`((sin 33^(@))/( sin 11^(@) sin 49^(@) sin 71^(@)))^(2) + (( cos 33^(@))/(cos 11^(@) cos 49^(@) cos 71^(@)))^(2) ` is equal to ______.

Explanation

To solve the expression

\[ \left(\frac{\sin 33^\circ}{\sin 11^\circ \sin 49^\circ \sin 71^\circ}\right)^2 + \left(\frac{\cos 33^\circ}{\cos 11^\circ \cos 49^\circ \cos 71^\circ}\right)^2, \]

we will follow these steps:

### Step 1: Rewrite the Sine Terms
We can use the identity that relates sine functions. Notice that:

\[ \sin 49^\circ = \sin(60^\circ - 11^\circ) \quad \text{and} \quad \sin 71^\circ = \sin(60^\circ + 11^\circ). \]

Thus, we can rewrite the sine terms as:

\[ \sin 11^\circ \sin 49^\circ \sin 71^\circ = \sin 11^\circ \sin(60^\circ - 11^\circ) \sin(60^\circ + 11^\circ). \]

### Step 2: Apply the Product-to-Sum Identity
Using the identity

\[ \sin A \sin B = \frac{1}{2} [\cos(A - B) - \cos(A + B)], \]

we can simplify:

\[ \sin(60^\circ - 11^\circ) \sin(60^\circ + 11^\circ) = \frac{1}{2} [\cos(11^\circ) - \cos(71^\circ)]. \]

### Step 3: Substitute Back into the Expression
Now we substitute this back into our expression:

\[ \sin 11^\circ \cdot \frac{1}{2} [\cos(11^\circ) - \cos(71^\circ)]. \]

### Step 4: Use the Identity for Sine
We know that:

\[ \sin 33^\circ = \sin(3 \times 11^\circ) = 3 \sin 11^\circ - 4 \sin^3 11^\circ. \]

### Step 5: Simplify the Cosine Terms
Similarly, we can express the cosine terms using:

\[ \cos 11^\circ \cos 49^\circ \cos 71^\circ = \cos 11^\circ \cdot \frac{1}{4} \cos(3 \times 11^\circ). \]

### Step 6: Substitute and Simplify
Now substituting these into our original expression, we get:

\[ \left(\frac{3 \sin 11^\circ - 4 \sin^3 11^\circ}{\frac{1}{4} \sin(3 \times 11^\circ)}\right)^2 + \left(\frac{\cos(3 \times 11^\circ)}{\frac{1}{4} \cos(3 \times 11^\circ)}\right)^2. \]

### Step 7: Final Calculation
After simplifying, we find that both terms will yield a common factor, leading us to:

\[ \left(4\right)^2 + \left(4\right)^2 = 16 + 16 = 32. \]

Thus, the final answer is

\[ \boxed{32}. \]