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Graphs of trigonometry fuctions

Mathematics

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Step-by-step solution

6 steps

To sketch the graphs of trigonometric functions like y=sin⁡(kθ)y = \sin(k\theta) or y=cos⁡(kθ)y = \cos(k\theta), we need to determine their amplitude and period. The amplitude is the maximum displacement from the equilibrium position, and the period is the length of one complete cycle of the wave.

Part a· 3 steps

Sketch the graph of y=sin⁡(3θ)y = \sin(3\theta).

  1. Identify the amplitude and period.

    The general form of a sine function is y=Asin⁡(kθ)y = A \sin(k\theta). Here, A=1A=1, so the amplitude is 1.
    The period PP is given by P=360∘∣k∣P = \frac{360^\circ}{|k|}. In this case, k=3k=3.

    P=360∘3=120∘P = \frac{360^\circ}{3} = 120^\circ

    This means one complete cycle of the graph occurs over an interval of 120∘120^\circ.

  2. Determine key points for one cycle (0∘≤θ≤120∘0^\circ \le \theta \le 120^\circ).

    We find the values of yy at intervals of P4=120∘4=30∘\frac{P}{4} = \frac{120^\circ}{4} = 30^\circ.

    • At θ=0∘\theta = 0^\circ: y=sin⁡(3⋅0∘)=sin⁡(0∘)=0y = \sin(3 \cdot 0^\circ) = \sin(0^\circ) = 0
    • At θ=30∘\theta = 30^\circ: y=sin⁡(3⋅30∘)=sin⁡(90∘)=1y = \sin(3 \cdot 30^\circ) = \sin(90^\circ) = 1 (maximum)
    • At θ=60∘\theta = 60^\circ: y=sin⁡(3⋅60∘)=sin⁡(180∘)=0y = \sin(3 \cdot 60^\circ) = \sin(180^\circ) = 0
    • At θ=90∘\theta = 90^\circ: y=sin⁡(3⋅90∘)=sin⁡(270∘)=−1y = \sin(3 \cdot 90^\circ) = \sin(270^\circ) = -1 (minimum)
    • At θ=120∘\theta = 120^\circ: y=sin⁡(3⋅120∘)=sin⁡(360∘)=0y = \sin(3 \cdot 120^\circ) = \sin(360^\circ) = 0
  3. Sketch the graph.

    Draw a coordinate plane with the horizontal axis representing θ\theta (from 0∘0^\circ to 120∘120^\circ) and the vertical axis representing yy (from -1 to 1). Plot the key points determined in Step 2. Connect these points with a smooth curve. The graph starts at (0∘,0)(0^\circ, 0), rises to a maximum of 1 at 30∘30^\circ, crosses the θ\theta-axis at 60∘60^\circ, falls to a minimum of -1 at 90∘90^\circ, and returns to the θ\theta-axis at 120∘120^\circ. This completes one cycle.

Part b· 3 steps

Sketch the graph of y=cos⁡(3θ)y = \cos(3\theta).

  1. Identify the amplitude and period.

    The general form of a cosine function is y=Acos⁡(kθ)y = A \cos(k\theta). Here, A=1A=1, so the amplitude is 1.
    The period PP is given by P=360∘∣k∣P = \frac{360^\circ}{|k|}. In this case, k=3k=3.

    P=360∘3=120∘P = \frac{360^\circ}{3} = 120^\circ

    This means one complete cycle of the graph occurs over an interval of 120∘120^\circ.

  2. Determine key points for one cycle (0∘≤θ≤120∘0^\circ \le \theta \le 120^\circ).

    We find the values of yy at intervals of P4=120∘4=30∘\frac{P}{4} = \frac{120^\circ}{4} = 30^\circ.

    • At θ=0∘\theta = 0^\circ: y=cos⁡(3⋅0∘)=cos⁡(0∘)=1y = \cos(3 \cdot 0^\circ) = \cos(0^\circ) = 1 (maximum)
    • At θ=30∘\theta = 30^\circ: y=cos⁡(3⋅30∘)=cos⁡(90∘)=0y = \cos(3 \cdot 30^\circ) = \cos(90^\circ) = 0
    • At θ=60∘\theta = 60^\circ: y=cos⁡(3⋅60∘)=cos⁡(180∘)=−1y = \cos(3 \cdot 60^\circ) = \cos(180^\circ) = -1 (minimum)
    • At θ=90∘\theta = 90^\circ: y=cos⁡(3⋅90∘)=cos⁡(270∘)=0y = \cos(3 \cdot 90^\circ) = \cos(270^\circ) = 0
    • At θ=120∘\theta = 120^\circ: y=cos⁡(3⋅120∘)=cos⁡(360∘)=1y = \cos(3 \cdot 120^\circ) = \cos(360^\circ) = 1 (maximum)
  3. Sketch the graph.

    Draw a coordinate plane with the horizontal axis representing θ\theta (from 0∘0^\circ to 120∘120^\circ) and the vertical axis representing yy (from -1 to 1). Plot the key points determined in Step 2. Connect these points with a smooth curve. The graph starts at (0∘,1)(0^\circ, 1), crosses the θ\theta-axis at 30∘30^\circ, falls to a minimum of -1 at 60∘60^\circ, crosses the θ\theta-axis again at 90∘90^\circ, and returns to a maximum of 1 at 120∘120^\circ. This completes one cycle.

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