Mathbench

Trigonometric functions

The graphs of sine, cosine and tangent — period, amplitude and equilibrium — and the inverse functions arcsin, arccos and arctan, with what each one gives back. Set for OMPT-B, OMPT-D and OMPT-G. A label on a question, such as “Not on OMPT-G”, marks a topic that not all of these levels examine.

All OMPT worksheets

Questions

Sheet length
1

Give the amplitude of f(x)=−5sin⁡(x2−1)−4.f(x) = -5\sin\left(\frac{x}{2} - 1\right) - 4\mathclose{.}

Answer:

2

Give the equilibrium (the yy-value of the middle line) of f(x)=5cos⁡(4x−1)−4.f(x) = 5\cos\left(4 x - 1\right) - 4\mathclose{.}

Answer:

3

Give the exact value, in radians: arctan⁡(−1)\arctan\left(-1\right)

Answer:

4

Give the exact value of arctan⁡(tan⁡(4π3)).\arctan\left(\tan\left(\frac{4\pi}{3}\right)\right)\mathclose{.}

Answer:

5

Give the period of f(x)=−4cos⁡(x2+1)−2.f(x) = -4\cos\left(\frac{x}{2} + 1\right) - 2\mathclose{.}

Answer:

6Not on OMPT-G

Give the period of f(x)=−2tan⁡(3x).f(x) = -2\tan\left(3 x\right)\mathclose{.}

Answer:

7

Give the amplitude of f(x)=3cos⁡(x3)−1.f(x) = 3\cos\left(\frac{x}{3}\right) - 1\mathclose{.}

Answer:

8

Give the equilibrium (the yy-value of the middle line) of f(x)=3cos⁡(x3)−2.f(x) = 3\cos\left(\frac{x}{3}\right) - 2\mathclose{.}

Answer:

9

Give the exact value, in radians: arccos⁡(−12)\arccos\left(- \frac{1}{2}\right)

Answer:

10

Give the exact value of arccos⁡(cos⁡(−3π4)).\arccos\left(\cos\left(-\frac{3\pi}{4}\right)\right)\mathclose{.}

Answer:

11

Give the period of f(x)=−3sin⁡(4x+2)−2.f(x) = -3\sin\left(4 x + 2\right) - 2\mathclose{.}

Answer:

12Not on OMPT-G

Give the period of f(x)=5tan⁡(x2).f(x) = 5\tan\left(\frac{x}{2}\right)\mathclose{.}

Answer:

13

Give the amplitude of f(x)=−3cos⁡(x2)+4.f(x) = -3\cos\left(\frac{x}{2}\right) + 4\mathclose{.}

Answer:

14

Give the equilibrium (the yy-value of the middle line) of f(x)=−2sin⁡(x3)−4.f(x) = -2\sin\left(\frac{x}{3}\right) - 4\mathclose{.}

Answer:

15

Give the exact value, in radians: arcsin⁡(22)\arcsin\left(\frac{\sqrt{2}}{2}\right)

Answer:

16

Give the exact value of arccos⁡(cos⁡(7π6)).\arccos\left(\cos\left(\frac{7\pi}{6}\right)\right)\mathclose{.}

Answer:

17

Give the period of f(x)=5sin⁡(2x−1)+2.f(x) = 5\sin\left(2 x - 1\right) + 2\mathclose{.}

Answer:

18Not on OMPT-G

Give the period of f(x)=−2tan⁡(2x).f(x) = -2\tan\left(2 x\right)\mathclose{.}

Answer:

19

Give the amplitude of f(x)=−2cos⁡(x2)−4.f(x) = -2\cos\left(\frac{x}{2}\right) - 4\mathclose{.}

Answer:

20

Give the equilibrium (the yy-value of the middle line) of f(x)=−5cos⁡(4x+1)+4.f(x) = -5\cos\left(4 x + 1\right) + 4\mathclose{.}

Answer:

21

Give the exact value, in radians: arctan⁡(3)\arctan\left(\sqrt{3}\right)

Answer:

22

Give the exact value of arctan⁡(tan⁡(2π3)).\arctan\left(\tan\left(\frac{2\pi}{3}\right)\right)\mathclose{.}

Answer:

23

Give the period of f(x)=3sin⁡(4x+1)−4.f(x) = 3\sin\left(4 x + 1\right) - 4\mathclose{.}

Answer:

24Not on OMPT-G

Give the period of f(x)=2tan⁡(4x).f(x) = 2\tan\left(4 x\right)\mathclose{.}

Answer:

25

Give the amplitude of f(x)=5sin⁡(x3+1)+4.f(x) = 5\sin\left(\frac{x}{3} + 1\right) + 4\mathclose{.}

Answer:

26

Give the equilibrium (the yy-value of the middle line) of f(x)=3sin⁡(3x)−1.f(x) = 3\sin\left(3 x\right) - 1\mathclose{.}

Answer:

27

Give the exact value, in radians: arccos⁡(−22)\arccos\left(- \frac{\sqrt{2}}{2}\right)

Answer:

28

Give the exact value of arctan⁡(tan⁡(5π6)).\arctan\left(\tan\left(\frac{5\pi}{6}\right)\right)\mathclose{.}

Answer:

29

Give the period of f(x)=−2cos⁡(3x−1)−3.f(x) = -2\cos\left(3 x - 1\right) - 3\mathclose{.}

Answer:

30Not on OMPT-G

Give the period of f(x)=4tan⁡(4x).f(x) = 4\tan\left(4 x\right)\mathclose{.}

Answer:

Answers

Every answer below was re-derived independently before this page was built.

1

Answer: 55

The amplitude is the size of the number in front, ∣−5∣=5:|-5| = 5\mathclose{:} the graph swings 55 above and below its middle line.

Common mistakes, and the answer each one gives:

  • Gave the vertical shift instead → −4-4
  • Gave the coefficient of xx instead → 12\frac{1}{2}
  • Kept the minus sign; an amplitude is a distance → −5-5
2

Answer: −4-4

The graph swings evenly about the line y=−4,y = -4\mathclose{,} its equilibrium (middle) line; the amplitude 55 is how far it goes either side.

Common mistakes, and the answer each one gives:

  • Gave the amplitude instead → 55
  • Gave the maximum value instead → 11
3

Answer: −π4-\frac{\pi}{4}

arctan⁡\arctan gives the ONE angle yy with −π2<y<π2-\frac{\pi}{2} < y < \frac{\pi}{2} whose tangent is −1:-1\mathclose{:} that is −π4.-\frac{\pi}{4}\mathclose{.}

Common mistakes, and the answer each one gives:

  • Gave a solution of tan⁡ y=−1\tan\,y = -1 that lies outside the range of arctan⁡\arctan → 3π4\frac{3\pi}{4}
  • Gave the angle in degrees → −45-45
4

Answer: π3\frac{\pi}{3}

tan⁡(4π3)=3,\tan\left(\frac{4\pi}{3}\right) = \sqrt{3}\mathclose{,} and arctan⁡\arctan gives back the ONE angle in (−π2,π2)(-\frac{\pi}{2}, \frac{\pi}{2}) with that value: π3.\frac{\pi}{3}\mathclose{.} That is not 4π3,\frac{4\pi}{3}\mathclose{,} which lies outside the range.

Common mistakes, and the answer each one gives:

  • Cancelled arctan⁡\arctan against tan⁡\tan → 4π3\frac{4\pi}{3}
  • Gave the value of tan⁡,\tan\mathclose{,} not an angle → 3\sqrt{3}
5

Answer: 4π4\pi

The period of cos⁡(Bx+C)\cos(Bx + C) is 2πB,\dfrac{2\pi}{B}\mathclose{,} here 2π12=4π.\dfrac{2\pi}{\frac{1}{2}} = 4\pi\mathclose{.}

Common mistakes, and the answer each one gives:

  • Multiplied 2π2\pi by the coefficient of xx instead of dividing → π\pi
  • Ignored the coefficient of xx → 2π2\pi
6Not on OMPT-G

Answer: π3\frac{\pi}{3}

tan⁡(x)\tan(x) repeats every π\pi — not 2π2\pi like sine and cosine — so tan⁡(Bx)\tan(Bx) repeats every πB,\dfrac{\pi}{B}\mathclose{,} which here is π3.\frac{\pi}{3}\mathclose{.}

Common mistakes, and the answer each one gives:

  • Used 2π,2\pi\mathclose{,} the period of sine and cosine → 2π3\frac{2\pi}{3}
  • Multiplied π\pi by the coefficient of xx instead of dividing → 3π3\pi
7

Answer: 33

The amplitude is the size of the number in front, ∣3∣=3:|3| = 3\mathclose{:} the graph swings 33 above and below its middle line.

Common mistakes, and the answer each one gives:

  • Gave the vertical shift instead → −1-1
  • Gave the coefficient of xx instead → 13\frac{1}{3}
8

Answer: −2-2

The graph swings evenly about the line y=−2,y = -2\mathclose{,} its equilibrium (middle) line; the amplitude 33 is how far it goes either side.

Common mistakes, and the answer each one gives:

  • Gave the amplitude instead → 33
  • Gave the maximum value instead → 11
9

Answer: 2π3\frac{2\pi}{3}

arccos⁡\arccos gives the ONE angle yy with 0≤y≤π0 \le y \le \pi whose cosine is −12:- \frac{1}{2}\mathclose{:} that is 2π3.\frac{2\pi}{3}\mathclose{.}

Common mistakes, and the answer each one gives:

  • Gave a solution of cos⁡ y=−12\cos\,y = - \frac{1}{2} that lies outside the range of arccos⁡\arccos → 4π3\frac{4\pi}{3}
  • Gave the angle in degrees → 120120
10

Answer: 3π4\frac{3\pi}{4}

cos⁡(−3π4)=−22,\cos\left(-\frac{3\pi}{4}\right) = - \frac{\sqrt{2}}{2}\mathclose{,} and arccos⁡\arccos gives back the ONE angle in [0,π][0, \pi] with that value: 3π4.\frac{3\pi}{4}\mathclose{.} That is not −3π4,-\frac{3\pi}{4}\mathclose{,} which lies outside the range.

Common mistakes, and the answer each one gives:

  • Cancelled arccos⁡\arccos against cos⁡\cos → −3π4-\frac{3\pi}{4}
  • Gave the value of cos⁡,\cos\mathclose{,} not an angle → −22- \frac{\sqrt{2}}{2}
11

Answer: π2\frac{\pi}{2}

The period of sin⁡(Bx+C)\sin(Bx + C) is 2πB,\dfrac{2\pi}{B}\mathclose{,} here 2π4=π2.\dfrac{2\pi}{4} = \frac{\pi}{2}\mathclose{.}

Common mistakes, and the answer each one gives:

  • Multiplied 2π2\pi by the coefficient of xx instead of dividing → 8π8\pi
  • Ignored the coefficient of xx → 2π2\pi
12Not on OMPT-G

Answer: 2π2\pi

tan⁡(x)\tan(x) repeats every π\pi — not 2π2\pi like sine and cosine — so tan⁡(Bx)\tan(Bx) repeats every πB,\dfrac{\pi}{B}\mathclose{,} which here is 2π.2\pi\mathclose{.}

Common mistakes, and the answer each one gives:

  • Used 2π,2\pi\mathclose{,} the period of sine and cosine → 4π4\pi
  • Multiplied π\pi by the coefficient of xx instead of dividing → π2\frac{\pi}{2}
13

Answer: 33

The amplitude is the size of the number in front, ∣−3∣=3:|-3| = 3\mathclose{:} the graph swings 33 above and below its middle line.

Common mistakes, and the answer each one gives:

  • Gave the vertical shift instead → 44
  • Gave the coefficient of xx instead → 12\frac{1}{2}
  • Kept the minus sign; an amplitude is a distance → −3-3
14

Answer: −4-4

The graph swings evenly about the line y=−4,y = -4\mathclose{,} its equilibrium (middle) line; the amplitude 22 is how far it goes either side.

Common mistakes, and the answer each one gives:

  • Gave the amplitude instead → 22
  • Gave the maximum value instead → −2-2
15

Answer: π4\frac{\pi}{4}

arcsin⁡\arcsin gives the ONE angle yy with −π2≤y≤π2-\frac{\pi}{2} \le y \le \frac{\pi}{2} whose sine is 22:\frac{\sqrt{2}}{2}\mathclose{:} that is π4.\frac{\pi}{4}\mathclose{.}

Common mistakes, and the answer each one gives:

  • Gave a solution of sin⁡ y=22\sin\,y = \frac{\sqrt{2}}{2} that lies outside the range of arcsin⁡\arcsin → 3π4\frac{3\pi}{4}
  • Gave the angle in degrees → 4545
16

Answer: 5π6\frac{5\pi}{6}

cos⁡(7π6)=−32,\cos\left(\frac{7\pi}{6}\right) = - \frac{\sqrt{3}}{2}\mathclose{,} and arccos⁡\arccos gives back the ONE angle in [0,π][0, \pi] with that value: 5π6.\frac{5\pi}{6}\mathclose{.} That is not 7π6,\frac{7\pi}{6}\mathclose{,} which lies outside the range.

Common mistakes, and the answer each one gives:

  • Cancelled arccos⁡\arccos against cos⁡\cos → 7π6\frac{7\pi}{6}
  • Gave the value of cos⁡,\cos\mathclose{,} not an angle → −32- \frac{\sqrt{3}}{2}
17

Answer: π\pi

The period of sin⁡(Bx+C)\sin(Bx + C) is 2πB,\dfrac{2\pi}{B}\mathclose{,} here 2π2=π.\dfrac{2\pi}{2} = \pi\mathclose{.}

Common mistakes, and the answer each one gives:

  • Multiplied 2π2\pi by the coefficient of xx instead of dividing → 4π4\pi
  • Ignored the coefficient of xx → 2π2\pi
18Not on OMPT-G

Answer: π2\frac{\pi}{2}

tan⁡(x)\tan(x) repeats every π\pi — not 2π2\pi like sine and cosine — so tan⁡(Bx)\tan(Bx) repeats every πB,\dfrac{\pi}{B}\mathclose{,} which here is π2.\frac{\pi}{2}\mathclose{.}

Common mistakes, and the answer each one gives:

  • Used 2π,2\pi\mathclose{,} the period of sine and cosine → π\pi
  • Multiplied π\pi by the coefficient of xx instead of dividing → 2π2\pi
19

Answer: 22

The amplitude is the size of the number in front, ∣−2∣=2:|-2| = 2\mathclose{:} the graph swings 22 above and below its middle line.

Common mistakes, and the answer each one gives:

  • Gave the vertical shift instead → −4-4
  • Gave the coefficient of xx instead → 12\frac{1}{2}
  • Kept the minus sign; an amplitude is a distance → −2-2
20

Answer: 44

The graph swings evenly about the line y=4,y = 4\mathclose{,} its equilibrium (middle) line; the amplitude 55 is how far it goes either side.

Common mistakes, and the answer each one gives:

  • Gave the amplitude instead → 55
  • Gave the maximum value instead → 99
21

Answer: π3\frac{\pi}{3}

arctan⁡\arctan gives the ONE angle yy with −π2<y<π2-\frac{\pi}{2} < y < \frac{\pi}{2} whose tangent is 3:\sqrt{3}\mathclose{:} that is π3.\frac{\pi}{3}\mathclose{.}

Common mistakes, and the answer each one gives:

  • Gave a solution of tan⁡ y=3\tan\,y = \sqrt{3} that lies outside the range of arctan⁡\arctan → 4π3\frac{4\pi}{3}
  • Gave the angle in degrees → 6060
22

Answer: −π3-\frac{\pi}{3}

tan⁡(2π3)=−3,\tan\left(\frac{2\pi}{3}\right) = - \sqrt{3}\mathclose{,} and arctan⁡\arctan gives back the ONE angle in (−π2,π2)(-\frac{\pi}{2}, \frac{\pi}{2}) with that value: −π3.-\frac{\pi}{3}\mathclose{.} That is not 2π3,\frac{2\pi}{3}\mathclose{,} which lies outside the range.

Common mistakes, and the answer each one gives:

  • Cancelled arctan⁡\arctan against tan⁡\tan → 2π3\frac{2\pi}{3}
  • Gave the value of tan⁡,\tan\mathclose{,} not an angle → −3- \sqrt{3}
23

Answer: π2\frac{\pi}{2}

The period of sin⁡(Bx+C)\sin(Bx + C) is 2πB,\dfrac{2\pi}{B}\mathclose{,} here 2π4=π2.\dfrac{2\pi}{4} = \frac{\pi}{2}\mathclose{.}

Common mistakes, and the answer each one gives:

  • Multiplied 2π2\pi by the coefficient of xx instead of dividing → 8π8\pi
  • Ignored the coefficient of xx → 2π2\pi
24Not on OMPT-G

Answer: π4\frac{\pi}{4}

tan⁡(x)\tan(x) repeats every π\pi — not 2π2\pi like sine and cosine — so tan⁡(Bx)\tan(Bx) repeats every πB,\dfrac{\pi}{B}\mathclose{,} which here is π4.\frac{\pi}{4}\mathclose{.}

Common mistakes, and the answer each one gives:

  • Used 2π,2\pi\mathclose{,} the period of sine and cosine → π2\frac{\pi}{2}
  • Multiplied π\pi by the coefficient of xx instead of dividing → 4π4\pi
25

Answer: 55

The amplitude is the size of the number in front, ∣5∣=5:|5| = 5\mathclose{:} the graph swings 55 above and below its middle line.

Common mistakes, and the answer each one gives:

  • Gave the vertical shift instead → 44
  • Gave the coefficient of xx instead → 13\frac{1}{3}
26

Answer: −1-1

The graph swings evenly about the line y=−1,y = -1\mathclose{,} its equilibrium (middle) line; the amplitude 33 is how far it goes either side.

Common mistakes, and the answer each one gives:

  • Gave the amplitude instead → 33
  • Gave the maximum value instead → 22
27

Answer: 3π4\frac{3\pi}{4}

arccos⁡\arccos gives the ONE angle yy with 0≤y≤π0 \le y \le \pi whose cosine is −22:- \frac{\sqrt{2}}{2}\mathclose{:} that is 3π4.\frac{3\pi}{4}\mathclose{.}

Common mistakes, and the answer each one gives:

  • Gave a solution of cos⁡ y=−22\cos\,y = - \frac{\sqrt{2}}{2} that lies outside the range of arccos⁡\arccos → 5π4\frac{5\pi}{4}
  • Gave the angle in degrees → 135135
28

Answer: −π6-\frac{\pi}{6}

tan⁡(5π6)=−33,\tan\left(\frac{5\pi}{6}\right) = - \frac{\sqrt{3}}{3}\mathclose{,} and arctan⁡\arctan gives back the ONE angle in (−π2,π2)(-\frac{\pi}{2}, \frac{\pi}{2}) with that value: −π6.-\frac{\pi}{6}\mathclose{.} That is not 5π6,\frac{5\pi}{6}\mathclose{,} which lies outside the range.

Common mistakes, and the answer each one gives:

  • Cancelled arctan⁡\arctan against tan⁡\tan → 5π6\frac{5\pi}{6}
  • Gave the value of tan⁡,\tan\mathclose{,} not an angle → −33- \frac{\sqrt{3}}{3}
29

Answer: 2π3\frac{2\pi}{3}

The period of cos⁡(Bx+C)\cos(Bx + C) is 2πB,\dfrac{2\pi}{B}\mathclose{,} here 2π3=2π3.\dfrac{2\pi}{3} = \frac{2\pi}{3}\mathclose{.}

Common mistakes, and the answer each one gives:

  • Multiplied 2π2\pi by the coefficient of xx instead of dividing → 6π6\pi
  • Ignored the coefficient of xx → 2π2\pi
30Not on OMPT-G

Answer: π4\frac{\pi}{4}

tan⁡(x)\tan(x) repeats every π\pi — not 2π2\pi like sine and cosine — so tan⁡(Bx)\tan(Bx) repeats every πB,\dfrac{\pi}{B}\mathclose{,} which here is π4.\frac{\pi}{4}\mathclose{.}

Common mistakes, and the answer each one gives:

  • Used 2π,2\pi\mathclose{,} the period of sine and cosine → π2\frac{\pi}{2}
  • Multiplied π\pi by the coefficient of xx instead of dividing → 4π4\pi