Prove that the function f(x)=x/{sin(x)} is an increasing function in the interval ]0,π/2[

To prove that the function \( f(x) = \frac{x}{\sin(x)} \) is increasing on the interval \( (0, \frac{\pi}{2}) \), we need to show that its derivative \( f'(x) \) is positive for all \( x \in (0, \frac{\pi}{2}) \).

### Step 1: Compute the derivative of \( f(x) \)

Using the quotient rule for derivatives, we differentiate \( f(x) = \frac{x}{\sin(x)} \):

\[
f'(x) = \frac{\sin(x) \cdot \frac{d}{dx}(x) - x \cdot \frac{d}{dx}(\sin(x))}{\sin^2(x)}
\]

The derivatives of \( x \) and \( \sin(x) \) are:

\[
\frac{d}{dx}(x) = 1, \quad \frac{d}{dx}(\sin(x)) = \cos(x)
\]

Substituting these into the derivative expression:

\[
f'(x) = \frac{\sin(x) \cdot 1 - x \cdot \cos(x)}{\sin^2(x)}
\]

Simplifying this expression:

\[
f'(x) = \frac{\sin(x) - x \cos(x)}{\sin^2(x)}
\]

### Step 2: Show that \( f'(x) > 0 \) on \( (0, \frac{\pi}{2}) \)

We now need to prove that \( f'(x) > 0 \) for all \( x \in (0, \frac{\pi}{2}) \). This means we need to check whether the numerator of \( f'(x) \), \( \sin(x) - x \cos(x) \), is positive in this interval.

Define \( g(x) = \sin(x) - x \cos(x) \). We need to show that \( g(x) > 0 \) for \( x \in (0, \frac{\pi}{2}) \).

1. **At \( x = 0 \):**
   \[
   g(0) = \sin(0) - 0 \cdot \cos(0) = 0
   \]
   The function starts at 0 at \( x = 0 \).

2. **At \( x = \frac{\pi}{2} \):**
   \[
   g\left(\frac{\pi}{2}\right) = \sin\left(\frac{\pi}{2}\right) - \frac{\pi}{2} \cos\left(\frac{\pi}{2}\right) = 1 - 0 = 1
   \]
   The function is positive at \( x = \frac{\pi}{2} \).

To analyze the behavior of \( g(x) \) for all \( x \in (0, \frac{\pi}{2}) \), note that \( \sin(x) \) increases slower than \( x \cos(x) \) in this interval, but not fast enough for \( \sin(x) - x \cos(x) \) to turn negative. Thus, \( g(x) \) remains positive.

Since \( g(x) > 0 \) for all \( x \in (0, \frac{\pi}{2}) \), we conclude that \( f'(x) > 0 \) in this interval, proving that \( f(x) = \frac{x}{\sin(x)} \) is increasing on \( (0, \frac{\pi}{2}) \).

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### Would you like more details or have any questions?

Here are 5 related questions:
1. What happens to \( f(x) \) as \( x \to 0^+ \)?
2. What is the behavior of \( f(x) \) as \( x \to \frac{\pi}{2}^- \)?
3. Can we apply the same reasoning for \( f(x) \) on the interval \( (\frac{\pi}{2}, \pi) \)?
4. How would the behavior of \( f(x) \) change if we extended the interval to \( (0, \pi) \)?
5. What is the second derivative of \( f(x) = \frac{x}{\sin(x)} \), and what does it tell us?

**Tip:** Always verify that the derivative remains positive when proving that a function is increasing in a given interval.

Prove that the function f(x)=x/sin(x) is an increasing function in the interval (0,π/2).

Explore the detailed solution proving that the function f(x)=x/sin(x) is increasing on the interval (0,π/2). Using calculus, we apply the quotient rule and analyze the first derivative to confirm the function's behavior. Ideal for advanced high school or early university students.

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