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Algebra - Graphing trigonometric functions-extended

Grade 9IB

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The sine function y=asin⁡(bx)+dy = a \sin(bx) + d represents a wave starting at the principal axis. The parameter aa affects the amplitude, bb determines the period, and dd shifts the graph vertically.

A standard sine wave starting at (0,0) with a peak at (90,1) and a trough at (270,-1).
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The cosine function y=acos⁡(bx)+dy = a \cos(bx) + d starts at its maximum value (when a>0a > 0). It follows the same periodic behavior as sine but is shifted 90∘90^{\circ} horizontally.

A standard cosine wave starting at (0,1) with a trough at (180,-1).
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The period of a function is the horizontal distance required to complete one full cycle. For y=sin⁡(bx)y = \sin(bx) or y=cos⁡(bx)y = \cos(bx), the period is calculated as 360∘b\frac{360^{\circ}}{b}.

A sine wave with period 180 degrees showing two full cycles within 360 degrees.
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Vertical translation dd shifts the midline of the graph. The maximum value is d+∣a∣d + |a| and the minimum value is d−∣a∣d - |a|.

📐Formulae

Amplitude=∣a∣=Maximum−Minimum2\text{Amplitude} = |a| = \frac{\text{Maximum} - \text{Minimum}}{2}

Period=360∘b (in degrees) or 2πb (in radians)\text{Period} = \frac{360^{\circ}}{b} \text{ (in degrees) or } \frac{2\pi}{b} \text{ (in radians)}

Principal Axis (Midline):y=d=Maximum+Minimum2\text{Principal Axis (Midline)}: y = d = \frac{\text{Maximum} + \text{Minimum}}{2}

Maximum Value=d+∣a∣\text{Maximum Value} = d + |a|

Minimum Value=d−∣a∣\text{Minimum Value} = d - |a|

💡Examples

Problem 1:

Determine the amplitude, period, and midline of the function y=3sin⁡(2x)+5y = 3 \sin(2x) + 5.

Solution:

The amplitude is ∣a∣=3|a| = 3. The value of bb is 22, so the period is 360∘2=180∘\frac{360^{\circ}}{2} = 180^{\circ}. The midline is y=5y = 5.

Explanation:

Identify a=3,b=2,d=5a=3, b=2, d=5 from the standard form y=asin⁡(bx)+dy = a \sin(bx) + d. The amplitude is the coefficient of the sine term, the period is 360∘360^{\circ} divided by bb, and the midline is the constant added to the function.

Problem 2:

Find the equation of a cosine function that has a maximum value of 1010, a minimum value of −2-2, and a period of 90∘90^{\circ}.

Solution:

  1. Midline d=10+(−2)2=4d = \frac{10 + (-2)}{2} = 4.
  2. Amplitude a=10−4=6a = 10 - 4 = 6.
  3. Period factor b=360∘90∘=4b = \frac{360^{\circ}}{90^{\circ}} = 4. Equation: y=6cos⁡(4x)+4y = 6 \cos(4x) + 4.

Explanation:

First find the vertical center (midline) and the distance from the center to the peak (amplitude). Then calculate bb using the period formula. Substitute these into the general cosine equation.

Problem 3:

Describe the transformations required to map y=cos⁡(x)y = \cos(x) onto y=cos⁡(x−45∘)−2y = \cos(x - 45^{\circ}) - 2.

Solution:

The graph is translated 45∘45^{\circ} to the right (horizontal shift) and 22 units downwards (vertical shift).

Explanation:

In the form y=cos⁡(x−c)+dy = \cos(x - c) + d, the value c=45∘c = 45^{\circ} indicates a horizontal shift right, and d=−2d = -2 indicates a vertical translation down.

Problem 4:

Identify the equation of the trigonometric function shown in the graph, which has a maximum at y=4y=4, a minimum at y=0y=0, and completes one cycle at x=120∘x=120^{\circ}.

Graph of y = 2sin(3x) + 2 showing a period of 120 and range [0, 4].

Solution:

  1. Find the amplitude: a=4−02=2a = \frac{4 - 0}{2} = 2.
  2. Find the midline: d=4+02=2d = \frac{4 + 0}{2} = 2.
  3. Find the period: The graph completes a cycle at 120∘120^{\circ}, so 360∘b=120∘  ⟹  b=3\frac{360^{\circ}}{b} = 120^{\circ} \implies b = 3.
  4. Since the graph starts at the midline and goes up, it is a sine function: y=2sin⁡(3x)+2y = 2 \sin(3x) + 2.

Explanation:

The amplitude is half the distance between peak and trough. The midline is the average of the peak and trough. The frequency factor bb is found by dividing the standard period by the observed period.

Problem 5:

Sketch the graph of y=3cos⁡(x)−1y = 3 \cos(x) - 1 for 0∘≤x≤360∘0^{\circ} \le x \le 360^{\circ} and determine its range.

Graph of y = 3cos(x) - 1 showing the wave starting at y=2 and dropping to y=-4 at 180 degrees.

Solution:

  1. Amplitude is ∣3∣=3|3| = 3.
  2. Midline is y=−1y = -1.
  3. Maximum value =−1+3=2= -1 + 3 = 2.
  4. Minimum value =−1−3=−4= -1 - 3 = -4.
  5. Range is [−4,2][-4, 2].
  6. The graph starts at (0,2)(0, 2), crosses the midline at 90∘90^{\circ}, hits the minimum at 180∘180^{\circ}, and returns to the maximum at 360∘360^{\circ}.

Explanation:

To sketch a cosine graph, plot the five key points: the start (max), first intercept (midline), middle (min), second intercept (midline), and end (max).