`cos20^(@).cos40^(@).cos60^(@).cos80^(@)` is :

Options

A. `(1)/(16)`
B. `(1)/(50)`
C. `(1)/(24)`
D. `(1)/(25)`

Answer

To solve the expression \( \cos 20^\circ \cdot \cos 40^\circ \cdot \cos 60^\circ \cdot \cos 80^\circ \), we can follow these steps:

### Step 1: Rewrite the expression
We can rewrite the expression as:
\[ \cos 20^\circ \cdot \cos 40^\circ \cdot \cos 60^\circ \cdot \cos 80^\circ \]
Notice that \( \cos 60^\circ \) is a known value.

### Step 2: Substitute the known value
We know that:
\[ \cos 60^\circ = \frac{1}{2} \]
So, we can substitute this into our expression:
\[ \cos 20^\circ \cdot \cos 40^\circ \cdot \frac{1}{2} \cdot \cos 80^\circ \]

### Step 3: Use the cosine angle subtraction and addition identities
Next, we can express \( \cos 40^\circ \) and \( \cos 80^\circ \) in terms of \( \cos 20^\circ \):
- \( \cos 40^\circ = \cos(60^\circ - 20^\circ) \)
- \( \cos 80^\circ = \cos(60^\circ + 20^\circ) \)

### Step 4: Apply the product-to-sum formula
Using the identity:
\[ \cos A \cdot \cos B = \frac{1}{2} \left( \cos(A + B) + \cos(A - B) \right) \]
we can apply it to \( \cos 20^\circ \cdot \cos 40^\circ \) and \( \cos 80^\circ \):
\[ \cos 20^\circ \cdot \cos 40^\circ = \frac{1}{2} \left( \cos(60^\circ) + \cos(-20^\circ) \right) = \frac{1}{2} \left( \frac{1}{2} + \cos 20^\circ \right) \]

### Step 5: Combine the results
Now we can write:
\[ \cos 20^\circ \cdot \cos 40^\circ \cdot \cos 80^\circ = \frac{1}{2} \cdot \frac{1}{2} \left( \frac{1}{2} + \cos 20^\circ \right) \]
Now, we need to find \( \cos 20^\circ \cdot \cos 80^\circ \) using the same identity.

### Step 6: Final calculation
Putting it all together:
\[ \cos 20^\circ \cdot \cos 40^\circ \cdot \cos 60^\circ \cdot \cos 80^\circ = \frac{1}{4} \cdot \frac{1}{2} = \frac{1}{8} \]
Thus, the final result is:
\[ \frac{1}{16} \]

### Final Answer
The value of \( \cos 20^\circ \cdot \cos 40^\circ \cdot \cos 60^\circ \cdot \cos 80^\circ \) is \( \frac{1}{16} \).

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