The expression `sin27^@cos57^@sin87^@` simplifies to

Options

A. `(sin 9^(@))/(4)`
B. `(cos9^(@))/(4)`
C. `(sin9^(@))/(2)`
D. `(cos9^(@))/(2)`

Answer

To simplify the expression \( P = \sin 27^\circ \cdot \cos 57^\circ \cdot \sin 87^\circ \), we can follow these steps:

### Step 1: Rewrite the trigonometric functions
We know that:
- \( \cos 57^\circ = \sin(90^\circ - 57^\circ) = \sin 33^\circ \)
- \( \sin 87^\circ = \sin(90^\circ - 3^\circ) = \cos 3^\circ \)

Thus, we can rewrite \( P \):
\[ P = \sin 27^\circ \cdot \sin 33^\circ \cdot \cos 3^\circ \]

### Step 2: Use the double angle identity
To simplify \( \sin 27^\circ \cdot \sin 33^\circ \), we can use the identity:
\[ 2 \sin A \sin B = \cos(A - B) - \cos(A + B) \]
Let \( A = 27^\circ \) and \( B = 33^\circ \):
\[ 2 \sin 27^\circ \sin 33^\circ = \cos(27^\circ - 33^\circ) - \cos(27^\circ + 33^\circ) \]
Calculating the angles:
\[ 27^\circ - 33^\circ = -6^\circ \quad \text{and} \quad 27^\circ + 33^\circ = 60^\circ \]
Thus:
\[ 2 \sin 27^\circ \sin 33^\circ = \cos(-6^\circ) - \cos(60^\circ) \]
Since \( \cos(-\theta) = \cos(\theta) \):
\[ 2 \sin 27^\circ \sin 33^\circ = \cos 6^\circ - \frac{1}{2} \]
Now, we can express \( P \):
\[ P = \frac{1}{2}(\cos 6^\circ - \frac{1}{2}) \cdot \cos 3^\circ \]

### Step 3: Distribute and simplify
\[ P = \frac{1}{2} \cos 6^\circ \cdot \cos 3^\circ - \frac{1}{4} \cos 3^\circ \]

### Step 4: Use the cosine product identity
We can again use the identity:
\[ 2 \cos A \cos B = \cos(A + B) + \cos(A - B) \]
Let \( A = 6^\circ \) and \( B = 3^\circ \):
\[ 2 \cos 6^\circ \cos 3^\circ = \cos(6^\circ + 3^\circ) + \cos(6^\circ - 3^\circ) = \cos 9^\circ + \cos 3^\circ \]
Thus:
\[ \cos 6^\circ \cos 3^\circ = \frac{1}{2}(\cos 9^\circ + \cos 3^\circ) \]
Substituting back into \( P \):
\[ P = \frac{1}{2} \cdot \frac{1}{2}(\cos 9^\circ + \cos 3^\circ) - \frac{1}{4} \cos 3^\circ \]
\[ P = \frac{1}{4} \cos 9^\circ + \frac{1}{4} \cos 3^\circ - \frac{1}{4} \cos 3^\circ \]
The \( \cos 3^\circ \) terms cancel out:
\[ P = \frac{1}{4} \cos 9^\circ \]

### Final Answer
Thus, the expression \( \sin 27^\circ \cdot \cos 57^\circ \cdot \sin 87^\circ \) simplifies to:
\[ \boxed{\frac{1}{4} \cos 9^\circ} \]