`(2cos40^(@)-cos20^(@))/(sin20^(@))=?`

Options

A. 0
B. `sqrt(3)`
C. `-1`
D. 2

Answer

To solve the expression \((2\cos 40^\circ - \cos 20^\circ) / \sin 20^\circ\), we can follow these steps:

### Step 1: Rewrite the expression
We start with the expression:
\[ \frac{2\cos 40^\circ - \cos 20^\circ}{\sin 20^\circ} \]

### Step 2: Use the identity for \(2\cos A\)
We can rewrite \(2\cos 40^\circ\) as \(\cos 40^\circ + \cos 40^\circ\):
\[ \frac{\cos 40^\circ + \cos 40^\circ - \cos 20^\circ}{\sin 20^\circ} \]

### Step 3: Group the cosines
Now, we can group the cosines:
\[ \frac{(\cos 40^\circ - \cos 20^\circ) + \cos 40^\circ}{\sin 20^\circ} \]

### Step 4: Apply the cosine subtraction formula
Using the formula for \(\cos A - \cos B\):
\[ \cos A - \cos B = -2 \sin\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right) \]
where \(A = 40^\circ\) and \(B = 20^\circ\):
\[ \cos 40^\circ - \cos 20^\circ = -2 \sin\left(\frac{40^\circ + 20^\circ}{2}\right) \sin\left(\frac{40^\circ - 20^\circ}{2}\right) = -2 \sin(30^\circ) \sin(10^\circ) \]
Since \(\sin(30^\circ) = \frac{1}{2}\), we have:
\[ \cos 40^\circ - \cos 20^\circ = -2 \cdot \frac{1}{2} \sin(10^\circ) = -\sin(10^\circ) \]

### Step 5: Substitute back into the expression
Now substituting this back into our expression:
\[ \frac{-\sin(10^\circ) + \cos 40^\circ}{\sin 20^\circ} \]

### Step 6: Substitute \(\cos 40^\circ\)
We can also express \(\cos 40^\circ\) as \(\sin(50^\circ)\) (since \(\cos(90^\circ - x) = \sin x\)):
\[ \frac{-\sin(10^\circ) + \sin(50^\circ)}{\sin 20^\circ} \]

### Step 7: Apply the sine subtraction formula
Now we can apply the sine subtraction formula:
\[ \sin A - \sin B = 2 \cos\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right) \]
where \(A = 50^\circ\) and \(B = 10^\circ\):
\[ \sin(50^\circ) - \sin(10^\circ) = 2 \cos(30^\circ) \sin(20^\circ) \]
Since \(\cos(30^\circ) = \frac{\sqrt{3}}{2}\), we have:
\[ \sin(50^\circ) - \sin(10^\circ) = 2 \cdot \frac{\sqrt{3}}{2} \sin(20^\circ) = \sqrt{3} \sin(20^\circ) \]

### Step 8: Substitute back into the expression
Substituting this back into our expression gives:
\[ \frac{\sqrt{3} \sin(20^\circ)}{\sin 20^\circ} \]

### Step 9: Simplify
Now we can simplify:
\[ \sqrt{3} \]

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