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Trigonometric Functions (Introduction) — High School Algebra 2 practice questions

18 multiple-choice questions on Trigonometric Functions (Introduction), about 12% of the High School Algebra 2 bank. Every one carries a written rationale.

Written and maintained by Nick Burton · last updated 2026-08-22 · how we write and review questions

What this chapter covers

Trigonometric Functions (Introduction) is one of 9 chapters in CoStudy's High School Algebra 2 bank, and it holds 18 of the bank's 150 multiple-choice questions — roughly 12% of the total. That proportion is not arbitrary: chapters follow the certifying body's published exam outline, and the number of questions in each is set by that domain's published weight, so the share of your practice time this chapter takes matches the share of the real exam it accounts for.

Studying by chapter is worth doing once you have a diagnostic score. A single overall percentage tells you whether you are close; it does not tell you which domain is dragging. Working a weak chapter in isolation, and re-testing it in isolation, is the fastest way to move a score that has stalled — and it is why the mock exams in CoStudy report by domain rather than as one number.

Free Trigonometric Functions (Introduction) practice questions

7 questions drawn from this chapter, with the full rationale shown — the controlling principle behind the right answer, and why each wrong option tempts and fails.

Find the inverse of f(x) = 2x + 3:

  1. f⁻¹(x) = 2x - 3
  2. f⁻¹(x) = -2x-3
  3. f⁻¹(x) = (x+3)/2
  4. f⁻¹(x) = (x - 3)/2 (swap x↔y, solve for y: x = 2y+3 → y = (x-3)/2)
  5. f⁻¹(x) = 1/(2x+3)

Answer: D — f⁻¹(x) = (x - 3)/2 (swap x↔y, solve for y: x = 2y+3 → y = (x-3)/2)

Inverse procedure: write y = 2x+3, swap variables (x = 2y+3), solve for y: y = (x-3)/2. Verify: f(f⁻¹(x)) = 2((x-3)/2) + 3 = x-3+3 = x ✓.

Use the change-of-base formula to evaluate log₃(20):

  1. 20/3
  2. 3·log(20)
  3. log(20)/log(3) ≈ 2.727 (change-of-base: log_a(b) = log(b)/log(a) using any base)
  4. 20·log(3)
  5. 17

Answer: C — log(20)/log(3) ≈ 2.727 (change-of-base: log_a(b) = log(b)/log(a) using any base)

Change-of-base: log_a(b) = log_c(b)/log_c(a). Use base 10 or e. log₃(20) = log(20)/log(3) ≈ 1.301/0.477 ≈ 2.727. Enables calculator evaluation of any-base log.

Solve 2 sin(x) = 1 on [0, 2π).

  1. x = π/6 only
  2. x = π/3 and 2π/3
  3. x = π/6 and 5π/6
  4. x = π/2

Answer: C — x = π/6 and 5π/6

sin(x) = 1/2 → reference angle π/6. In [0, 2π), sine is positive in QI and QII: π/6 and π − π/6 = 5π/6. A) Forgot second-quadrant solution. B) Used wrong reference angle. D) Used sin = 1.

Evaluate sin(7π/6).

  1. 1/2
  2. −1/2
  3. √3/2
  4. −√3/2

Answer: B — −1/2

A) Missed the negative sign for the third quadrant. B) 7π/6 = 180° + 30° lies in quadrant III where sine is negative. Reference angle π/6 = 30° gives sin(π/6) = 1/2, so sin(7π/6) = −1/2. C) Confused with cosine reference value. D) Cosine sign error in quadrant III.

Divide: (x³ - 2x² + x - 5) by (x - 1) (using synthetic division)

  1. quotient x² - x with remainder -5
  2. quotient x - 1 with remainder 0
  3. quotient x² + x with remainder 5
  4. quotient x² with remainder 0
  5. quotient x² - x with remainder -5 (synthetic: 1 | 1 -2 1 -5; 1, -1, 0, -5; quotient x² - x, remainder -5)

Answer: E — quotient x² - x with remainder -5 (synthetic: 1 | 1 -2 1 -5; 1, -1, 0, -5; quotient x² - x, remainder -5)

Synthetic division with root c=1: bring down 1; 1·1 = 1, add to -2: -1; -1·1 = -1, add to 1: 0; 0·1 = 0, add to -5: -5. Coefficients 1, -1, 0 → x²-x; remainder -5.

Standard form of a parabola opening right with vertex (2, 3) and 4p = 8:

  1. (y-3)² = 8(x-2)
  2. y² = x
  3. (x-2)² = 8(y-3)
  4. x² = 4y
  5. (y-3)² = 8(x-2) (horizontal parabola opening right: (y-k)² = 4p(x-h))

Answer: E — (y-3)² = 8(x-2) (horizontal parabola opening right: (y-k)² = 4p(x-h))

Conic forms: parabola (y-k)² = 4p(x-h) opens right if 4p > 0. Vertex (h,k) = (2,3), 4p = 8. So (y-3)² = 8(x-2). Distinguish horizontal vs. vertical parabolas.

Solve: x² - 6x + 13 = 0

  1. x = 3 ± 2i (discriminant 36-52 = -16; x = (6 ± √-16)/2 = (6 ± 4i)/2 = 3 ± 2i)
  2. x = 3 only
  3. No solution
  4. x = -3 ± 2
  5. x = 6 ± 13

Answer: A — x = 3 ± 2i (discriminant 36-52 = -16; x = (6 ± √-16)/2 = (6 ± 4i)/2 = 3 ± 2i)

Quadratic formula with negative discriminant gives complex roots. a=1, b=-6, c=13. D = 36 - 52 = -16. x = (6 ± √-16)/2 = (6 ± 4i)/2 = 3 ± 2i. Complex conjugate pair.

Practise the full chapter

These are a sample. The full Trigonometric Functions (Introduction) chapter runs 18 items with per-chapter progress tracking, on the web and in the iOS app.

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