Question

Show that the function \(f(x) = \log \sin x\) is increasing in the interval \((0, \frac{\pi}{2})\) and decreasing in \((\frac{\pi}{2}, \pi)\).

Hint:

The sign of the first derivative determines whether a function is increasing or decreasing. A helpful way to remember the signs of trigonometric functions is the "All Students Take Calculus" (ASTC) rule for quadrants I, II, III, and IV, respectively.

Answer 1

Step 1: Understanding the Concept:
To determine the intervals where a function is increasing or decreasing, we use the first derivative test. A differentiable function \(f(x)\) is: - Increasing on an interval if its first derivative, \(f'(x)\), is positive (\(f'(x)>0\)) for all x in that interval. - Decreasing on an interval if its first derivative, \(f'(x)\), is negative (\(f'(x)<0\)) for all x in that interval.
Step 2: Key Formula or Approach:
1. Find the first derivative of the function, \(f'(x)\).
2. Determine the sign of \(f'(x)\) in the interval \((0, \frac{\pi}{2})\).
3. Determine the sign of \(f'(x)\) in the interval \((\frac{\pi}{2}, \pi)\).
4. Conclude whether the function is increasing or decreasing based on the sign of the derivative in each interval.
Step 3: Detailed Explanation:
The given function is \(f(x) = \log(\sin x)\). The domain of this function requires \(\sin x>0\), which is true for \(x \in (0, \pi)\).
1. Find the first derivative, \(f'(x)\):
Using the chain rule, \(\frac{d}{dx}(\log u) = \frac{1}{u} \cdot \frac{du}{dx}\). \[ f'(x) = \frac{d}{dx}(\log(\sin x)) = \frac{1}{\sin x} \cdot \frac{d}{dx}(\sin x) \] \[ f'(x) = \frac{1}{\sin x} \cdot \cos x = \frac{\cos x}{\sin x} = \cot x \] 2. Analyze the sign of \(f'(x)\) in \((0, \frac{\pi}{2})\):
The interval \((0, \frac{\pi}{2})\) corresponds to the first quadrant.
In the first quadrant, both \(\sin x\) and \(\cos x\) are positive.
Therefore, \(f'(x) = \cot x = \frac{\cos x}{\sin x}\) is positive for all \(x \in (0, \frac{\pi}{2})\).
Since \(f'(x)>0\), the function \(f(x)\) is increasing in the interval \((0, \frac{\pi}{2})\).
3. Analyze the sign of \(f'(x)\) in \((\frac{\pi}{2}, \pi)\):
The interval \((\frac{\pi}{2}, \pi)\) corresponds to the second quadrant.
In the second quadrant, \(\sin x\) is positive, but \(\cos x\) is negative.
Therefore, \(f'(x) = \cot x = \frac{\cos x}{\sin x} = \frac{\text{negative}}{\text{positive}}\) is negative for all \(x \in (\frac{\pi}{2}, \pi)\).
Since \(f'(x)<0\), the function \(f(x)\) is decreasing in the interval \((\frac{\pi}{2}, \pi)\).
Step 4: Final Answer:
We have shown that \(f'(x)>0\) in \((0, \frac{\pi}{2})\) and \(f'(x)<0\) in \((\frac{\pi}{2}, \pi)\). Therefore, the function \(f(x) = \log \sin x\) is increasing in the first interval and decreasing in the second.