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MCR3U Trigonometric Functions
Grade 11 Unit 6 — practice questions with full solutions

Special angles and exact values, the sine and cosine laws, trigonometric graphs, and proving identities.

Free MCR3U trigonometric functions practice for Grade 11 students in Ontario. These 5 questions cover the same material as a typical unit test or exam review on this topic, and every one comes with a complete worked solution — not just an answer key. Useful whether you are preparing for a MCR3U unit test, catching up on a lesson, or reviewing before the final exam.
MCR3U · Grade 11 Trigonometric Functions 4 core 1 challenge
1Core
11√245°45°1√3260°30°45°-45°-90°30°-60°-90°
Use the special triangles to state exact values for (a) sin 45°, (b) cos 30°, (c) tan 60°, and (d) sin 30°.
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  1. (a) In the 45-45-90 triangle, sin 45° = opposite ÷ hypotenuse = 1 ÷ √2, usually written √2⁄2.
  2. (b) In the 30-60-90 triangle, the side adjacent to 30° is √3 and the hypotenuse is 2, so cos 30° = √3⁄2.
  3. (c) tan 60° = opposite ÷ adjacent = √3 ÷ 1 = √3.
  4. (d) sin 30° = opposite ÷ hypotenuse = 1 ÷ 2.
Final answer(a) √2⁄2   (b) √3⁄2   (c) √3   (d) 1⁄2
2Core
4590135180225270315360-4-3-2-11234xymax
From the graph, determine (a) the amplitude, (b) the period, (c) the equation in the form y = a sin(kx), and (d) the number of complete cycles between 0° and 720°.
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  1. (a) The curve runs from −3 to +3, so amplitude = (max − min) ÷ 2 = 3.
  2. (b) One full cycle finishes at 180°, so the period is 180°.
  3. (c) Period = 360° ÷ k, so 180 = 360 ÷ k gives k = 2.
  4. The equation is y = 3 sin(2x).
  5. (d) 720° ÷ 180° = 4 complete cycles.
Final answer(a) 3   (b) 180°   (c) y = 3 sin(2x)   (d) 4 cycles
3Core
In triangle ABC, a = 11 cm, b = 14 cm and c = 17 cm. Find angle C to one decimal place.
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  1. Three sides are known, so rearrange the cosine law to find an angle.
  2. cos C = (a² + b² − c²) ÷ (2ab).
  3. cos C = (121 + 196 − 289) ÷ (2 × 11 × 14) = 28 ÷ 308 ≈ 0.090909.
  4. C = cos⁻¹(0.090909) ≈ 84.8°.
Final answerC ≈ 84.8°
4Core
Prove the identity:   (1 − cos²θ) ÷ (sin θ cos θ) = tan θ
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  1. Work with the left side only. Use the Pythagorean identity sin²θ + cos²θ = 1.
  2. Rearranged, 1 − cos²θ = sin²θ.
  3. Substitute: sin²θ ÷ (sin θ cos θ).
  4. Cancel one factor of sin θ: sin θ ÷ cos θ.
  5. By definition that is tan θ, which matches the right side. ∎
Final answerLS = sin²θ ÷ (sinθ cosθ) = sinθ ÷ cosθ = tanθ = RS
5Challenge
Solve for 0° ≤ θ ≤ 360°:   2 sin²θ − sin θ − 1 = 0
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  1. Treat this as a quadratic in sin θ. Let s = sin θ: 2s² − s − 1 = 0.
  2. Factor: (2s + 1)(s − 1) = 0, so s = −½ or s = 1.
  3. For sin θ = 1: θ = 90°.
  4. For sin θ = −½: the related acute angle is 30°, and sine is negative in quadrants 3 and 4.
  5. That gives θ = 180° + 30° = 210° and θ = 360° − 30° = 330°.
Final answerθ = 90°, 210°, 330°

MCR3U Trigonometric Functions — common questions

Short answers to the things students ask most about this unit.

What are the special triangles used for?

They give exact trigonometric values for 30°, 45° and 60° without a calculator — the values most often required in exact-answer questions.

How do you prove a trigonometric identity?

Work on one side only, usually the more complicated one, and use known identities such as sin²θ + cos²θ = 1 until it matches the other side.

More MCR3U practice — grade 11 functions

MCR3U Functions & TransformationsMCR3U Quadratic FunctionsMCR3U Exponential FunctionsMCR3U Sequences & SeriesAll MCR3U units
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