3.08E - Solving Sinusoidal Equations | High School

University:
High School
Course:
High School Physics
Academic year:

2024
Views:

272

Pages:

12
Author:

1 1

Chapter 3: Trigonometry Lessons 3.08E Solving Sinusoidal Equations Logic Period Solving for 𝑦 given 𝑥 • 2 𝜋 𝑓 𝑥 = − sin 𝑥−3 3 4 +7 2 𝜋 𝑓 5 = − sin 5−3 3 4 +7 Find 𝑓 5 2 𝜋 𝑓 5 = − sin 2 3 4 +7 2 𝜋 𝑓 5 = − sin +7 3 2 2 𝑓 5 = − (1) + 7 3 2 𝑓 5 =− +7 3 𝑓 5 = 19 3 Solving for Zeros of Sinusoidal Functions Steps for Solving for Zeros of Sinusoidal Function 1. Identify the Period of the Function 𝑃 = 2. 3. 4. Isolate base function: sin 𝑏 𝑥 − ℎ 2𝜋 𝑏 or cos 𝑏 𝑥 − ℎ Let θ = 𝑏 𝑥 − ℎ Solve for 𝜃1 and 𝜃2 using: - Unit Circle for both, or calculator for 𝜃1 and formulas for 𝜃2 5. Solve for 𝑥 − values using 𝜃1 = 𝑏 𝑥1 − ℎ and 𝜃2 = 𝑏(𝑥2 − ℎ) 6. Extend over domain, or generalize using period 𝑃𝑛 𝑥1 + 𝑃𝑛, 𝑥2 + 𝑃𝑛 𝑤ℎ𝑒𝑟𝑒 𝑛 ∈ ℤ Example 1: 𝑓 𝑥 = 2 cos(𝜋𝑥) + 2 𝑉 3.4 2𝜋 𝑃= =2 𝜋 0 = 2 cos(𝜋𝑥) + 2 2 − = cos(𝜋𝑥) 2 θ = 𝜋𝑥 2 − = cos(θ) 2 3𝜋 𝜃1 = 4 5𝜋 𝜃2 = 4 0 −0.6 𝜃1 = 𝑏 𝑥1 − ℎ and 𝜃2 = 𝑏(𝑥2 − ℎ) 3𝜋 5𝜋 = 𝜋𝑥1 and = 𝜋𝑥2 4 4 3 5 = 𝑥1 and = 𝑥2 4 4 3 5 + 2𝑛, + 2𝑛 𝑤ℎ𝑒𝑟𝑒 𝑛 ∈ ℤ 4 4 Example 2: 𝑓 𝑥 = 3 sin 𝜋 𝑥 − 1 6 +3 𝑉 0 2𝜋 𝑃= =2 𝜋 0 = 3 sin 𝜋 𝑥 − 1 −1 = sin 𝜋 𝑥 − 1 θ=𝜋 𝑥−1 +3 𝜃 =𝑏 𝑥−ℎ 𝜋 − =𝜋 𝑥−1 2 1 2 =𝑥 −1 = sin(θ) 𝜋 𝜃=− 2 1 + 2𝑛 , 𝑛 ∈ ℤ 2 Example 3: 𝑓 𝑥 = −3 sin(𝑥) + 2 5 𝑉 2𝜋 𝑃= = 2𝜋 1 0 = −3 sin 𝑥 + 2 𝜃1 ≈ 0.7297 2 −1 𝜃2 = 𝜋 − 𝜃1 𝜃2 ≈ 𝜋 − 0.7297 2 = sin(𝑥) 3 𝜃2 ≈ 2.4119 θ=𝑥 2 = sin(θ) 3 θ= sin−1 2 3 𝑥1 ≈ 0.7297 and 𝑥2 ≈ 2.4119 0.7297 + 2𝜋𝑛, 2.4119 + 2𝜋𝑛 𝑤ℎ𝑒𝑟𝑒 𝑛 ∈ ℤ 0 𝑉 Example 4: 𝑓 𝑥 = −3 cos 𝑥 − 1 1 + 2 3.5 2𝜋 𝑃= = 2𝜋 1 1 0 = −3 cos 𝑥 − 1 + 2 1 = cos(𝑥 − 1) 6 θ=𝑥−1 1 = cos(θ) 6 θ= cos −1 1 6 𝜃1 ≈ 1.4033 −2.5 𝜃2 = −𝜃1 𝜃2 ≈ −1.4033 𝑥1 − 1 ≈ 1.4033 and 𝑥2 − 1 ≈ −1.4033 𝑥1 ≈ 2.4033 and 𝑥2 ≈ −0.4033 −0.4033 + 2𝜋𝑛, 2.4033 + 2𝜋𝑛 𝑤ℎ𝑒𝑟𝑒 𝑛 ∈ ℤ Example: −2 = 5 cos 𝑥 2𝜋 𝑃= = 2𝜋 1 𝑉 −2 3𝜋 0 = cos(𝑥 + ) 2 3𝜋 θ=𝑥+ 2 0 = cos(θ) 3π θ2 = 2 3 −2 over [0,4π] 3𝜋 −2 = 5 cos 𝑥 + −2 2 π θ1 = 2 3𝜋 + 2 −7 3𝜋 𝜋 𝑥1 + = 2 2 𝑥1 = −𝜋 3𝜋 3𝜋 𝑥2 + = 2 2 and 𝑥2 = 0 But also need solutions up to 4π 0, 𝜋, 2𝜋, 3𝜋, 4𝜋 HW Visions 2 P118: 1ace, 2ace, 5bcd A = (3.6-2.4)/2 = 0.6 m a = -0.6 m k = (3.6+2.4)/2 = 3.0 m P = 12 -> b = 2π/12 = π/6 Vertex = min -> cosine opens up h = 3.5 f(x) = -0.6 cos (π/6(x – 3.5)) + 3 2.7 = -0.6 cos (π/6(x – 3.5)) + 3 -0.3 = -0.6 cos (π/6(x – 3.5)) 1/2 = cos (π/6(x – 3.5)) Two answers! π/3 and 5π/3 (quad I and IV have positive cosine) A = (3.6-2.4)/2 = 0.6 m a = -0.6 m k = (3.6+2.4)/2 = 3.0 m P = 12 -> b = 2π/12 = π/6 Vertex = min -> cosine opens up h = 3.5 1: π/3 = π/6(x – 3.5) 2: 5π/3 = π/6(x – 3.5) 2 = x – 3.5 10 = x – 3.5 x = 5.5 hrs (5:30 am) But also!!! 5:30 pm and 1:30 am (P = 12h) x = 13.5 hrs (1:30 pm) 1:30 am 5:30 am 1:30 pm 5:30 pm