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Section 4.5 Graphs of Sine and Cosine. Sine Curve Key Points:0 Value: 0 100 π 2π2π π 2π2π 1.

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Presentation on theme: "Section 4.5 Graphs of Sine and Cosine. Sine Curve Key Points:0 Value: 0 100 π 2π2π π 2π2π 1."— Presentation transcript:

1 Section 4.5 Graphs of Sine and Cosine

2 Sine Curve Key Points:0 Value: 0 100 π 2π2π π 2π2π 1

3 Cosine Curve Key Points:0 Value: 1 0 0 1 π 2π2π π 2π2π 1

4 Equations For the rest of this section, we will be graphing: y = a Sin (bx – c) + d y = a Cos (bx – c) + d y = Sin x a = 1 b = 1 c = 0 d = 0

5 Graph the equation y = 2 Sin x π 2π2π 1 -2 Key Points:0 Value: 0 20-20 π 2π2π 2

6 Amplitude (a) Half the distance between the maximum and minimum values of the function Given by the value of │a │ Graph the functions: y = 4 Sin x y = ½ Cos x y = -2 Sin x

7 π 2π2π 4 -4 y = 4 Sin x y = ½Cos x 3 2 1 -2 -3 y = -2Sin x

8 y = a Sin (bx – c) + d b gives us the period of the curve Period = y = 4 Sin 2x Amplitude = Period = 4 = π

9 Key Points Would having a period of π change the key points of the curve? π 2π2π 1

10 Finding Key Points In GeneralFor Y = 4Sin 2x 1) Find the period of the curve 2) Divide the period by 4 3) From your starting point, add this distance 4 times for each period 1) Period = π 2) Distance = 3) 0,,,,

11 π 1 y = 4Sin 2x 4 -4

12 Graph the following curves y = 4 Cos 8x y = ½ Cos 2 π x y = -2 Sin 6x

13 y = 4Cos 8x Amplitude =4b =8 → Period = → Distance = 4 -4

14 y = ½Cos 2 π x Amplitude =½b = 2π2π → Period = → Distance = ½ - ½

15 y = -2Sin 6x Amplitude =2b =6 → Period = → Distance = 2 - 2

16 y = a Sin (bx – c) + d a = b = c = amplitude Find the period → Find the “phase shift” → horizontal shift →

17 y = ½ Sin (x - ) a = b = c = ½ 1 → Period = → P. S. =

18 y = -3 Cos (2 π x + 4 π ) a = b = c = 3 2π2π → Period = → P. S. =

19 y = a Sin (bx – c) + d a = b = c = d = amplitude Find the period → Find the “phase shift” → Vertical Shift

20 y = a = b = c = d = 2 → Period = → P. S. = 3

21 y = a = b = c = d = 4 → Period = → P. S. = -2

22 1 4 2 -6 -2 y = -2


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