Solutions — Transformed Trigonometric Functions and Equations
-
Q1 — Parameters
(a) Amplitude=4, period=2π, phase shift=0, vertical shift=0
(b) Amplitude=1, period=2π/3, phase shift=0, vertical shift=0
(c) Rewrite: y=2sin(x−π/4)+3. Amplitude=2, period=2π, phase shift=π/4 right, vertical shift=3 up
(d) Rewrite: y=−3cos(2(x+π/4))−1. Amplitude=3, period=π, phase shift=π/4 left, vertical shift=1 down, reflected.
-
Q2 — Max, min and midline
(a) Midline y=−2, max=3, min=−7
(b) a=−3 (reflected), k=4. max=4+3=7, min=4−3=1. Midline y=4
(c) Midline y=1, max=3, min=−1
(d) Midline y=−3, max=1, min=−7
-
Q3 — Basic equations over [0, 2π]
(a) sin(x)=1/2: ref angle π/6. Q1,Q2: x=π/6, 5π/6
(b) cos(x)=−1/2: ref angle π/3. Q2,Q3: x=2π/3, 4π/3
(c) tan(x)=1: ref angle π/4. Q1,Q3: x=π/4, 5π/4
(d) sin(x)=−√3/2: ref angle π/3. Q3,Q4: x=4π/3, 5π/3
-
Q4 — Writing equations
(a) y=3sin(x)
(b) Period=π → b=2. y=sin(2x)
(c) Period=4π → b=1/2. y=2sin(x/2)+5
(d) Reflected, shift right π/2. y=−sin(x−π/2)
-
Q5 — Solving with rearrangement
(a) sin(x)=√3/2. Q1,Q2: x=π/3, 2π/3
(b) cos(x)=−1/2. Q2,Q3: x=2π/3, 4π/3
(c) tan(x)=1/√3. Q1,Q3: x=π/6, 7π/6
(d) sin(x)=−1/2. Q3,Q4: x=7π/6, 11π/6
-
Q6 — Equations with b≠1
(a) sin(2x)=1/2. Let u=2x, u∈[0,4π]. sin(u)=1/2: u=π/6, 5π/6, π/6+2π=13π/6, 5π/6+2π=17π/6. x=u/2: x=π/12, 5π/12, 13π/12, 17π/12
(b) cos(2x)=−√2/2. u=2x∈[0,2π]. cos(u)=−√2/2: u=3π/4, 5π/4. x=3π/8, 5π/8
(c) sin(x/2)=1/2. u=x/2, u∈[0,2π]. sin(u)=1/2: u=π/6, 5π/6. x=2u: x=π/3, 5π/3
-
Q7 — Sketching
(a) y=2sin(x)−1: midline y=−1, max=1 at x=π/2, min=−3 at x=3π/2, x-intercepts where sin(x)=1/2: x=π/6 and 5π/6.
(b) y=3cos(2x)+1: midline y=1, amplitude=3, period=π. Max=4 at x=0, π; min=−2 at x=π/2.
-
Q8 — Equation from description
(a) a=4, b=2π/π=2, k=2. y=4sin(2x)+2
(b) k=(5+(−1))/2=2, a=(5−(−1))/2=3. Period=4π → b=1/2. y=3cos(x/2)+2
(c) Reflected (starts going down), amplitude=3, period=2π. y=−3sin(x)
-
Q9 — Tide modelling
(a) Amplitude=3 m (tidal range is 6 m from midline). Period=2π/(π/6)=12 hours (two high/low tides per day). Midline=5 m (average tide height).
(b) Max = 5+3 = 8 m. sin(πt/6)=1 when πt/6=π/2 → t=3 hours after midnight = 3:00 am.
(c) 6.5=3sin(πt/6)+5 → sin(πt/6)=0.5. πt/6=π/6 or 5π/6 → t=1 or t=5. Adding period (12 h): t=13 and t=17. All times: 1:00 am, 5:00 am, 1:00 pm, 5:00 pm.
-
Q10 — Ferris wheel
(a) Diameter=20m → radius=10m. Lowest point=2m → centre=12m. k=12, a=−10 (starts at bottom → cosine reflected). Period=40s → b=2π/40=π/20. h = −10cos(πt/20) + 12
(b) t=10: h=−10cos(π/2)+12=0+12=12 m. t=25: h=−10cos(5π/4)+12=−10(−√2/2)+12=5√2+12≈19.1 m.
(c) 17=−10cos(πt/20)+12 → cos(πt/20)=−0.5. πt/20=2π/3 or 4π/3 → t=40/3≈13.3s or t=80/3≈26.7s. In first 40s: t ≈ 13.3 s and t ≈ 26.7 s.