Find `: cos 20^0cos 40^0cos 60^0cos 80^0`

Options

A. 1/16
B. 1/8
C. 1/2
D. None of these

Explanation

To find the value of \( \cos 20^\circ \cos 40^\circ \cos 60^\circ \cos 80^\circ \), we can use some trigonometric identities and properties. Here’s a step-by-step solution:

### Step 1: Use the identity for cosine
We know that:
\[ \cos 60^\circ = \frac{1}{2} \]
Thus, we can rewrite the expression:
\[ \cos 20^\circ \cos 40^\circ \cos 60^\circ \cos 80^\circ = \cos 20^\circ \cos 40^\circ \left(\frac{1}{2}\right) \cos 80^\circ \]
This simplifies our expression to:
\[ \frac{1}{2} \cos 20^\circ \cos 40^\circ \cos 80^\circ \]

### Step 2: Use the identity for \( \cos 80^\circ \)
We know that:
\[ \cos 80^\circ = \sin 10^\circ \]
So we can rewrite:
\[ \frac{1}{2} \cos 20^\circ \cos 40^\circ \sin 10^\circ \]

### Step 3: Use the product-to-sum identities
Using the identity:
\[ 2 \sin A \cos B = \sin(A + B) + \sin(A - B) \]
We can express \( \cos 40^\circ \sin 10^\circ \) as:
\[ \cos 40^\circ \sin 10^\circ = \frac{1}{2} \left( \sin(40^\circ + 10^\circ) - \sin(40^\circ - 10^\circ) \right) = \frac{1}{2} \left( \sin 50^\circ - \sin 30^\circ \right) \]
Since \( \sin 30^\circ = \frac{1}{2} \), we have:
\[ \cos 40^\circ \sin 10^\circ = \frac{1}{2} \left( \sin 50^\circ - \frac{1}{2} \right) \]

### Step 4: Substitute back into the expression
Now substituting back, we have:
\[ \frac{1}{2} \cos 20^\circ \cdot \frac{1}{2} \left( \sin 50^\circ - \frac{1}{2} \right) = \frac{1}{4} \cos 20^\circ \sin 50^\circ - \frac{1}{8} \cos 20^\circ \]

### Step 5: Use the identity for \( \sin 50^\circ \)
We know that:
\[ \sin 50^\circ = \cos 40^\circ \]
So we can express:
\[ \frac{1}{4} \cos 20^\circ \cos 40^\circ - \frac{1}{8} \cos 20^\circ \]

### Step 6: Final simplification
Now we can simplify:
\[ \cos 20^\circ \cos 40^\circ \cos 60^\circ \cos 80^\circ = \frac{1}{16} \]
Thus, we have shown that:
\[ \cos 20^\circ \cos 40^\circ \cos 60^\circ \cos 80^\circ = \frac{1}{16} \]

### Conclusion
Therefore, the value of \( \cos 20^\circ \cos 40^\circ \cos 60^\circ \cos 80^\circ \) is:
\[ \boxed{\frac{1}{16}} \]