Trigonometry - Graphing transformations of sin and cos

This paragraph introduces the concept of transformations of sine and cosine functions. The instructor explains that starting with a basic sine or cosine graph, various changes can be made to the function's equation, which in turn alter the graph. The focus is on understanding the key values within the equation that indicate changes in amplitude, period, phase shift, and vertical shift. The amplitude is determined by the absolute value of the coefficient 'a' in front of the trigonometric function, and whether the graph is flipped over the x-axis depends on the sign of 'a'. The period is adjusted by the coefficient 'b' next to the variable x, with the new period calculated as 2π divided by the absolute value of 'b'. The phase shift, indicated by 'c', is found by dividing 'c' by 'b', and it tells us how far the graph has been shifted left or right. Lastly, 'd' represents the vertical shift, showing how much the graph has moved up or down. The paragraph sets the stage for examples that will demonstrate these concepts in action.

The second paragraph delves into the practical application of the concepts introduced earlier, starting with the function y = sin(5x). The instructor begins by recalling the basic sine wave with an amplitude of 1 and a period of 2π. For the given function, 'a' is 1, indicating no change in amplitude, and 'b' is 5, which affects the period. The new period is calculated as 2π/5, meaning the function completes a full cycle in 2π/5 instead of the usual 2π. Since there are no values for 'c' and 'd', there is no phase shift or vertical shift, and the graph remains in its standard position. The instructor then sketches one complete cycle of the sine wave, emphasizing the importance of understanding the basic shape and how the given function alters it. Key values such as the halfway points are also discussed to provide a more detailed understanding of the graph's behavior.

In this paragraph, the instructor presents a more complex example, y = -cos(x) + 2, which involves both amplitude and vertical shift transformations. The amplitude is determined by the absolute value of 'a', which is 1, but the negative sign indicates that the graph is flipped over the x-axis. The period remains unchanged at 2π because 'b' is 1. There is no phase shift since 'c' is not present, but there is a vertical shift indicated by 'd', which is 2, moving the graph up. The instructor then constructs the graph, starting with the vertical shift, followed by marking the amplitude peaks at 3 and -1 (accounting for the negative amplitude). The function's graph is flipped to create an upside-down U shape, and key values are identified at π/2, 3π/2, and other points within the cycle to complete the graph.

The third example tackled in the script is y = 4sin(x + 3π/2), which includes an amplitude change and a phase shift. The amplitude is 4, significantly larger than the basic sine wave. There is no 'b' value affecting the period, so it remains at 2π. The phase shift is determined by 'c', which is 3π/2, and when divided by the 'b' value (which is 1), it indicates a phase shift of -3π/2. This means the graph starts not at 0 but at -3π/2. The instructor calculates where the graph should stop after one complete period, which is at π/2, and then breaks down the cycle into quarters to find key values such as -2π/2, -π/2, 0, and π/2. The graph is drawn accordingly, starting from the phase shift point and completing one period, with the sine wave's peak and trough marked according to the new amplitude of 4.

The final paragraph in the script discusses the function y = 3cos(4x - π/2), which includes an amplitude of 3 and a phase shift. The instructor begins by noting the amplitude and the 'b' value of 4, which affects the period. The new period is π/2, as calculated by 2π divided by the absolute value of 4. The phase shift is determined by 'c', which is π/2, and when divided by 'b', it results in a phase shift of π/8. There is no vertical shift, as indicated by the absence of a 'd' value. The instructor then graphs the function, marking the amplitude peaks at 3 and -3, and starting the graph at the phase shift point, π/8. The complete cycle is drawn within π/2, and key values are identified at 3π/8, 2π/8, and other points within the cycle. The instructor emphasizes the continuous nature of the graph beyond the drawn period.