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

Options

A. `2sqrt3`
B. `2sqrt3`
C. `1sqrt3`
D. `sqrt3`

Answer

To solve the expression \((\cos 40^\circ - \cos 140^\circ) / (\sin 80^\circ + \sin 20^\circ)\), we can follow these steps:

### Step 1: Use the Cosine Difference Identity
We know that:
\[ \cos A - \cos B = -2 \sin\left(\frac{A + B}{2}\right) \sin\left(\frac{A - B}{2}\right) \]
In our case, let \(A = 40^\circ\) and \(B = 140^\circ\):
\[ \cos 40^\circ - \cos 140^\circ = -2 \sin\left(\frac{40^\circ + 140^\circ}{2}\right) \sin\left(\frac{40^\circ - 140^\circ}{2}\right) \]
Calculating the averages:
\[ \frac{40^\circ + 140^\circ}{2} = \frac{180^\circ}{2} = 90^\circ \]
\[ \frac{40^\circ - 140^\circ}{2} = \frac{-100^\circ}{2} = -50^\circ \]
Thus, we have:
\[ \cos 40^\circ - \cos 140^\circ = -2 \sin(90^\circ) \sin(-50^\circ) \]
Since \(\sin(90^\circ) = 1\) and \(\sin(-50^\circ) = -\sin(50^\circ)\):
\[ \cos 40^\circ - \cos 140^\circ = -2(1)(-\sin(50^\circ)) = 2\sin(50^\circ) \]

### Step 2: Simplify the Denominator
Next, we simplify the denominator \(\sin 80^\circ + \sin 20^\circ\). We can use the sine sum identity:
\[ \sin A + \sin B = 2 \sin\left(\frac{A + B}{2}\right) \cos\left(\frac{A - B}{2}\right) \]
Let \(A = 80^\circ\) and \(B = 20^\circ\):
\[ \sin 80^\circ + \sin 20^\circ = 2 \sin\left(\frac{80^\circ + 20^\circ}{2}\right) \cos\left(\frac{80^\circ - 20^\circ}{2}\right) \]
Calculating the averages:
\[ \frac{80^\circ + 20^\circ}{2} = \frac{100^\circ}{2} = 50^\circ \]
\[ \frac{80^\circ - 20^\circ}{2} = \frac{60^\circ}{2} = 30^\circ \]
Thus, we have:
\[ \sin 80^\circ + \sin 20^\circ = 2 \sin(50^\circ) \cos(30^\circ) \]
Since \(\cos(30^\circ) = \frac{\sqrt{3}}{2}\):
\[ \sin 80^\circ + \sin 20^\circ = 2 \sin(50^\circ) \cdot \frac{\sqrt{3}}{2} = \sqrt{3} \sin(50^\circ) \]

### Step 3: Substitute Back into the Original Expression
Now substituting back into the original expression:
\[ \frac{\cos 40^\circ - \cos 140^\circ}{\sin 80^\circ + \sin 20^\circ} = \frac{2\sin(50^\circ)}{\sqrt{3} \sin(50^\circ)} \]
The \(\sin(50^\circ)\) terms cancel out:
\[ = \frac{2}{\sqrt{3}} \]

### Final Answer
Thus, the value of the expression \((\cos 40^\circ - \cos 140^\circ) / (\sin 80^\circ + \sin 20^\circ)\) is:
\[ \frac{2}{\sqrt{3}} \]