What is the value of
`({[[4cos (90-A)sin^3 (90+A)]-], [[4sin(90+A)cos^3(90-A)]]})/(cos((180+8A)/(2)))`?

Options

A. 1
B. -1
C. 0
D. 2

Answer

To solve the expression

\[ \frac{4\cos(90^\circ - A) \sin^3(90^\circ + A) - 4\sin(90^\circ + A) \cos^3(90^\circ - A)}{\cos\left(\frac{180^\circ + 8A}{2}\right)}, \]

we will simplify the numerator and denominator step by step.

### Step 1: Simplify the trigonometric functions
Using the trigonometric identities:
- \(\cos(90^\circ - A) = \sin A\)
- \(\sin(90^\circ + A) = \cos A\)

We can rewrite the terms in the numerator:

\[ 4\cos(90^\circ - A) \sin^3(90^\circ + A) = 4\sin A \cos^3 A \]

And for the second term:

\[ 4\sin(90^\circ + A) \cos^3(90^\circ - A) = 4\cos A \sin^3 A \]

Thus, the numerator becomes:

\[ 4\sin A \cos^3 A - 4\cos A \sin^3 A \]

### Step 2: Factor the numerator
We can factor out \(4\sin A \cos A\):

\[ 4\sin A \cos A (\cos^2 A - \sin^2 A) \]

### Step 3: Simplify the denominator
Now, simplify the denominator:

\[ \cos\left(\frac{180^\circ + 8A}{2}\right) = \cos(90^\circ + 4A) = -\sin(4A) \]

### Step 4: Combine the results
Now we can rewrite the entire expression:

\[ \frac{4\sin A \cos A (\cos^2 A - \sin^2 A)}{-\sin(4A)} \]

### Step 5: Use the identity for \(\sin(4A)\)
Recall that:

\[ \sin(4A) = 2\sin(2A)\cos(2A) = 2(2\sin A \cos A)(\cos^2 A - \sin^2 A) \]

So we can replace \(\sin(4A)\) in the denominator:

\[ \frac{4\sin A \cos A (\cos^2 A - \sin^2 A)}{-2(2\sin A \cos A)(\cos^2 A - \sin^2 A)} \]

### Step 6: Cancel common terms
The \(\sin A \cos A (\cos^2 A - \sin^2 A)\) terms cancel out:

\[ \frac{4}{-2} = -2 \]

### Final Answer
Thus, the value of the expression is:

\[ \boxed{-2} \]