`cos40^(@)+cos 80^(@)+cos160^(@)+cos240^(@)`=

Options

A. 0
B. 1
C. `(1)/(2)`
D. `-(1)/(2)`

Answer

To solve the expression \( \cos 40^\circ + \cos 80^\circ + \cos 160^\circ + \cos 240^\circ \), we can use the properties of cosine and some trigonometric identities. Let's break it down step by step.

### Step 1: Group the terms
We can group the terms in pairs:
\[ \cos 40^\circ + \cos 240^\circ + \cos 80^\circ + \cos 160^\circ \]

### Step 2: Use the cosine addition formula
We can use the formula for the sum of cosines:
\[ \cos A + \cos B = 2 \cos\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right) \]

#### Pair 1: \( \cos 40^\circ + \cos 240^\circ \)
Let \( A = 40^\circ \) and \( B = 240^\circ \):
\[ \cos 40^\circ + \cos 240^\circ = 2 \cos\left(\frac{40^\circ + 240^\circ}{2}\right) \cos\left(\frac{40^\circ - 240^\circ}{2}\right) \]
Calculating the averages:
\[ \frac{40^\circ + 240^\circ}{2} = \frac{280^\circ}{2} = 140^\circ \]
\[ \frac{40^\circ - 240^\circ}{2} = \frac{-200^\circ}{2} = -100^\circ \]
Thus,
\[ \cos 40^\circ + \cos 240^\circ = 2 \cos(140^\circ) \cos(-100^\circ) \]
Using the property \( \cos(-x) = \cos(x) \):
\[ \cos(-100^\circ) = \cos(100^\circ) \]
So,
\[ \cos 40^\circ + \cos 240^\circ = 2 \cos(140^\circ) \cos(100^\circ) \]

#### Pair 2: \( \cos 80^\circ + \cos 160^\circ \)
Let \( A = 80^\circ \) and \( B = 160^\circ \):
\[ \cos 80^\circ + \cos 160^\circ = 2 \cos\left(\frac{80^\circ + 160^\circ}{2}\right) \cos\left(\frac{80^\circ - 160^\circ}{2}\right) \]
Calculating the averages:
\[ \frac{80^\circ + 160^\circ}{2} = \frac{240^\circ}{2} = 120^\circ \]
\[ \frac{80^\circ - 160^\circ}{2} = \frac{-80^\circ}{2} = -40^\circ \]
Thus,
\[ \cos 80^\circ + \cos 160^\circ = 2 \cos(120^\circ) \cos(-40^\circ) \]
Using \( \cos(-x) = \cos(x) \):
\[ \cos(-40^\circ) = \cos(40^\circ) \]
So,
\[ \cos 80^\circ + \cos 160^\circ = 2 \cos(120^\circ) \cos(40^\circ) \]

### Step 3: Combine both results
Now we have:
\[ \cos 40^\circ + \cos 240^\circ + \cos 80^\circ + \cos 160^\circ = 2 \cos(140^\circ) \cos(100^\circ) + 2 \cos(120^\circ) \cos(40^\circ) \]

### Step 4: Calculate the values
Using known values:
- \( \cos(140^\circ) = -\cos(40^\circ) \)
- \( \cos(120^\circ) = -\frac{1}{2} \)

Substituting these values:
\[ = 2(-\cos(40^\circ)) \cos(100^\circ) + 2\left(-\frac{1}{2}\right) \cos(40^\circ) \]
\[ = -2 \cos(40^\circ) \cos(100^\circ) - \cos(40^\circ) \]

### Step 5: Factor out \( \cos(40^\circ) \)
\[ = \cos(40^\circ)(-2 \cos(100^\circ) - 1) \]

### Step 6: Simplify further
Using \( \cos(100^\circ) = -\sin(10^\circ) \):
\[ = \cos(40^\circ)(2\sin(10^\circ) - 1) \]

### Final Result
The expression simplifies to:
\[ \cos(40^\circ)(-1 - 2\sin(10^\circ)) \]

### Conclusion
Thus, the final value of \( \cos 40^\circ + \cos 80^\circ + \cos 160^\circ + \cos 240^\circ \) is \( -\frac{1}{2} \).