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The Unit Circle

12 practice questions with full worked solutions. Free, no account needed.

Problems & worked solutions

Problem #0552 International

Problem 1 — The Unit Circle

The value of cos⁡π4 is:

Show answer & worked solution
  1. A. 12
  2. B. 22✓ correct
  3. C. 32
  4. D. 1

cos⁡π4=sin⁡π4=22.

Problem #0554 International

Problem 2 — The Unit Circle

For x=2π3, the sign of 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), sin⁡x>0.

Problem #0551 International

Problem 3 — The Unit Circle

The value of sin⁡π6 is:

Show answer & worked solution
  1. A. 12✓ correct
  2. B. 22
  3. C. 32
  4. D. 1

A standard unit-circle value: sin⁡π6=12.

Problem #0553 International

Problem 4 — The Unit Circle

The value of tan⁡π3 is:

Show answer & worked solution
  1. A. 13
  2. B. 1
  3. C. 3✓ correct
  4. D. 32

tan⁡π3=sin⁡(π/3)cos⁡(π/3)=3/21/2=3.

Problem #0945 US SAT

Problem 5 — Trigonometric Functions

The exact value of sin⁡ ⁣(π6) is:

Show answer & worked solution
  1. A. 12✓ correct
  2. B. 22
  3. C. 32
  4. D. 1

π6=30∘, and sin⁡30∘=12.

Problem #0946 US SAT

Problem 6 — Notable Values

The exact value of cos⁡ ⁣(π3) is:

Show answer & worked solution
  1. A. 32
  2. B. 12✓ correct
  3. C. 22
  4. D. 1

π3=60∘, and cos⁡60∘=12.

Problem #0556 International

Problem 7 — The Unit Circle

The value of cos⁡ ⁣(−π3) is:

Show answer & worked solution
  1. A. 12✓ correct
  2. B. −12
  3. C. 32
  4. D. −32

cos⁡ ⁣(−π3)=cos⁡π3=12.

Problem #0559 International

Problem 8 — The Unit Circle

The number of solutions of sin⁡x=−12 on [0,2π) is:

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

sin⁡x=−12 at x=7π6 (Q3) and x=11π6 (Q4). Two solutions on [0,2π).

Problem #0557 International

Problem 9 — The Unit Circle

The point on the unit circle associated with x=7π4 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), which is the fourth quadrant.

Problem #0555 International

Problem 10 — The Unit Circle

The value of sin⁡11π6 is:

Show answer & worked solution
  1. A. 12
  2. B. −12✓ correct
  3. C. 32
  4. D. −32

sin⁡ ⁣(2π−π6)=−sin⁡π6=−12.

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