Question Video: Recognizing the Effect of a Phase Shift in Degrees on the Graph of the Sine Function | Nagwa Question Video: Recognizing the Effect of a Phase Shift in Degrees on the Graph of the Sine Function | Nagwa

Question Video: Recognizing the Effect of a Phase Shift in Degrees on the Graph of the Sine Function Mathematics • Second Year of Secondary School

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Which of the following is the graph of 𝑦 = sin (π‘₯ βˆ’ 90)? [A] Graph A [B] Graph B [C] Graph C [D] Graph D [E] Graph E

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Video Transcript

Which of the following is the graph of 𝑦 equals sin of π‘₯ minus 90?

We will begin by recalling some of the key features of the main sine function graph. The function 𝑓 of π‘₯ equals sin of π‘₯ has a period length of two πœ‹ radians or 360 degrees. All five graphs provided show the scaling along the π‘₯-axis in 90-degree increments. The main sine function has an amplitude of one, which is the distance from the midline to the peak or the trough of the curve. All five graphs provided show sine functions with an amplitude of one.

When graphing the main sine function, we usually start with the point zero, zero then plot four more important points between zero and 360. These coordinate points include 90, one; 180, zero; 270, negative one; and 360, zero. This shape repeats itself every period of 360 degrees to the left and to the right. The graph of 𝑓 of π‘₯ minus 90 is found through a transformation of the main sine function. We recall that a horizontal translation or phase shift of negative 𝑏 is represented by 𝑓 of π‘₯ plus 𝑏.

In our example, the value of 𝑏 is negative 90. This means our horizontal shift of negative 𝑏 is negative negative 90, which equals positive 90 and translates all points to the right 90 degrees. A horizontal translation does not change the amplitude or the period of the sine function. So, we are looking for the five points of the main sine function simply shifted 90 to the right, with no period or amplitude change.

Let’s take a closer look at graph (A). We can trace one period of the sine function, starting with the point zero, one. We notice this graph has been shifted vertically exactly one unit up. A vertical translation is represented by adding a value to the function instead of adding a value directly to π‘₯. This appears to be the graph of sin of π‘₯ plus one. So, we eliminate option (A).

Moving on to graph (B), we notice that the period length has been changed to 180, which is half the period we are looking for. This sort of transformation is referred to as a period change or horizontal stretch by a scale factor of one over 𝑑, where the function is written as 𝑓 of 𝑑 times π‘₯. To take half the period, we use 𝑑 equals two. So, this is the function of sin of two π‘₯. Thus, we eliminate option (B) and move on to graph (C).

After highlighting one period of graph (C), we notice it is the main sine function of 𝑓 of π‘₯ equals sin of π‘₯. Since graph (C) shows no transformations at all, we eliminate this option. Graph (D) has a period change. Instead of 360 degrees, it seems this graph has a period of about 120 degrees. This graph also appears to have a horizontal shift to the left 30. We are looking for a sine graph with a period of 360 and a horizontal shift to the right 90. So, we eliminate option (D).

This only leaves us with graph (E) to consider. We begin by tracing one period of sine. This graph has been shifted to the right 90 and still has a period length of 360 and an amplitude of one. Therefore, we have demonstrated that option (E) is the graph of the function 𝑦 equals sin of π‘₯ minus 90.

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