Question

what is the period?
-3π/2 -π -π/2 0 π/2 π 3π/2
?π /

Explanation:

Step1: Recall Period Definition

The period of a periodic function is the horizontal length over which the function repeats its pattern. For a trigonometric graph (like sine or cosine), we can find the period by measuring the distance between two consecutive peaks (or troughs).

Step2: Analyze the Graph's Intervals

Looking at the x - axis labels: from \(-\frac{3\pi}{2}\) to \(-\pi\), \(-\pi\) to \(-\frac{\pi}{2}\), \(-\frac{\pi}{2}\) to \(0\), \(0\) to \(\frac{\pi}{2}\), \(\frac{\pi}{2}\) to \(\pi\), etc. Let's take two consecutive peaks or troughs. Notice that between \(-\pi\) and \(-\frac{\pi}{2}\) is an interval of \(\frac{\pi}{2}\), but wait, actually, let's check the distance between two identical points. Let's take the distance between \(-\pi\) and \(-\frac{\pi}{2}\)? No, wait, let's see the pattern. Wait, the graph repeats every \(\frac{\pi}{2}\)? Wait, no, let's count the number of periods between two points. Wait, from \(-\frac{3\pi}{2}\) to \(-\pi\): the length is \(-\pi - (-\frac{3\pi}{2})=\frac{\pi}{2}\). But let's check the standard way. The period \(T\) is the length of one full cycle. Let's look at the x - axis. Let's take two consecutive points where the graph repeats. For example, from \(-\pi\) to \(-\frac{\pi}{2}\) is not a full cycle. Wait, maybe the graph is a sine or cosine - like wave. Wait, the distance between \(\frac{\pi}{2}\) and \(\pi\) is \(\pi-\frac{\pi}{2}=\frac{\pi}{2}\). Wait, no, let's see the number of cycles. Wait, between \(-\pi\) and \(\pi\), how many periods? Wait, the graph has a period such that from \(0\) to \(\frac{\pi}{2}\) is a half - period? No, wait, let's look at the x - axis ticks. The distance between \(-\frac{\pi}{2}\) and \(0\) is \(\frac{\pi}{2}\), between \(0\) and \(\frac{\pi}{2}\) is \(\frac{\pi}{2}\), but the graph repeats every \(\frac{\pi}{2}\)? Wait, no, let's think again. Wait, the period of a function \(y = A\sin(Bx + C)+D\) is \(T=\frac{2\pi}{|B|}\). But from the graph, let's see the distance between two consecutive peaks. Let's take a peak at, say, \(x = -\frac{3\pi}{4}\) (mid - point of \(-\pi\) and \(-\frac{\pi}{2}\)) and the next peak at \(x = -\frac{\pi}{4}\) (mid - point of \(-\frac{\pi}{2}\) and \(0\)). The distance between these two peaks is \(-\frac{\pi}{4}-(-\frac{3\pi}{4})=\frac{\pi}{2}\). Wait, no, that can't be. Wait, maybe I made a mistake. Wait, let's look at the x - axis intervals. The labels are \(-\frac{3\pi}{2},-\pi,-\frac{\pi}{2},0,\frac{\pi}{2},\pi,\frac{3\pi}{2}\). Let's see the number of periods between \(-\pi\) and \(\pi\). From \(-\pi\) to \(0\) is \(\pi\) units, and how many periods are there? Let's count the number of cycles. From \(-\pi\) to \(0\), the graph completes 2 cycles? Wait, no, the graph shown has a period of \(\frac{\pi}{2}\)? Wait, no, let's check the distance between two consecutive identical points. Let's take the point where the graph crosses the x - axis at \(-\pi\) and the next x - axis crossing at \(-\frac{\pi}{2}\). The distance is \(-\frac{\pi}{2}-(-\pi)=\frac{\pi}{2}\). But for a sine wave, the distance between two consecutive x - axis crossings (in the same direction) is half the period. Wait, if the distance between two consecutive x - axis crossings (going from negative to positive, for example) is \(\frac{\pi}{2}\), then the period \(T\) would be \(\pi\)? No, wait, no. Wait, let's take a standard sine function \(y=\sin(x)\) has a period of \(2\pi\). If we have \(y = \sin(2x)\), the period is \(\pi\) (since \(T=\frac{2\pi}{2}=\pi\)), and \(y=\sin(4x)\) has a period of \(\frac{\pi}{2}\) (since \(T = \frac{2\pi}{4}=\frac{\pi}{2}\)). Looking at the graph, the number of cycles between \(-\pi\) and \(\pi\): from \(-\pi\) to \(\pi\) is \(2\pi\) units. How many periods are there? Let's count the number of peaks. From \(-\frac{3\pi}{2}\) to \(\frac{3\pi}{2}\), let's see: at \(x = -\frac{3\pi}{2}\), \(x=-\pi\), \(x = -\frac{\pi}{2}\), \(x = 0\), \(x=\frac{\pi}{2}\), \(x=\pi\), \(x=\frac{3\pi}{2}\). The graph repeats every \(\frac{\pi}{2}\)? Wait, no, let's count the number of cycles in the interval from \(0\) to \(\pi\). From \(0\) to \(\pi\), the graph completes 2 cycles. So the length of the interval is \(\pi\), and the number of cycles is 2, so the period \(T=\frac{\pi}{2}\) (since \(T=\frac{\text{length of interval}}{\text{number of cycles}}=\frac{\pi}{2}\)). Wait, that makes sense. So the period is \(\frac{\pi}{2}\), which can be written as \(\frac{1\pi}{2}\). So the numerator is \(1\) and the denominator is \(2\).

Answer:

The period is \(\frac{1\pi}{2}\), so the value of "?" is \(1\) and the denominator is \(2\). So the period is \(\frac{\pi}{2}\), and in the given format \(\frac{?\pi}{[\text{denominator}]}\), \(? = 1\) and the denominator is \(2\). So the answer for the numerator (the "?") is \(1\) and the denominator is \(2\). But the question is asking for the period in the form \(\frac{?\pi}{[\text{box}]}\), so the period is \(\frac{\pi}{2}\), so the numerator is \(1\) and the denominator is \(2\). So the value of "?" is \(1\) and the denominator is \(2\). If we are to fill in the "?", the answer is \(1\) (and the denominator is \(2\)).