The following in the form A + i B `( cos 2 theta + i sin 2 theta )^(-5) ( cos 3 theta - i sin 3 theta)^(6) ( sin theta - i cos theta)^(3)` is

Options

A. `( cos 25 theta + isin 25 theta)`
B. `i(cos 25 theta +isin 25 theta)`
C. `i(cos 25 theta-isin25 theta)`
D. `(cos 25 theta-isin 25 theta)`

Answer

To solve the expression \((\cos 2\theta + i \sin 2\theta)^{-5} (\cos 3\theta - i \sin 3\theta)^{6} (\sin \theta - i \cos \theta)^{3}\) in the form \(A + iB\), we can follow these steps:

### Step 1: Rewrite the Terms Using Euler's Formula
Using Euler's formula, we can express the terms in the expression:
- \(\cos x + i \sin x = e^{ix}\)
- \(\cos x - i \sin x = e^{-ix}\)

Thus, we can rewrite:
- \((\cos 2\theta + i \sin 2\theta)^{-5} = (e^{i2\theta})^{-5} = e^{-i10\theta}\)
- \((\cos 3\theta - i \sin 3\theta)^{6} = (e^{-i3\theta})^{6} = e^{-i18\theta}\)
- \((\sin \theta - i \cos \theta)^{3} = (i(\cos \theta + i \sin \theta))^{3} = i^{3}(\cos \theta + i \sin \theta)^{3} = -i(\cos 3\theta + i \sin 3\theta) = -i e^{i3\theta}\)

### Step 2: Combine the Terms
Now we can combine the rewritten terms:
\[ e^{-i10\theta} \cdot e^{-i18\theta} \cdot (-i e^{i3\theta}) = -i e^{-i(10\theta + 18\theta - 3\theta)} = -i e^{-i25\theta} \]

### Step 3: Simplify the Expression
Now we simplify:
\[ -i e^{-i25\theta} = -i(\cos 25\theta + i \sin 25\theta) = -i \cos 25\theta - \sin 25\theta \]
This can be expressed as:
\[ -\sin 25\theta - i \cos 25\theta \]

### Step 4: Write in the Form \(A + iB\)
Thus, we can write the expression in the form \(A + iB\):
\[ A = -\sin 25\theta, \quad B = -\cos 25\theta \]

### Final Answer
The expression in the form \(A + iB\) is:
\[ -\sin 25\theta - i \cos 25\theta \]