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 can follow these steps:

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

### Step 2: Factor the Left Side
Factor out the common terms:
\[ \sin \alpha \cos \alpha (\cos^2 \alpha - \sin^2 \alpha) < 0 \]

### Step 3: Use Trigonometric Identities
Recognize that \( \cos^2 \alpha - \sin^2 \alpha = \cos 2\alpha \):
\[ \sin \alpha \cos \alpha \cos 2\alpha < 0 \]

### Step 4: Simplify Further
Using the identity \( 2 \sin \alpha \cos \alpha = \sin 2\alpha \), we can rewrite the inequality:
\[ \frac{1}{2} \sin 2\alpha \cos 2\alpha < 0 \]

### Step 5: Analyze the Inequality
The product \( \sin 2\alpha \cos 2\alpha < 0 \) implies that one of the factors must be negative while the other is positive.

### Step 6: Determine the Intervals
1. **For \( \sin 2\alpha < 0 \)**: This occurs when \( 2\alpha \) is in the interval \( (\pi, 2\pi) \), which translates to \( \alpha \) in \( \left(\frac{\pi}{2}, \pi\right) \). However, this is outside our interval of interest \( (0, \frac{\pi}{2}) \).

2. **For \( \cos 2\alpha < 0 \)**: This occurs when \( 2\alpha \) is in the interval \( \left(\frac{\pi}{2}, \frac{3\pi}{2}\right) \), which translates to \( \alpha \) in \( \left(\frac{\pi}{4}, \frac{3\pi}{4}\right) \). But again, we are interested only in \( (0, \frac{\pi}{2}) \).

### Step 7: Combine the Results
Thus, we need to find the values of \( \alpha \) in the interval \( (0, \frac{\pi}{2}) \) such that:
\[ \frac{\pi}{4} < \alpha < \frac{\pi}{2} \]

### Final Result
The values of \( \alpha \) that satisfy the inequality in the interval \( (0, \frac{\pi}{2}) \) are:
\[ \alpha \in \left(\frac{\pi}{4}, \frac{\pi}{2}\right) \]