For what and only what values of `alpha` lying between `0 and pi/2` is the inequality is `sin alphacos^(3)alpha ltsin^(3) alpha cos alpha` valid?

Options

A. `alpha in(0,pi//4)`
B. `alpha in(0,pi//2)`
C. `alpha in(pi//4,pi//2)`
D. none of these

Answer

To solve the inequality \( \sin \alpha \cos^3 \alpha < \sin^3 \alpha \cos \alpha \) for values of \( \alpha \) in the interval \( (0, \frac{\pi}{2}) \), we will follow these steps:

### Step 1: Rearranging the Inequality
Start by rearranging the given inequality:
\[ \sin \alpha \cos^3 \alpha - \sin^3 \alpha \cos \alpha < 0 \]

### Step 2: Factoring Out Common Terms
Factor out \( \sin \alpha \cos \alpha \):
\[ \sin \alpha \cos \alpha \left( \cos^2 \alpha - \sin^2 \alpha \right) < 0 \]

### Step 3: Identifying the Factors
Now we have two factors:
1. \( \sin \alpha \cos \alpha \)
2. \( \cos^2 \alpha - \sin^2 \alpha \)

### Step 4: Analyzing the First Factor
The first factor, \( \sin \alpha \cos \alpha \), is positive in the interval \( (0, \frac{\pi}{2}) \) because both sine and cosine are positive in this interval.

### Step 5: Analyzing the Second Factor
Next, we analyze the second factor, \( \cos^2 \alpha - \sin^2 \alpha \):
\[ \cos^2 \alpha - \sin^2 \alpha < 0 \]
This can be rewritten as:
\[ \cos^2 \alpha < \sin^2 \alpha \]
or
\[ \tan^2 \alpha > 1 \]
which implies:
\[ \tan \alpha > 1 \]

### Step 6: Finding the Critical Point
The inequality \( \tan \alpha > 1 \) holds when:
\[ \alpha > \frac{\pi}{4} \]

### Step 7: Combining the Results
Since \( \sin \alpha \cos \alpha > 0 \) for \( \alpha \in (0, \frac{\pi}{2}) \) and \( \cos^2 \alpha - \sin^2 \alpha < 0 \) for \( \alpha > \frac{\pi}{4} \), we conclude that the inequality holds for:
\[ \frac{\pi}{4} < \alpha < \frac{\pi}{2} \]

### Final Answer
Thus, the values of \( \alpha \) for which the inequality is valid are:
\[ \alpha \in \left( \frac{\pi}{4}, \frac{\pi}{2} \right) \]