Graph 1. The sine function: y = a sinb (x + p) + q
e.g. 1
Sketch the graph of y = sin x for x
Solution
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To ensure that all the critical values are indicated on the graph, we have to use the correct x-values.
If y = a sinbx, then 90º/b will give us the intervals we have to use from 0. In our example b = 1, therefore 90º/1 = 90º.
Therefore we will use x-values of (00,90º, 180º, 270o, 360o etc) If we use a calculator, we will use 90 as the “step”.
Domain: all the possible x values on the graph
Range: all the possible y-values on the graph
Amplitude: the maximum distance from the equilibrium position
Period: number of degrees to complete a wave or a cycle.
Activity 1
Use the graph y = sin x above to answer these questions:
Solutions | ||
y = sinx | ||
1 | Maximum Values | 1 , at x = –270° and 90° |
Minimum Values | –1, at x = –90° and 270° (2) | |
2 | Domain | x ∈ [− 360°;360°], x ∈ ℝ |
Range | [–1; 1] y ∈ ℝ (4) | |
3 | x-intercepts | –360°, –180°, 0°, 180° and 360° (2) |
4 | Amplitude | 1(1) |
5 | Period | 360° (1) |
[10] |
Graph 2. The cosine function:
y = a cos b(x + p) + q
e.g. 2
Sketch the graph of y = cos x for x ∈ [–360°;360°]
x | –360° | –270° | –180° | –90° | 0° | 90° | 180° | 270° | 360° |
y | 1 | 0 | -1 | 0 | 1 | 0 | -1 | 0 | 1 |
To ensure that all the critical values are indicated on the graph, we have to use the correct x-values.
If y = acosbx, then 90º/b will give us the intervals we have to use from 0°. In our example b = 1, therefore 90º/1 = 900.
Therefore we will use x-values of (00,900, 1800, 270o, 360o etc)
If we use a calculator, we will use 900 as the “step”.
y = cos x | ||
1 | Maximum Values | 1, at x = 0° and 360° |
2 | Minimum Values | –1, at x = –180° and 180° |
3 | x-intercepts | –270°, –90°, 90° and 270°. |
4 | Amplitude | 1 |
5 | Period | 360° |
6 | Domain | x ∈ [− 360°;360°], x ∈ R |
7 | Range | [–1; 1] y ∈ R |
Graph 3. The tangent function:
y = a tan b(x + p) + q
e.g.3
Sketch the graph of y = tan x for x ∈ [− 180°;180°]
Solution | |||||||||||||
x | –180° | –135° | –90° | –45° | 0° | 45° | 90° | 135° | 180° | 225° | 270° | 315° | 360° |
y | 0 | 1 | undefined | -1 | 0 | 1 | undefined | -1 | 0 | 1 | undefined | -1 | 0 |
y = tanx | ||
1 | Asymptotes | x = –90°, x = 90° and x = 270° |
2 | x-intercepts | –180°, 0°, 180° and 360°. |
3 | Period | 180° |
4 | Domain | x ∈ [− 180°;360°], x ∈ R |
5 | Range | (–∞; ∞). y ∈ R |
To ensure that all the critical values are indicated on the graph, we have to use the correct x-values.
If y = atanbx, then 45º/b will give us the intervals we have to use from 0. In our example b = 1, therefore 45º/1 = 45º.
Therefore we will use x-values of (0º,45º, 90º, 135º etc) If we use a calculator, we will use 45º as the “step”
e.g.4
y = -1sinx ….
Amplitude = 1
y = 2sinx….Amplitude = 2
y = sinx…Amplitude = 1
y = asinbx…..Amplitude = a
(The amplitude value is always positive, irrespective if a is negative..
Example: if a = -2, then the amplitude is 2.
The parameter a changes the amplitude of the graph.
y = –2 cosx ...
Amplitude = 2
y = ½cosx ... Amplitude = ½
y = cosx ... Amplitude = 1
y = acosbx ... Amplitude = a
The parameter a changes the amplitude of the graph.
Conclusion
The parameter a changes the amplitude of the graph in y = asinbx and y = acosbx.
The graph y = atanbx has no maximum or maximum value. The value does not change the amplitude of y = atanbx as there is no amplitude.
The value of a affects the y-value of each point.
Each y-value is multiplied by a.
Consider the graphs of y = sin x + q, y = cos x + q and y = tan x + q.
e.g.5
1. Sketch the following graphs on the same set of axes for the domain [–360°; 360°]:
Conclusion
The parameter q shifts the whole graph up or down by q units.
Consider the graphs of y = sin bx, y = cos bx and y = tan bx.
e.g.6
Conclusion
Consider the graphs of the form y = sin (x + p), y = cos (x + p) and y = tan (x + p).
e.g.7
Conclusion
For graphs of the form y = sin (x + p), y = cos (x + p) and y = tan (x + p),
p affects the horizontal shift of the graph.
Equation | b=1 | b=2 | b=3 | b=½ |
y = sinbx or y = cosbx | From 0°, use intervals of 90° Interval = 90º/b Period = 360º/b | From 0°, use intervals of 45° Interval = 90º/b Period = 360º/b | From 0°, use intervals of 30° Interval = 90º/b Period = 360º/b | From 0°, use intervals of 180° Interval = 90º/b Period = 360º/b |
y = tanbx | From 0°, use intervals of 45° Interval = 45º/b Period = 180º/b | From 0°, use intervals of 22,5° Interval = 45º/b Period = 180º/b | From 0°, use intervals of 15° Interval = 45º/b Period = 180º/b | From 0°, use intervals of 90° Interval = 45º/b Period = 180º/b |
y = sin(x+p) or y = cos(x+p) | From 0°, use intervals of (90°– p°). with p > 0 The intervals for y = sin(x – 30) and y = sin(x + 30) will be the same. The intervals will be 90 – 30=60 | |||
y = tan(x+p) | From 0°, use intervals of (45°– p°). with p>0 The intervals for y = tan(x – 30) and y = tan(x + 30) will be the same. The intervals will be 45 – 30 = 15. |
Activity 2
Solutions
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What you need to be able to do
In Unit 10, we will discuss the solutions to trigonometric equations. You will be shown how to determine the solution of 2cosx=sin(x+30) algebraically. In this question the solutions can be read off the graphs.