The value of sin `10^(@)` sin `30^(@)` sin `50^(@)` sin `70^(@)` is

Options

A. `(1)/(36)`
B. `(1)/(32)`
C. `(1)/(16)`
D. `(1)/(18)`

Explanation

To find the value of \( \sin 10^\circ \sin 30^\circ \sin 50^\circ \sin 70^\circ \), we can follow these steps:

### Step 1: Write down the expression
We start with the expression:
\[ \sin 10^\circ \sin 30^\circ \sin 50^\circ \sin 70^\circ \]

### Step 2: Use the identity for sine
Recall that \( \sin(90^\circ - x) = \cos x \). We can rewrite \( \sin 50^\circ \) and \( \sin 70^\circ \):
\[ \sin 50^\circ = \cos 40^\circ \quad \text{and} \quad \sin 70^\circ = \cos 20^\circ \]
Thus, we can rewrite the expression as:
\[ \sin 10^\circ \sin 30^\circ \cos 40^\circ \cos 20^\circ \]

### Step 3: Substitute known values
We know that \( \sin 30^\circ = \frac{1}{2} \). Therefore, we can substitute this into our expression:
\[ \sin 10^\circ \cdot \frac{1}{2} \cdot \cos 40^\circ \cdot \cos 20^\circ \]
This simplifies to:
\[ \frac{1}{2} \sin 10^\circ \cos 40^\circ \cos 20^\circ \]

### Step 4: Multiply and divide by \( \cos 10^\circ \)
To simplify further, we can multiply and divide by \( \cos 10^\circ \):
\[ \frac{1}{2} \cdot \frac{\sin 10^\circ \cos 10^\circ}{\cos 10^\circ} \cdot \cos 40^\circ \cdot \cos 20^\circ \]
Using the identity \( 2 \sin x \cos x = \sin 2x \), we have:
\[ \frac{1}{4} \sin 20^\circ \cdot \cos 40^\circ \cdot \frac{1}{\cos 10^\circ} \]

### Step 5: Use the identity again
Now, we can apply the identity again:
\[ \sin 20^\circ \cos 40^\circ = \frac{1}{2} \sin 40^\circ \]
Thus, we can rewrite our expression as:
\[ \frac{1}{8} \cdot \frac{\sin 40^\circ}{\cos 10^\circ} \]

### Step 6: Simplify further
We can again use the identity:
\[ \sin 40^\circ = 2 \sin 20^\circ \cos 20^\circ \]
Substituting this back gives:
\[ \frac{1}{16} \cdot \frac{\sin 80^\circ}{\cos 10^\circ} \]
Since \( \sin 80^\circ = \cos 10^\circ \), we have:
\[ \frac{1}{16} \]

### Final Answer
Thus, the value of \( \sin 10^\circ \sin 30^\circ \sin 50^\circ \sin 70^\circ \) is:
\[ \boxed{\frac{1}{16}} \]