What is the real part of `( sin x + icos x )^(3)`, where `i= sqrt( -1)` ?

Options

A. `-cos 3x`
B. `- sin3x`
C. `sin 3x`
D. `cos 3x`

Answer

To find the real part of the expression \((\sin x + i \cos x)^3\), we can follow these steps:

### Step 1: Use the Binomial Expansion Formula
We can use the binomial expansion for \((a + b)^3\), which is given by:
\[ a^3 + 3a^2b + 3ab^2 + b^3 \]
Here, let \(a = \sin x\) and \(b = i \cos x\).

### Step 2: Calculate Each Term
Now, we calculate each term in the expansion:
1. \(a^3 = (\sin x)^3 = \sin^3 x\)
2. \(b^3 = (i \cos x)^3 = i^3 (\cos x)^3 = -i \cos^3 x\) (since \(i^3 = -i\))
3. \(3a^2b = 3(\sin x)^2(i \cos x) = 3i \sin^2 x \cos x\)
4. \(3ab^2 = 3(\sin x)(i \cos x)^2 = 3 \sin x (i^2 \cos^2 x) = 3 \sin x (-1) \cos^2 x = -3 \sin x \cos^2 x\)

### Step 3: Combine the Terms
Now, we combine all the terms:
\[ (\sin x + i \cos x)^3 = \sin^3 x - i \cos^3 x + 3i \sin^2 x \cos x - 3 \sin x \cos^2 x \]

### Step 4: Separate Real and Imaginary Parts
Now, we can separate the real and imaginary parts:
- Real part: \(\sin^3 x - 3 \sin x \cos^2 x\)
- Imaginary part: \(-\cos^3 x + 3 \sin^2 x \cos x\)

### Step 5: Simplify the Real Part
The real part can be simplified further:
\[ \text{Real part} = \sin^3 x - 3 \sin x \cos^2 x \]
Using the identity \(3 \sin x \cos^2 x = \frac{3}{4} \sin(3x)\), we can express the real part as:
\[ \text{Real part} = \sin^3 x - 3 \sin x \cos^2 x = \frac{1}{4} (3 \sin x - \sin(3x)) \]

### Final Answer
Thus, the real part of \((\sin x + i \cos x)^3\) is:
\[ \sin^3 x - 3 \sin x \cos^2 x \]