{2} = 27^\circ \]

### Step 4: Substitute back into the equation
Substituting back into the equation gives:
\[ \sin 72^\circ - \sin 18^\circ = 2 \cos 45^\circ \sin 27^\circ \]
Since \( \cos 45^\circ = \frac{1}{\sqrt{2}} \):
\[ \sin 72^\circ - \sin 18^\circ = 2 \cdot \frac{1}{\sqrt{2}} \sin 27^\circ = \sqrt{2} \sin 27^\circ \]

### Step 5: Set the equation equal to \( \sqrt{n} \sin 27^\circ \)
Now we can set this equal to the right side of the original equation:
\[ \sqrt{2} \sin 27^\circ = \sqrt{n} \sin 27^\circ \]

### Step 6: Cancel \( \sin 27^\circ \) (assuming \( \sin 27^\circ \neq 0 \))
We can divide both sides by \( \sin 27^\circ \):
\[ \sqrt{2} = \sqrt{n} \]

### Step 7: Square both sides to solve for \( n \)
Squaring both sides gives:
\[ 2 = n \]

### Final Answer
Thus, the value of \( n \) is:
\[ \boxed{2} \]