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Grade 11 · MCR3U · Functions

MCR3U: Transformations of Sine and Cosine

Every sinusoidal graph in MCR3U is y=sin⁡xy = \sin x or y=cos⁡xy = \cos x stretched, shifted or reflected. Read four numbers from y=asin⁡(k(x−d))+cy = a\sin(k(x - d)) + c and you can sketch the graph, or read them from a graph and write its equation.

How It Works

The parent graphs. y=sin⁡xy = \sin x starts at (0,0)(0, 0), peaks at (90∘,1)(90^\circ, 1), crosses the axis at (180∘,0)(180^\circ, 0), bottoms out at (270∘,−1)(270^\circ, -1) and is back to (360∘,0)(360^\circ, 0). y=cos⁡xy = \cos x has the same shape but starts at its maximum, (0,1)(0, 1). Both have period 360∘360^\circ, amplitude 11 and midline y=0y = 0.

Amplitude. ∣a∣|a| stretches the graph vertically. If a<0a < 0, the graph is also reflected in the midline, so a sine curve goes down first.

Period. kk compresses the graph horizontally by a factor of 1∣k∣\frac{1}{|k|}, so the period is 360∘∣k∣\frac{360^\circ}{|k|}. Factor kk out of the brackets before you read the shift: y=sin⁡(2x−60∘)y = \sin(2x - 60^\circ) is y=sin⁡(2(x−30∘))y = \sin(2(x - 30^\circ)), a shift of 30∘30^\circ, not 60∘60^\circ.

Phase shift. dd moves the graph right when you see x−dx - d and left when you see x+dx + d.

Midline. cc moves the graph up or down, so the midline is y=cy = c. The maximum is c+∣a∣c + |a| and the minimum is c−∣a∣c - |a|.

The mapping rule. Each point (x,y)(x, y) on the parent graph goes to (xk+d,ay+c)\left(\frac{x}{k} + d, ay + c\right). Map the five key points, then join them with a smooth wave. The stretches come before the shifts, which is why you divide by kk before adding dd.

5 Worked Examples

Easy first, test level last. Try each one on paper, then open the solution.

  1. Easy

    State the amplitude, period, phase shift and midline of y=3sin⁡(2(x−45∘))+1y = 3\sin(2(x - 45^\circ)) + 1, and its maximum and minimum.
    Show solution
    1. a=3a = 3, k=2k = 2, d=45∘d = 45^\circ and c=1c = 1.
    2. Period: 360∘2=180∘\frac{360^\circ}{2} = 180^\circ.
    3. Maximum: 1+3=41 + 3 = 4. Minimum: 1−3=−21 - 3 = -2.

    Answer: Amplitude 33, period 180∘180^\circ, 45∘45^\circ right, midline y=1y = 1; maximum 44, minimum −2-2.

  2. Easy

    For y=−2cos⁡x+5y = -2\cos x + 5 with 0∘≤x≤360∘0^\circ \le x \le 360^\circ, find the maximum and minimum and where they happen.
    Show solution
    1. a=−2a = -2: amplitude 22, reflected in the midline y=5y = 5. The graph starts at a minimum instead of a maximum.
    2. Minimum 5−2=35 - 2 = 3 where cos⁡x=1\cos x = 1, at x=0∘x = 0^\circ and x=360∘x = 360^\circ.
    3. Maximum 5+2=75 + 2 = 7 where cos⁡x=−1\cos x = -1, at x=180∘x = 180^\circ.

    Answer: Maximum 77 at 180∘180^\circ; minimum 33 at 0∘0^\circ and 360∘360^\circ.

  3. Medium

    Use the mapping rule to find the images of the five key points of y=sin⁡xy = \sin x under y=2sin⁡(3(x+30∘))−1y = 2\sin(3(x + 30^\circ)) - 1.
    Show solution
    1. a=2a = 2, k=3k = 3, d=−30∘d = -30^\circ, c=−1c = -1, so (x,y)→(x3−30∘,2y−1)(x, y) \to \left(\frac{x}{3} - 30^\circ, 2y - 1\right).
    2. (0,0)→(−30∘,−1)(0, 0) \to (-30^\circ, -1) and (90∘,1)→(0∘,1)(90^\circ, 1) \to (0^\circ, 1).
    3. (180∘,0)→(30∘,−1)(180^\circ, 0) \to (30^\circ, -1), (270∘,−1)→(60∘,−3)(270^\circ, -1) \to (60^\circ, -3) and (360∘,0)→(90∘,−1)(360^\circ, 0) \to (90^\circ, -1).
    4. Check: one cycle runs from −30∘-30^\circ to 90∘90^\circ, which is 120∘=360∘3120^\circ = \frac{360^\circ}{3}.

    Answer: (−30∘,−1)(-30^\circ, -1), (0∘,1)(0^\circ, 1), (30∘,−1)(30^\circ, -1), (60∘,−3)(60^\circ, -3), (90∘,−1)(90^\circ, -1)

  4. Medium

    Write y=4cos⁡(2x−120∘)y = 4\cos(2x - 120^\circ) so you can read the phase shift, then state the period and the phase shift.
    Show solution
    1. Factor 22 out of the brackets: 2x−120∘=2(x−60∘)2x - 120^\circ = 2(x - 60^\circ).
    2. y=4cos⁡(2(x−60∘))y = 4\cos(2(x - 60^\circ)), so d=60∘d = 60^\circ.
    3. Period: 360∘2=180∘\frac{360^\circ}{2} = 180^\circ.

    Answer: y=4cos⁡(2(x−60∘))y = 4\cos(2(x - 60^\circ)): period 180∘180^\circ, phase shift 60∘60^\circ right.

  5. Test level

    A sinusoidal graph has a maximum at (30∘,7)(30^\circ, 7) and the next minimum at (120∘,1)(120^\circ, 1). Write a cosine equation and a sine equation for it.
    Show solution
    1. Midline: c=7+12=4c = \frac{7 + 1}{2} = 4. Amplitude: a=7−12=3a = \frac{7 - 1}{2} = 3.
    2. Maximum to minimum is half a cycle: 120∘−30∘=90∘120^\circ - 30^\circ = 90^\circ, so the period is 180∘180^\circ and k=360∘180∘=2k = \frac{360^\circ}{180^\circ} = 2.
    3. Cosine starts at a maximum, so d=30∘d = 30^\circ: y=3cos⁡(2(x−30∘))+4y = 3\cos(2(x - 30^\circ)) + 4.
    4. Sine reaches its maximum a quarter cycle, 45∘45^\circ, after it starts on the midline, so it starts at 30∘−45∘=−15∘30^\circ - 45^\circ = -15^\circ: y=3sin⁡(2(x+15∘))+4y = 3\sin(2(x + 15^\circ)) + 4.
    5. Check the sine equation at x=30∘x = 30^\circ: 3sin⁡(90∘)+4=73\sin(90^\circ) + 4 = 7.

    Answer: y=3cos⁡(2(x−30∘))+4y = 3\cos(2(x - 30^\circ)) + 4 or y=3sin⁡(2(x+15∘))+4y = 3\sin(2(x + 15^\circ)) + 4

Learn It on Martian Lab

Martian Lab teaches this in the Graphing Trignometric Functions lesson of its MCR3U Trigonometric Functions unit. Watch it, then take the Trigonometric Functions practice tests to see if it stuck.

Watch the Graphing Trignometric Functions lesson free

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