What is the value of `((cos40^(@)-cos140^(@)))/((sin80^(@)+sin20^(@)))` ?

Options

A. `2sqrt(3)`
B. `(2)/(sqrt(3))`
C. `(1)/(sqrt(3))`
D. `sqrt(3)`

Answer

To solve the expression \(\frac{\cos 40^\circ - \cos 140^\circ}{\sin 80^\circ + \sin 20^\circ}\), we will use trigonometric identities.

### Step 1: Apply the Cosine Difference Formula
We can use the formula for the difference of cosines:
\[ \cos A - \cos B = -2 \sin\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right) \]
In our case, let \(A = 40^\circ\) and \(B = 140^\circ\):
\[ \cos 40^\circ - \cos 140^\circ = -2 \sin\left(\frac{40^\circ + 140^\circ}{2}\right) \sin\left(\frac{40^\circ - 140^\circ}{2}\right) \]

### Step 2: Calculate the Averages
Calculating the averages:
\[ \frac{40^\circ + 140^\circ}{2} = \frac{180^\circ}{2} = 90^\circ \]
\[ \frac{40^\circ - 140^\circ}{2} = \frac{-100^\circ}{2} = -50^\circ \]

### Step 3: Substitute Back into the Expression
Now substituting back into the expression:
\[ \cos 40^\circ - \cos 140^\circ = -2 \sin(90^\circ) \sin(-50^\circ) \]
Since \(\sin(90^\circ) = 1\) and \(\sin(-50^\circ) = -\sin(50^\circ)\), we have:
\[ \cos 40^\circ - \cos 140^\circ = -2(1)(-\sin(50^\circ)) = 2\sin(50^\circ) \]

### Step 4: Simplify the Denominator
Next, we simplify the denominator \(\sin 80^\circ + \sin 20^\circ\) using the sine addition formula:
\[ \sin A + \sin B = 2 \sin\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right) \]
Let \(A = 80^\circ\) and \(B = 20^\circ\):
\[ \sin 80^\circ + \sin 20^\circ = 2 \sin\left(\frac{80^\circ + 20^\circ}{2}\right) \cos\left(\frac{80^\circ - 20^\circ}{2}\right) \]

### Step 5: Calculate the Averages for Sine
Calculating the averages:
\[ \frac{80^\circ + 20^\circ}{2} = \frac{100^\circ}{2} = 50^\circ \]
\[ \frac{80^\circ - 20^\circ}{2} = \frac{60^\circ}{2} = 30^\circ \]

### Step 6: Substitute Back into the Denominator
Now substituting back into the expression:
\[ \sin 80^\circ + \sin 20^\circ = 2 \sin(50^\circ) \cos(30^\circ) \]
Since \(\cos(30^\circ) = \frac{\sqrt{3}}{2}\), we have:
\[ \sin 80^\circ + \sin 20^\circ = 2 \sin(50^\circ) \cdot \frac{\sqrt{3}}{2} = \sqrt{3} \sin(50^\circ) \]

### Step 7: Combine the Results
Now we can substitute everything back into our original expression:
\[ \frac{\cos 40^\circ - \cos 140^\circ}{\sin 80^\circ + \sin 20^\circ} = \frac{2 \sin(50^\circ)}{\sqrt{3} \sin(50^\circ)} \]

### Step 8: Simplify the Expression
The \(\sin(50^\circ)\) terms cancel out:
\[ = \frac{2}{\sqrt{3}} \]

### Final Answer
Thus, the value of the expression is:
\[ \frac{2}{\sqrt{3}} \]