The value of `cos12^(@)+cos84^(@)+cos156^(@)+cos132^(@)` is

Options

A. `(1)/(2)`
B. 1
C. `-(1)/(2)`
D. `(1)/(8)`

Answer

To find the value of \( \cos 12^\circ + \cos 84^\circ + \cos 156^\circ + \cos 132^\circ \), we can use the cosine addition formulas. Let's break it down step by step.

### Step 1: Pair the cosines
We can pair the cosines to make the calculations easier:
\[ \cos 12^\circ + \cos 132^\circ \quad \text{and} \quad \cos 84^\circ + \cos 156^\circ \]

### Step 2: Use the cosine addition formula
The cosine addition formula states that:
\[ \cos A + \cos B = 2 \cos\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right) \]

#### For \( \cos 12^\circ + \cos 132^\circ \):
- Here, \( A = 12^\circ \) and \( B = 132^\circ \).
- Calculate \( A + B = 12^\circ + 132^\circ = 144^\circ \)
- Calculate \( A - B = 12^\circ - 132^\circ = -120^\circ \)

Now applying the formula:
\[ \cos 12^\circ + \cos 132^\circ = 2 \cos\left(\frac{144^\circ}{2}\right) \cos\left(\frac{-120^\circ}{2}\right) \]
\[ = 2 \cos(72^\circ) \cos(-60^\circ) \]
Since \( \cos(-\theta) = \cos(\theta) \):
\[ = 2 \cos(72^\circ) \cos(60^\circ) \]
We know that \( \cos(60^\circ) = \frac{1}{2} \):
\[ = 2 \cos(72^\circ) \cdot \frac{1}{2} = \cos(72^\circ) \]

#### For \( \cos 84^\circ + \cos 156^\circ \):
- Here, \( A = 84^\circ \) and \( B = 156^\circ \).
- Calculate \( A + B = 84^\circ + 156^\circ = 240^\circ \)
- Calculate \( A - B = 84^\circ - 156^\circ = -72^\circ \)

Now applying the formula:
\[ \cos 84^\circ + \cos 156^\circ = 2 \cos\left(\frac{240^\circ}{2}\right) \cos\left(\frac{-72^\circ}{2}\right) \]
\[ = 2 \cos(120^\circ) \cos(-36^\circ) \]
Since \( \cos(120^\circ) = -\frac{1}{2} \):
\[ = 2 \left(-\frac{1}{2}\right) \cos(36^\circ) = -\cos(36^\circ) \]

### Step 3: Combine the results
Now we combine the results from the two pairs:
\[ \cos 12^\circ + \cos 132^\circ + \cos 84^\circ + \cos 156^\circ = \cos(72^\circ) - \cos(36^\circ) \]

### Step 4: Use known values
We know that:
\[ \cos(72^\circ) = \sin(18^\circ) \quad \text{and} \quad \cos(36^\circ) = \sin(54^\circ) \]
Thus, we can express:
\[ \cos(72^\circ) - \cos(36^\circ) = \sin(18^\circ) - \sin(54^\circ) \]

### Final Result
Using the sine subtraction formula:
\[ \sin(18^\circ) - \sin(54^\circ) = -\sin(36^\circ) \]
Thus, the final value is:
\[ \cos 12^\circ + \cos 84^\circ + \cos 156^\circ + \cos 132^\circ = 0 \]