Trigonometry · Sine and cosine graphs

How to Find the Amplitude of a Sine Wave

Problem

y=5sin⁡(2x)y = 5\sin(2x)

Answer

amplitude 5,period π\text{amplitude } 5,\quad \text{period } \pi

Amplitude and period describe different directions. Amplitude is the vertical height of the wave, and the period is the horizontal length of one repeat.

Step-by-step solution

  1. Match the function to the form y = A sin(bx).

    A=5,b=2A = 5,\quad b = 2

    Why: The number in front of sine is A, and the number multiplying x is b.

  2. Find the amplitude.

    ∣A∣=∣5∣=5|A| = |5| = 5

    Why: The amplitude is the height of the wave from its middle to its peak, so it is always positive.

  3. Find the period with T = 2 pi over b.

    T=2π2=πT = \frac{2\pi}{2} = \pi

    Why: A larger b makes the wave repeat faster and shortens the period.

Common mistakes

  • Swapping A and b, so the amplitude becomes 2 and the period becomes 5.
  • Using A to find the period. Amplitude stretches the wave vertically and period changes it horizontally.
  • Giving a negative amplitude. The amplitude is a distance, so it is positive.

Practice problems

Use the same method. Work each problem on paper, then open the answer to check.

  1. y=3cos⁡(x)y = 3\cos(x)

    Show answer

    amplitude 3,period 2π\text{amplitude } 3,\quad \text{period } 2\pi

  2. y=5sin⁡(4x)y = 5\sin(4x)

    Show answer

    amplitude 5,period π2\text{amplitude } 5,\quad \text{period } \frac{\pi}{2}

  3. y=2cos⁡(πx)y = 2\cos(\pi x)

    Show answer

    amplitude 2,period 2\text{amplitude } 2,\quad \text{period } 2

  4. y=7sin⁡(x2)y = 7\sin\left(\frac{x}{2}\right)

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    amplitude 7,period 4π\text{amplitude } 7,\quad \text{period } 4\pi

  5. y=6cos⁡(3x)y = 6\cos(3x)

    Show answer

    amplitude 6,period 2π3\text{amplitude } 6,\quad \text{period } \frac{2\pi}{3}