Trigonometry Beyond Right Triangles · Lesson 5

Sine and cosine graphs

The graphs of y=sin⁡xy = \sin x and y=cos⁡xy = \cos x are waves that repeat every 2π2\pi radians (360∘360^\circ) and run between −1-1 and 11. The general form y=asin⁡(b(x−h))+ky = a\sin(b(x - h)) + k (or the same with cosine) stretches and slides that wave. Take b>0b > 0.

Reading the four features
FeatureFrom the equationFrom the graph
Amplitude∣a∣|a|Half of (maximum minus minimum)
Period2πb\frac{2\pi}{b} in radians, or 360∘b\frac{360^\circ}{b} in degreesDistance from one peak to the next
Midliney=ky = kHalfway between the maximum and the minimum
Maximum and minimumk+∣a∣k + |a| and k−∣a∣k - |a|The highest and lowest y-values

A negative aa flips the wave upside down but leaves the amplitude alone, since the amplitude is a distance. The shift hh slides the wave sideways and changes none of the four features in the table.

Reading a graph works in reverse: take half the difference of the highest and lowest y-values for ∣a∣|a| and their average for kk, then measure one full cycle for the period and solve period =2πb= \frac{2\pi}{b} (or 360∘b\frac{360^\circ}{b}) for bb.

Worked example Easy

What is the period of the function f(x)=4cos⁡(5x)−2f(x) = 4\cos(5x) - 2, where xx is in radians?

  1. 2π5\frac{2\pi}{5}Answer
  2. 55
  3. 2π2\pi
  4. 10π10\pi

How to solve it

  1. For y=acos⁡(bx)+ky = a\cos(bx) + k, the period is 2πb\frac{2\pi}{b}.
  2. Here b=5b = 5, so the period is 2π5\frac{2\pi}{5}.
  3. The 4 and the −2-2 change the height of the wave, not its period.

Why each choice is right or wrong

  • A. Correct. The period is 2πb=2π5\frac{2\pi}{b} = \frac{2\pi}{5}: the graph completes 5 cycles in every 2π2\pi.
  • B. Incorrect. 55 is bb itself. bb counts the cycles in each 2π2\pi; the period is 2π5\frac{2\pi}{5}.
  • C. Incorrect. 2π2\pi is the period of cos⁡x\cos x, which ignores the 5 inside. Multiplying xx by 5 squeezes each cycle to 2π5\frac{2\pi}{5}.
  • D. Incorrect. 10π10\pi multiplies by bb instead of dividing: 2π⋅52\pi \cdot 5. A larger bb makes the period shorter, not longer.

Worked example Medium

The graph of y = a\sin(bx) + k for 0 \le x \le 360, with x in degrees4590135180225270315360−2−1123456xy(45, 5)(135, −1)
The graph of y=asin⁡(bx)+ky = a\sin(bx) + k for 0≤x≤3600 \le x \le 360, with xx in degrees A sine curve for x from 0 to 360, with x in degrees. It starts at (0, 2), rises to a maximum of 5 at x = 45, falls through (90, 2) to a minimum of −1 at x = 135, and returns to (180, 2). The same cycle repeats from 180 to 360: a maximum of 5 at x = 225, (270, 2), a minimum of −1 at x = 315, and (360, 2).

The graph shows y=asin⁡(bx)+ky = a\sin(bx) + k for 0≤x≤3600 \le x \le 360, where xx is in degrees and aa and bb are positive. Which equation matches the graph?

  1. y=2sin⁡(2x)+3y = 2\sin(2x) + 3
  2. y=3sin⁡(12x)+2y = 3\sin(\frac{1}{2}x) + 2
  3. y=6sin⁡(2x)+2y = 6\sin(2x) + 2
  4. y=3sin⁡(2x)+2y = 3\sin(2x) + 2Answer

How to solve it

  1. The maximum is 5 and the minimum is −1-1, so a=5−(−1)2=3a = \frac{5 - (-1)}{2} = 3.
  2. The midline is k=5+(−1)2=2k = \frac{5 + (-1)}{2} = 2.
  3. One full cycle runs from x=0x = 0 to x=180x = 180, so the period is 180∘180^\circ.
  4. 360∘b=180∘\frac{360^\circ}{b} = 180^\circ gives b=2b = 2, so y=3sin⁡(2x)+2y = 3\sin(2x) + 2.

Why each choice is right or wrong

  • A. Incorrect. y=2sin⁡(2x)+3y = 2\sin(2x) + 3 swaps the amplitude and the midline. The midline is halfway between 5 and −1-1, which is 2, and the amplitude is the distance from it to the top, 3.
  • B. Incorrect. y=3sin⁡(12x)+2y = 3\sin(\frac{1}{2}x) + 2 solves the period formula upside down. A period of 180∘180^\circ means 360∘b=180∘\frac{360^\circ}{b} = 180^\circ, so b=2b = 2, not 12\frac{1}{2}.
  • C. Incorrect. y=6sin⁡(2x)+2y = 6\sin(2x) + 2 uses the full height, 5−(−1)=65 - (-1) = 6, as the amplitude. The amplitude is half of that, 3.
  • D. Correct. The amplitude is 5−(−1)2=3\frac{5 - (-1)}{2} = 3, the midline is y=2y = 2, and a 180∘180^\circ period gives b=2b = 2.

Trap The period is not $b$

bb counts how many cycles fit in 2π2\pi; it is not the period. For y=sin⁡(4x)y = \sin(4x) the period is 2π4=π2\frac{2\pi}{4} = \frac{\pi}{2}, not 4 and not 8π8\pi.

Trap The amplitude does not include $k$

The amplitude is ∣a∣|a| alone. In y=3cos⁡x+5y = 3\cos x + 5 the amplitude is 3 and the maximum is 5+3=85 + 3 = 8. A choice of 8 for the amplitude has added the midline in.

Calculator When a calculator is faster

Graphing is fast here. Type the function, such as y=4sin(3(x-pi/6))-1, into the built-in Desmos calculator if you test online, then click the peaks and troughs to read the maximum, the minimum, and the distance between peaks. Set the angle mode to match the question first.

On a handheld graphing calculator, choose a window that shows at least one full cycle, such as 0≤x≤2π0 \le x \le 2\pi in radian mode. A scientific or four-function calculator doesn't graph, so know the table above well enough to read the features straight from the equation.