Practice Maths

Topic Review — Trigonometric Functions — Solutions

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★ U1T4 — Trigonometric Functions Review

Covers: Radian measure • Arcs & sectors • Exact values • Trig graphs • Transformations • Solving equations

  1. Q1 — Radian conversion

    Convert each angle to radians (exact): (a) 60°   (b) 270°   (c) 135°   (d) 210°

    (a) 60 × π/180 = π/3

    (b) 270 × π/180 = 3π/2

    (c) 135 × π/180 = 3π/4

    (d) 210 × π/180 = 7π/6

  2. Q2 — Radian to degrees

    Convert to degrees: (a) π/4   (b) 5π/6   (c) 4π/3   (d) 11π/6

    (a) 45°   (b) 150°   (c) 240°   (d) 330°

  3. Q3 — Arc length and sector area

    A sector has radius 8 cm and angle 5π/6 radians. Find: (a) arc length   (b) sector area.

    (a) l = rθ = 8 × 5π/6 = 20π/3 cm ≈ 20.9 cm

    (b) A = ½r2θ = ½×64×5π/6 = 80π/3 cm2 ≈ 83.8 cm2

  4. Q4 — Exact values

    Evaluate exactly: (a) sin(π/6)   (b) cos(2π/3)   (c) tan(π/4)   (d) sin(5π/4)   (e) cos(11π/6)

    (a) 1/2   (b) −1/2   (c) 1   (d) −√2/2   (e) √3/2

  5. Q5 — Period and amplitude

    State the amplitude and period of: (a) y = 5cos(3x)   (b) y = −2sin(x/2)   (c) y = 4tan(2x)

    (a) Amplitude=5, period=2π/3

    (b) Amplitude=2, period=4π

    (c) Amplitude: undefined for tan, period=π/2

  6. Q6 — Key features of a transformed function

    For y = 3sin(2x − π/3) + 2, state the amplitude, period, phase shift, vertical shift, maximum and minimum values.

    Rewrite: y = 3sin(2(x−π/6))+2. Amplitude=3. Period=π. Phase shift=π/6 right. Vertical shift=2 up. Max=5, min=−1.

  7. Q7 — Solving basic equations

    Solve for x ∈ [0, 2π]: (a) cos(x) = √3/2   (b) tan(x) = −√3   (c) 2sin(x) + 1 = 0

    (a) x = π/6, 11π/6

    (b) ref angle=π/3, tan negative in Q2,Q4: x = 2π/3, 5π/3

    (c) sin(x)=−1/2, Q3,Q4: x = 7π/6, 11π/6

  8. Q8 — Equations with double angle

    Solve sin(2x) = √3/2 for x ∈ [0, 2π].

    u=2x ∈[0,4π]. sin(u)=√3/2: u=π/3, 2π/3, 7π/3, 8π/3. x=u/2: x=π/6, π/3, 7π/6, 4π/3

  9. Q9 — Find the equation from features

    A sine function has amplitude 4, period 3π, midline y = −1, and phase shift π/4 to the right. Write its equation.

    a=4, period=3π → b=2/3, h=π/4, k=−1. y = 4sin((2/3)(x − π/4)) − 1

  10. Q10 — Application: sprinkler

    A rotating sprinkler waters a circular region. The water reaches 5 m. The sprinkler turns through an angle of 2π/3 radians. Find: (a) the arc length of the watered edge   (b) the area of the sector watered.

    (a) l = 5 × 2π/3 = 10π/3 ≈ 10.5 m

    (b) A = ½×25×2π/3 = 25π/3 ≈ 26.2 m2

  11. Q11 — Application: daylight hours model

    The number of daylight hours in Brisbane is modelled by D = 1.5sin(2π(t − 81)/365) + 12.1, where t is the day of the year. Find: (a) the maximum and minimum daylight hours   (b) the day of maximum daylight   (c) when daylight hours equal 12.

    (a) Max = 12.1+1.5 = 13.6 hours. Min = 12.1−1.5 = 10.6 hours.

    (b) sin=1 when 2π(t−81)/365=π/2 → t−81=365/4=91.25 → t=172. Day 172 = approximately 21 June (summer solstice).

    (c) 12 = 1.5sin(…)+12.1 → sin(…) = −0.0667 → t ≈ 81±1.2 → approximately day 80 (21 March) and day 82 (23 March), near the autumn equinox.