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The Unit Circle practice questions — A-Level Maths

12 free multiple-choice problems on the unit circle, ordered to match A-Level Maths difficulty, each with a full worked solution. No account needed — and there's a fresh problem set every day.

Problems & worked solutions

Problem #0559 International

Problem 1 — The Unit Circle

The number of solutions of sin⁡x=−12\sin x = -\dfrac{1}{2} on [0,2π)[0, 2\pi) is:

Show answer & worked solution
  1. A. 11
  2. B. 22✓ correct
  3. C. 33
  4. D. 44

sin⁡x=−12\sin x = -\dfrac{1}{2} at x=7π6x = \dfrac{7\pi}{6} (Q3) and x=11π6x = \dfrac{11\pi}{6} (Q4). Two solutions on [0,2π)[0, 2\pi).

Problem #0558 International

Problem 2 — The Unit Circle

The value of tan⁡7π6\tan\dfrac{7\pi}{6} is:

Show answer & worked solution
  1. A. 13\dfrac{1}{\sqrt{3}}✓ correct
  2. B. −13-\dfrac{1}{\sqrt{3}}
  3. C. 3\sqrt{3}
  4. D. −3-\sqrt{3}

tan⁡ ⁣(π+π6)=tan⁡π6=13\tan\!\left(\pi + \dfrac{\pi}{6}\right) = \tan\dfrac{\pi}{6} = \dfrac{1}{\sqrt{3}}.

Problem #0557 International

Problem 3 — The Unit Circle

The point on the unit circle associated with x=7π4x = \dfrac{7\pi}{4} lies in:

Show answer & worked solution
  1. A. the first quadrant
  2. B. the second quadrant
  3. C. the third quadrant
  4. D. the fourth quadrant✓ correct

7π4∈ ⁣(3π2,2π)\dfrac{7\pi}{4} \in \!\left(\dfrac{3\pi}{2}, 2\pi\right), which is the fourth quadrant.

Problem #0556 International

Problem 4 — The Unit Circle

The value of cos⁡ ⁣(−π3)\cos\!\left(-\dfrac{\pi}{3}\right) is:

Show answer & worked solution
  1. A. 12\dfrac{1}{2}✓ correct
  2. B. −12-\dfrac{1}{2}
  3. C. 32\dfrac{\sqrt{3}}{2}
  4. D. −32-\dfrac{\sqrt{3}}{2}

cos⁡ ⁣(−π3)=cos⁡π3=12\cos\!\left(-\dfrac{\pi}{3}\right) = \cos\dfrac{\pi}{3} = \dfrac{1}{2}.

Problem #0555 International

Problem 5 — The Unit Circle

The value of sin⁡11π6\sin\dfrac{11\pi}{6} is:

Show answer & worked solution
  1. A. 12\dfrac{1}{2}
  2. B. −12-\dfrac{1}{2}✓ correct
  3. C. 32\dfrac{\sqrt{3}}{2}
  4. D. −32-\dfrac{\sqrt{3}}{2}

sin⁡ ⁣(2π−π6)=−sin⁡π6=−12\sin\!\left(2\pi - \dfrac{\pi}{6}\right) = -\sin\dfrac{\pi}{6} = -\dfrac{1}{2}.

Problem #0554 International

Problem 6 — The Unit Circle

For x=2π3x = \dfrac{2\pi}{3}, the sign of sin⁡x\sin x is:

Show answer & worked solution
  1. A. positive✓ correct
  2. B. negative
  3. C. zero
  4. D. undefined

In the second quadrant (π/2<x<π\pi/2 < x < \pi), sin⁡x>0\sin x > 0.

Problem #0552 International

Problem 7 — The Unit Circle

The value of cos⁡π4\cos\dfrac{\pi}{4} is:

Show answer & worked solution
  1. A. 12\dfrac{1}{2}
  2. B. 22\dfrac{\sqrt{2}}{2}✓ correct
  3. C. 32\dfrac{\sqrt{3}}{2}
  4. D. 11

cos⁡π4=sin⁡π4=22\cos\dfrac{\pi}{4} = \sin\dfrac{\pi}{4} = \dfrac{\sqrt{2}}{2}.

Problem #0946 US SAT

Problem 8 — The Unit Circle

The exact value of cos⁡ ⁣(π3)\cos\!\left(\dfrac{\pi}{3}\right) is:

Show answer & worked solution
  1. A. 32\dfrac{\sqrt{3}}{2}
  2. B. 12\dfrac{1}{2}✓ correct
  3. C. 22\dfrac{\sqrt{2}}{2}
  4. D. 11

π3=60∘\dfrac{\pi}{3} = 60^{\circ}, and cos⁡60∘=12\cos 60^{\circ} = \dfrac{1}{2}.

Problem #0553 International

Problem 9 — The Unit Circle

The value of tan⁡π3\tan\dfrac{\pi}{3} is:

Show answer & worked solution
  1. A. 13\dfrac{1}{\sqrt{3}}
  2. B. 11
  3. C. 3\sqrt{3}✓ correct
  4. D. 32\dfrac{\sqrt{3}}{2}

tan⁡π3=sin⁡(π/3)cos⁡(π/3)=3/21/2=3\tan\dfrac{\pi}{3} = \dfrac{\sin(\pi/3)}{\cos(\pi/3)} = \dfrac{\sqrt{3}/2}{1/2} = \sqrt{3}.

Problem #0551 International

Problem 10 — The Unit Circle

The value of sin⁡π6\sin\dfrac{\pi}{6} is:

Show answer & worked solution
  1. A. 12\dfrac{1}{2}✓ correct
  2. B. 22\dfrac{\sqrt{2}}{2}
  3. C. 32\dfrac{\sqrt{3}}{2}
  4. D. 11

A standard unit-circle value: sin⁡π6=12\sin\dfrac{\pi}{6} = \dfrac{1}{2}.

2 more The Unit Circle questions in the app

Also covered in The Unit Circle practice across every exam.

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