Find the amplitude and period of 3sin⁡sin⁡2x−23\sin\sin 2x - 2. Sketch the curve y=3sin⁡sin⁡2x−2y = 3\sin\sin 2x - 2 for 0∘≤x≤360∘0^\circ \le x \le 360^\circ.

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Worked solution

9(a)

1. State the amplitude

For y=3sin⁡2x−2y = 3\sin 2x - 2, the amplitude is the magnitude of the coefficient of sin⁡2x\sin 2x.
Amplitude=3\text{Amplitude} = 3

2. Find the period

For a sine function asin⁡(bx)+ca\sin(bx)+c, the period is 360∘b\dfrac{360^\circ}{b} with b=2b=2.
Period=360∘2=180∘\text{Period} = \frac{360^\circ}{2} = 180^\circ

Answer

Amplitude =3= 3, Period =180∘= 180^\circ

9(b)

1. Identify key values for the curve

From part (a): amplitude =3=3, period =360∘2=180∘=\dfrac{360^\circ}{2}=180^\circ.

Vertical shift −2-2 gives the centre line y=−2y=-2.

Maximum value =−2+3=1=-2+3=1, minimum value =−2−3=−5=-2-3=-5.

Number of complete cycles in 0∘≤x≤360∘=360∘180∘=20^\circ\le x\le360^\circ = \dfrac{360^\circ}{180^\circ}=2.

2. Plot key points and draw the curve

Mark the key points of one period and repeat:

  • x=0∘: y=−2x=0^\circ:\ y=-2
  • x=45∘: y=1x=45^\circ:\ y=1 (maximum)
  • x=90∘: y=−2x=90^\circ:\ y=-2
  • x=135∘: y=−5x=135^\circ:\ y=-5 (minimum)
  • x=180∘: y=−2x=180^\circ:\ y=-2

Repeat the same shape over 180∘≤x≤360∘180^\circ\le x\le360^\circ (max at 225∘225^\circ, min at 315∘315^\circ).

Join with a smooth sinusoidal curve, labelling the axes and the max/min values 11 and −5-5.

Answer

A smooth sine curve oscillating about the line y=−2y=-2, with maximum value 11 and minimum value −5-5, completing 2 full cycles over 0∘≤x≤360∘0^\circ \le x \le 360^\circ: starting at (0∘,−2)(0^\circ,-2), max at x=45∘x=45^\circ (y=1y=1), crossing y=−2y=-2 at 90∘90^\circ, min at x=135∘x=135^\circ (y=−5y=-5), back to −2-2 at 180∘180^\circ, and repeating the same pattern from 180∘180^\circ to 360∘360^\circ.

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