AP Precalculus · Lesson 2

Sinusoidal functions and inverse trig

Know Know it

A sinusoidal function has the form f(x)=asin⁡(b(x−c))+df(x) = a\sin(b(x - c)) + d or f(x)=acos⁡(b(x−c))+df(x) = a\cos(b(x - c)) + d. The amplitude is ∣a∣|a|, the midline is y=dy = d, the period is 2π∣b∣\frac{2\pi}{|b|}, and c is the phase shift.

From a graph or a story, amplitude is max⁡−min⁡2\frac{\max - \min}{2} and the midline is y=max⁡+min⁡2y = \frac{\max + \min}{2}. A maximum and the next minimum are half a period apart.

With a>0a > 0, cosine starts a cycle at its maximum, and sine starts on the midline going up. Pick the one that matches a known point, and use c to slide it there.

The tangent function has period π\pi and a vertical asymptote wherever cos⁡x=0\cos x = 0, such as x=π2x = \frac{\pi}{2}.

To undo sine, cosine or tangent, restrict each to a piece where it is one-to-one. The inverse functions then give one angle from that piece, as the table shows.

Inverse trig functions and the angles they return
FunctionInputsOutputs
arcsin⁡x\arcsin x−1≤x≤1-1 \le x \le 1−π2≤y≤π2-\frac{\pi}{2} \le y \le \frac{\pi}{2}
arccos⁡x\arccos x−1≤x≤1-1 \le x \le 10≤y≤π0 \le y \le \pi
arctan⁡x\arctan xall real numbers−π2<y<π2-\frac{\pi}{2} < y < \frac{\pi}{2}

Trap Factor b before you read the shift

In y=sin⁡(2x−π)y = \sin(2x - \pi) the shift is not π\pi. Factor first: sin⁡(2(x−π2))\sin\left(2\left(x - \frac{\pi}{2}\right)\right), so the graph of sin⁡(2x)\sin(2x) moves π2\frac{\pi}{2} to the right.

Calculator On the exam

On the exam, a graphing calculator in radian mode is required in multiple-choice Part B and free-response Part A, and it is not allowed in multiple-choice Part A or free-response Part B. The periodic modeling free-response question is in Part B, so learn to find trig and inverse trig values of special angles by hand.

Worked example Easy

What are the amplitude and the period of y=−4cos⁡(π3x)+2y = -4\cos\left(\frac{\pi}{3}x\right) + 2?

  1. Amplitude 4, period 6Answer
  2. Amplitude -4, period 6
  3. Amplitude 4, period π3\frac{\pi}{3}
  4. Amplitude 2, period 6

How to solve it

  1. Amplitude: ∣a∣=∣−4∣=4|a| = |-4| = 4. The minus sign only flips the graph.
  2. Period: 2π∣b∣=2ππ/3=6\frac{2\pi}{|b|} = \frac{2\pi}{\pi/3} = 6.

Why each choice is right or wrong

  • A. Correct. The amplitude is ∣−4∣=4|-4| = 4, and the period is 2ππ/3=6\frac{2\pi}{\pi/3} = 6.
  • B. Incorrect. Amplitude -4 keeps the sign of a, but amplitude is a distance, ∣a∣|a|, so it is 4.
  • C. Incorrect. Period π3\frac{\pi}{3} is the value of b itself; the period is 2πb\frac{2\pi}{b}, which is 6.
  • D. Incorrect. Amplitude 2 reads the vertical shift, which sets the midline y=2y = 2, not the amplitude.

Explain Explain it

Task verbs: determine, construct

To build a sinusoid from facts: get d and |a| from the maximum and minimum, get b from the period with b=2πperiodb = \frac{2\pi}{\text{period}}, then choose sine or cosine and a shift c so the graph passes through a known point.

To solve sin⁡x=k\sin x = k over one period, find one angle with the unit circle or the inverse function. The second angle with the same sine is π\pi minus the first. Add multiples of 2π2\pi if the interval is longer, and keep only the angles inside it.

Worked example Medium

A sinusoidal function f has a maximum value of 11 at x=2x = 2, and its next minimum value, 3, is at x=6x = 6. Which of these could be f(x)f(x)?

  1. 8cos⁡(π4(x−2))+38\cos\left(\frac{\pi}{4}(x - 2)\right) + 3
  2. 4cos⁡(π2(x−2))+74\cos\left(\frac{\pi}{2}(x - 2)\right) + 7
  3. 4cos⁡(π4(x−2))+74\cos\left(\frac{\pi}{4}(x - 2)\right) + 7Answer
  4. 4sin⁡(π4(x−2))+74\sin\left(\frac{\pi}{4}(x - 2)\right) + 7

How to solve it

  1. Amplitude: 11−32=4\frac{11 - 3}{2} = 4. Midline: 11+32=7\frac{11 + 3}{2} = 7.
  2. Maximum to next minimum is half a period: 6−2=46 - 2 = 4, so the period is 8 and b=2π8=π4b = \frac{2\pi}{8} = \frac{\pi}{4}.
  3. Cosine starts at its maximum, so shift it to x=2x = 2: f(x)=4cos⁡(π4(x−2))+7f(x) = 4\cos\left(\frac{\pi}{4}(x - 2)\right) + 7.

Why each choice is right or wrong

  • A. Incorrect. The choice 8cos⁡8\cos plus 3 uses the full gap of 8 as the amplitude and the minimum as the midline, so its maximum would be 11 but its minimum -5.
  • B. Incorrect. The factor π2\frac{\pi}{2} gives a period of 4, which treats the 4 units from maximum to minimum as a full cycle instead of half of one.
  • C. Correct. The amplitude is 4, the midline is 7, the period is 8 so b=π4b = \frac{\pi}{4}, and cosine shifted to x=2x = 2 peaks there.
  • D. Incorrect. The sine form gives f(2)=4sin⁡0+7=7f(2) = 4\sin 0 + 7 = 7, a midline value, but f must reach its maximum 11 at x=2x = 2.

Worked example Medium

Which values of x in the interval 0≤x<2π0 \le x < 2\pi solve 2sin⁡x=−12\sin x = -1?

  1. π6\frac{\pi}{6} and 5π6\frac{5\pi}{6}
  2. 7π6\frac{7\pi}{6} and 11π6\frac{11\pi}{6}Answer
  3. 4π3\frac{4\pi}{3} and 5π3\frac{5\pi}{3}
  4. −π6-\frac{\pi}{6} only

How to solve it

  1. Divide by 2: sin⁡x=−12\sin x = -\frac{1}{2}, with reference angle π6\frac{\pi}{6}.
  2. Sine is negative in quadrants III and IV: π+π6=7π6\pi + \frac{\pi}{6} = \frac{7\pi}{6} and 2π−π6=11π62\pi - \frac{\pi}{6} = \frac{11\pi}{6}.

Why each choice is right or wrong

  • A. Incorrect. π6\frac{\pi}{6} and 5π6\frac{5\pi}{6} have sine 12\frac{1}{2}, positive, but the equation needs sine equal to −12-\frac{1}{2}.
  • B. Correct. sin⁡x=−12\sin x = -\frac{1}{2} with reference angle π6\frac{\pi}{6} in quadrants III and IV gives 7π6\frac{7\pi}{6} and 11π6\frac{11\pi}{6}.
  • C. Incorrect. 4π3\frac{4\pi}{3} and 5π3\frac{5\pi}{3} have sine −32-\frac{\sqrt{3}}{2}, which solves 2sin⁡x=−32\sin x = -\sqrt{3} instead.
  • D. Incorrect. −π6-\frac{\pi}{6} is the value arcsin returns, but it lies outside the interval, and there are two solutions inside it.

Construct Your turn

A sinusoidal function g has midline y=5y = 5, amplitude 2 and period π\pi, and g(0)=7g(0) = 7. Construct a formula for g and say how each fact sets a number in it.

Write it, then compare. Nothing you type here is saved or graded.

Compare with a model answer

A full answer does this

  • Finds b=2b = 2 from 2πb=π\frac{2\pi}{b} = \pi
  • Writes a formula equal to 2cos⁡(2x)+52\cos(2x) + 5, such as that one, and says it fits because g(0)=7g(0) = 7 is the maximum, 5+25 + 2

Model answer

The midline gives d=5d = 5 and the amplitude gives a=2a = 2. The period is 2πb=π\frac{2\pi}{b} = \pi, so b=2b = 2. Since g(0)=7=5+2g(0) = 7 = 5 + 2, the graph is at its maximum when x=0x = 0, and cosine starts at its maximum, so no shift is needed: g(x)=2cos⁡(2x)+5g(x) = 2\cos(2x) + 5.

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