Question

Show that the given function \( f(x) = \cos x \) is increasing in \( (\pi, 2\pi) \).

Hint:

To determine if a function is increasing or decreasing, always analyze the sign of its first derivative. A positive derivative implies an increasing function, while a negative derivative implies a decreasing function. Visualizing the sine and cosine graphs or the unit circle can quickly help determine the sign of trigonometric functions in different intervals.

Answer 1

Step 1: Understanding the Concept:
A function is said to be increasing on an interval if its first derivative is positive throughout that interval. To show that \( f(x) = \cos x \) is increasing on \( (\pi, 2\pi) \), we need to find its derivative, \( f'(x) \), and show that \( f'(x)>0 \) for all \( x \) in the interval \( (\pi, 2\pi) \).
Step 2: Key Formula or Approach:
The condition for a function \( f(x) \) to be increasing is \( f'(x)>0 \).
The derivative of \( \cos x \) is \( -\sin x \).
Step 3: Detailed Explanation or Calculation:
The given function is:
\[ f(x) = \cos x \] First, we find the derivative of the function:
\[ f'(x) = \frac{d}{dx}(\cos x) = -\sin x \] Now, we need to determine the sign of \( f'(x) \) in the interval \( (\pi, 2\pi) \).
The interval \( (\pi, 2\pi) \) corresponds to the third and fourth quadrants of the unit circle.
- In the third quadrant, \( x \in (\pi, 3\pi/2) \), the value of \( \sin x \) is negative (\( \sin x<0 \)).
- In the fourth quadrant, \( x \in (3\pi/2, 2\pi) \), the value of \( \sin x \) is also negative (\( \sin x<0 \)).
Therefore, for the entire interval \( x \in (\pi, 2\pi) \), we have \( \sin x<0 \).
Now, let's look at the derivative \( f'(x) = -\sin x \).
Since \( \sin x \) is negative, \( -\sin x \) will be positive.
\[ f'(x) = -(\text{a negative value})>0 \] So, \( f'(x)>0 \) for all \( x \in (\pi, 2\pi) \).
Step 4: Final Answer:
Since the first derivative \( f'(x) = -\sin x \) is positive on the interval \( (\pi, 2\pi) \), the function \( f(x) = \cos x \) is increasing on this interval. Hence proved.