Draw the graph of y, where y is the greatest integer of sin x, for x in the interval from 0 to 2π.

To graph the function \( y = \lfloor \sin x \rfloor \) for \( x \) in the interval from \( 0 \) to \( 2\pi \), we first need to understand how the sine function behaves within this range.

Understanding the Sine Function

The sine function oscillates between -1 and 1. Specifically:

• At \( x = 0 \), \( \sin(0) = 0 \)
• At \( x = \frac{\pi}{2} \), \( \sin\left(\frac{\pi}{2}\right) = 1 \)
• At \( x = \pi \), \( \sin(\pi) = 0 \)
• At \( x = \frac{3\pi}{2} \), \( \sin\left(\frac{3\pi}{2}\right) = -1 \)
• At \( x = 2\pi \), \( \sin(2\pi) = 0 \)

Applying the Greatest Integer Function

The greatest integer function, denoted as \( \lfloor x \rfloor \), rounds down to the nearest whole number. Therefore, we can summarize the values of \( y \) as follows:

• For \( 0 \leq x < \frac{\pi}{2} \), \( \sin x \) ranges from 0 to 1, so \( y = 0 \).
• At \( x = \frac{\pi}{2} \), \( y = 1 \).
• For \( \frac{\pi}{2} < x < \frac{3\pi}{2} \), \( \sin x \) goes from 1 down to -1, so \( y = 0 \) for \( \frac{\pi}{2} < x < \pi \) and \( y = -1 \) for \( \pi < x < \frac{3\pi}{2} \).
• At \( x = \frac{3\pi}{2} \), \( y = -1 \).
• For \( \frac{3\pi}{2} < x < 2\pi \), \( \sin x \) increases from -1 to 0, so \( y = -1 \) until \( x = 2\pi \), where \( y = 0 \).

Graphing the Function

Now, we can sketch the graph:

• From \( 0 \) to \( \frac{\pi}{2} \): \( y = 0 \)
• At \( x = \frac{\pi}{2} \): \( y = 1 \)
• From \( \frac{\pi}{2} \) to \( \pi \): \( y = 0 \)
• From \( \pi \) to \( \frac{3\pi}{2} \): \( y = -1 \)
• At \( x = \frac{3\pi}{2} \): \( y = -1 \)
• From \( \frac{3\pi}{2} \) to \( 2\pi \): \( y = 0 \)

In summary, the graph will consist of horizontal lines at \( y = 0 \), \( y = 1 \), and \( y = -1 \), with jumps at the specified points. This creates a step-like appearance reflecting the behavior of the greatest integer function applied to the sine function.