Solutions — Trigonometric Graphs and Exact Values

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1. Q1 — Basic exact values

   (a) sin(π/6) = 1/2   (b) cos(π/4) = √2/2   (c) tan(π/3) = √3   (d) sin(π/2) = 1   (e) cos(0) = 1   (f) tan(0) = 0

2. Q2 — Related angles using ASTC

   (a) 2π/3 is in Q2, ref angle π/3, sin positive: sin(2π/3) = √3/2.   (b) 3π/4 is in Q2, ref angle π/4, cos negative: cos(3π/4) = −√2/2.   (c) 5π/6 is in Q2, ref angle π/6, tan negative: tan(5π/6) = −1/√3 = −√3/3.   (d) 7π/6 is in Q3, ref angle π/6, sin negative: sin(7π/6) = −1/2.   (e) 5π/3 is in Q4, ref angle π/3, cos positive: cos(5π/3) = 1/2.   (f) 7π/4 is in Q4, ref angle π/4, tan negative: tan(7π/4) = −1.

3. Q3 — Period, amplitude, domain and range

   (a) y=sin(x): Period 2π, Amplitude 1, Domain ℝ, Range [−1,1].   (b) y=cos(x): Period 2π, Amplitude 1, Domain ℝ, Range [−1,1].   (c) y=tan(x): Period π, no amplitude (unbounded), Domain x≠π/2+nπ, Range ℝ.   (d) y=3sin(x): Period 2π, Amplitude 3, Domain ℝ, Range [−3,3].   (e) y=cos(2x): Period π, Amplitude 1, Domain ℝ, Range [−1,1].

4. Q4 — x-values for sin(x) on [0, 2π]

   (a) sin(x)=0: x = 0, π, 2π.   (b) sin(x)=1: x = π/2.   (c) sin(x)=−1: x = 3π/2.   (d) sin(x)=1/2: ref angle π/6; x = π/6 (Q1) and 5π/6 (Q2).   (e) sin(x)=−√3/2: ref angle π/3; sin negative in Q3 and Q4: x = 4π/3 (Q3) and 5π/3 (Q4).

5. Q5 — Reading and interpreting trig graphs

   (a) cos(x)=0: x=π/2, 3π/2. cos(x)=1: x=0, 2π. cos(x)=−1: x=π.   (b) y=cos(x) is y=sin(x) shifted left by π/2: cos(x) = sin(x + π/2).   (c) Vertical asymptotes of y=tan(x) on [−π, 2π]: x = −π/2, π/2, 3π/2.   (d) sin(x) starts at y=0; cos(x) starts at y=1.

6. Q6 — Symmetry properties

   (a) sin(5π/6) = sin(π − π/6) = sin(π/6) = 1/2.   (b) cos(2π/3) = cos(π − π/3) = −cos(π/3) = −1/2.   (c) sin(−π/4) = −sin(π/4) = −√2/2.   (d) cos(−5π/6) = cos(5π/6) = −√3/2 (even function; cos(5π/6) is in Q2 with ref angle π/6).

7. Q7 — Exact values in expressions

   (a) sin²(π/4)+cos²(π/4) = (√2/2)²+(√2/2)² = 1/2+1/2 = 1. This confirms the Pythagorean identity sin²x+cos²x=1.   (b) sin(π/6)cos(π/3)+cos(π/6)sin(π/3) = (1/2)(1/2)+(√3/2)(√3/2) = 1/4+3/4 = 1. This is the addition formula sin(π/6+π/3) = sin(π/2) = 1.   (c) (√3/2)²+(1/2)² = 3/4+1/4 = 1 ✓.   (d) (tan(π/4))²+1 = 1+1 = 2. This confirms the identity tan²x+1 = sec²x (sec(π/4) = √2, so sec²(π/4)=2 ✓).

8. Q8 — Comparing functions over domains

   (a) Period = 2π; over [0, 6π]: 6π/(2π) = 3 full cycles.   (b) y=cos(x)=0 every π/2+nπ. On [0, 4π]: x = π/2, 3π/2, 5π/2, 7π/2. That’s 4 times. x-values: π/2, 3π/2, 5π/2, 7π/2.   (c) Asymptotes of tan(x) occur every π, starting at π/2. On [0, 3π]: x=π/2, 3π/2, 5π/2. That’s 3 asymptotes.   (d) Gradient of sin(x) is zero at the turning points: maximum at x=π/2 and minimum at x=3π/2.

9. Q9 — Ferris wheel problem

   (a) h = 12+10sin(π/2) = 12+10(1) = 22 m. This is the top of the Ferris wheel (maximum height).   (b) h = 12+10sin(π) = 12+10(0) = 12 m. The rider is at the same height as the centre (neither top nor bottom).   (c) Minimum when sin(θ) = −1, i.e. θ = 3π/2. Min height = 12+10(−1) = 2 m.   (d) h(π/6) = 12+10sin(π/6) = 12+10(1/2) = 12+5 = 17 m. h(π/3) = 12+10sin(π/3) = 12+10(√3/2) = 12+5√3 = (12+5√3) m ≈ 20.66 m.

10. Q10 — Proving exact values

    (a) 30°-60°-90° triangle, hypotenuse 2: opposite to 30° = 1, adjacent = √3. sin(30°) = 1/2 ✓, cos(30°) = √3/2 ✓.   (b) 45°-45°-90° triangle, legs = 1, hypotenuse = √2: sin(45°) = 1/√2 = √2/2 ✓, cos(45°) = 1/√2 = √2/2 ✓.   (c) sin²x = 1−cos²x = 1−1/9 = 8/9. sin(x) = √(8/9) = 2√2/3 (positive, Q1).   (d) tan(x) = sin(x)/cos(x) = (2√2/3)/(1/3) = 2√2.