The value of `(sin 1^(@) +sin 3^(@) + sin 5^(@) +sin 7^(@))/(cos 1^(@)*cos 2^(@)sin 4^(@))` is __________

Explanation

To solve the expression \((\sin 1^\circ + \sin 3^\circ + \sin 5^\circ + \sin 7^\circ) / (\cos 1^\circ \cdot \cos 2^\circ \cdot \sin 4^\circ)\), we will use trigonometric identities.

### Step 1: Group the Sine Terms
We can group the sine terms as follows:
\[ \sin 1^\circ + \sin 7^\circ + \sin 3^\circ + \sin 5^\circ \]
This can be rearranged to:
\[ (\sin 1^\circ + \sin 7^\circ) + (\sin 3^\circ + \sin 5^\circ) \]

### Step 2: Apply the Sine Addition Formula
Using the sine addition formula:
\[ \sin A + \sin B = 2 \sin\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right) \]
we can calculate each group.

For \(\sin 1^\circ + \sin 7^\circ\):
- \(A = 1^\circ\), \(B = 7^\circ\)
\[ \sin 1^\circ + \sin 7^\circ = 2 \sin\left(\frac{1^\circ + 7^\circ}{2}\right) \cos\left(\frac{1^\circ - 7^\circ}{2}\right) = 2 \sin(4^\circ) \cos(3^\circ) \]

For \(\sin 3^\circ + \sin 5^\circ\):
- \(A = 3^\circ\), \(B = 5^\circ\)
\[ \sin 3^\circ + \sin 5^\circ = 2 \sin\left(\frac{3^\circ + 5^\circ}{2}\right) \cos\left(\frac{3^\circ - 5^\circ}{2}\right) = 2 \sin(4^\circ) \cos(1^\circ) \]

### Step 3: Combine the Results
Now, we can combine the results:
\[ \sin 1^\circ + \sin 3^\circ + \sin 5^\circ + \sin 7^\circ = 2 \sin(4^\circ) \cos(3^\circ) + 2 \sin(4^\circ) \cos(1^\circ) \]
Factoring out \(2 \sin(4^\circ)\):
\[ = 2 \sin(4^\circ) (\cos(3^\circ) + \cos(1^\circ)) \]

### Step 4: Use the Cosine Addition Formula
Now we will use the cosine addition formula:
\[ \cos A + \cos B = 2 \cos\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right) \]
Applying this to \(\cos(3^\circ) + \cos(1^\circ)\):
\[ \cos(3^\circ) + \cos(1^\circ) = 2 \cos\left(\frac{3^\circ + 1^\circ}{2}\right) \cos\left(\frac{3^\circ - 1^\circ}{2}\right) = 2 \cos(2^\circ) \cos(1^\circ) \]

### Step 5: Substitute Back
Substituting back into our expression:
\[ \sin 1^\circ + \sin 3^\circ + \sin 5^\circ + \sin 7^\circ = 2 \sin(4^\circ) \cdot 2 \cos(2^\circ) \cos(1^\circ) = 4 \sin(4^\circ) \cos(2^\circ) \cos(1^\circ) \]

### Step 6: Substitute into the Original Expression
Now substituting this back into our original expression:
\[ \frac{4 \sin(4^\circ) \cos(2^\circ) \cos(1^\circ)}{\cos(1^\circ) \cos(2^\circ) \sin(4^\circ)} \]
The \(\sin(4^\circ)\), \(\cos(1^\circ)\), and \(\cos(2^\circ)\) cancel out:
\[ = 4 \]

### Final Answer
Thus, the value of the expression is:
\[ \boxed{4} \]