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MHF4U Trigonometric Functions
Grade 12 Unit 5 — practice questions with full solutions

Radians, transformations of sine and cosine, identities, and solving trigonometric equations over a given interval.

MHF4U · Grade 12 Trigonometric Functions 4 core 1 challenge
1Core
4590135180225270315360-3-2-112345xymidline y = 1(45°, 3)(135°, -1)
From the transformed sine curve shown, determine (a) the amplitude, (b) the period, (c) the equation of the midline, and (d) the equation in the form y = a sin(kx) + c.
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  1. (a) The curve runs from −1 up to 3, so amplitude = (3 − (−1)) ÷ 2 = 2.
  2. (b) One complete cycle finishes at 180°, so the period is 180°.
  3. (c) The midline sits halfway between max and min: (3 + (−1)) ÷ 2 = 1, so y = 1.
  4. (d) Period = 360° ÷ k, so k = 360 ÷ 180 = 2.
  5. Putting it together: y = 2 sin(2x) + 1.
Final answer(a) 2   (b) 180°   (c) y = 1   (d) y = 2 sin(2x) + 1
2Core
Convert: (a) 150° to radians in exact form, (b) 5π⁄6 radians to degrees, (c) 2 radians to degrees to one decimal place.
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  1. (a) Multiply by π ÷ 180: 150 × π ÷ 180 = 150π⁄180, which reduces to 5π⁄6.
  2. (b) Multiply by 180 ÷ π: 5π⁄6 × 180⁄π = 5 × 30 = 150°.
  3. (c) 2 × 180 ÷ π ≈ 360 ÷ 3.14159 ≈ 114.6°.
  4. Parts (a) and (b) confirm each other — they are inverse conversions. ✔
Final answer(a) 5π⁄6   (b) 150°   (c) ≈ 114.6°
3Core
State the amplitude, period, phase shift and vertical shift of y = −3 cos(2(x − π⁄4)) + 5.
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  1. Amplitude is |a| = 3; the negative sign reflects the curve in the midline rather than changing amplitude.
  2. Period = 2π ÷ |k| = 2π ÷ 2 = π.
  3. The bracket (x − π⁄4) gives a phase shift of π⁄4 to the right.
  4. The +5 shifts the whole curve up 5, so the midline is y = 5.
Final answeramplitude 3 (reflected), period π, shift π⁄4 right, up 5
4Core
Prove:   (1 + cos x)⁄sin x + sin x⁄(1 + cos x) = 2 csc x
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  1. Work on the left side with common denominator sin x(1 + cos x).
  2. Numerator becomes (1 + cos x)² + sin²x.
  3. Expand: 1 + 2cos x + cos²x + sin²x.
  4. Since cos²x + sin²x = 1, this simplifies to 2 + 2cos x = 2(1 + cos x).
  5. So the fraction is 2(1 + cos x) ÷ [sin x(1 + cos x)] = 2 ÷ sin x = 2 csc x. ∎
Final answerLS simplifies to 2 ÷ sin x = 2 csc x = RS
5Challenge
Solve for 0 ≤ x ≤ 2π:   2 cos²x + cos x − 1 = 0
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  1. Treat as a quadratic in cos x. Let c = cos x: 2c² + c − 1 = 0.
  2. Factor: (2c − 1)(c + 1) = 0, so c = ½ or c = −1.
  3. For cos x = ½: cosine is positive in quadrants 1 and 4, giving x = π⁄3 and x = 5π⁄3.
  4. For cos x = −1: this happens at x = π.
  5. All three lie within the required interval.
Final answerx = π⁄3, π, 5π⁄3

MHF4U Trigonometric Functions — common questions

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

How do you convert between degrees and radians?

Multiply degrees by π/180 to get radians, or multiply radians by 180/π to get degrees.

How do you find the period of a transformed sine function?

Divide the normal period by the size of k. In degrees that is 360 ÷ |k|; in radians it is 2π ÷ |k|.

More MHF4U practice — grade 12 advanced functions

MHF4U Polynomial FunctionsMHF4U Polynomial Equations & InequalitiesMHF4U Rational FunctionsMHF4U Exponential & Logarithmic FunctionsAll MHF4U units
← Unit 4: Exponential & Logarithmic FunctionsUnit 6: Rates of Change & Combining Functions →
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