Multiple Angle Trigonometry

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Multiple Angle Trigonometry · Grade 12 · Worksheet 1

1. A Ferris wheel with a diameter of 40 meters completes one full revolution every 2 minutes. The height of a passenger above the ground can be modeled by the function h(t) = 20 + 20sin(πt - π/2), where t is time in minutes after boarding. At what times during the first 4 minutes will the passenger be exactly 30 meters above the ground? Answer: t = 1/3, 5/3, 7/3, 11/3 minutes Solution: This problem involves solving trigonometric equations with multiple angles, which requires finding all solutions within a specified domain. Remember that sine functions repeat every 2π radians, and you'll need to consider the phase shift in the function.
   Full step-by-step solution

   This problem involves solving trigonometric equations with multiple angles, which requires finding all solutions within a specified domain. The key is to use inverse trigonometric functions to find principal solutions, then apply periodicity to find all solutions in the given interval. Remember that sine functions repeat every 2π radians, and you'll need to consider the phase shift in the function.

2. Charlotte is analyzing the path of a laser beam reflecting off a series of mirrors. The beam's angle relative to a reference line, measured in degrees, is given by the equation cos(3θ + 15°) = 0.5, where θ is the time in seconds. Find all times θ in the interval [0°, 360°) that satisfy this equation. Answer: θ = 15°, 105°, 135°, 225°, 255°, 345° Solution: Set up the equation: cos(3θ + 15°) = 0.5. The general solution for cos(u) = 0.5 is u = 60° + 360°n or u = 300° + 360°n, where n is an integer. So, 3θ + 15° = 60° + 360°n or 3θ + 15° = 300° + 360°n.
   Full step-by-step solution

   Step 1: Set up the equation: cos(3θ + 15°) = 0.5. Step 2: The general solution for cos(u) = 0.5 is u = 60° + 360°n or u = 300° + 360°n, where n is an integer. Step 3: So, 3θ + 15° = 60° + 360°n or 3θ + 15° = 300° + 360°n. Step 4: Solve the first equation: 3θ = 45° + 360°n → θ = 15° + 120°n. Step 5: Solve the second equation: 3θ = 285° + 360°n → θ = 95° + 120°n. Step 6: Find all solutions in [0°, 360°): For θ = 15° + 120°n: n=0 → 15°, n=1 → 135°, n=2 → 255°, n=3 → 375° (too large). For θ = 95° + 120°n: n=0 → 95°, n=1 → 215°, n=2 → 335°, n=3 → 455° (too large). Step 7: The solutions are θ = 15°, 95°, 135°, 215°, 255°, 335°. The answer is θ = 15°, 95°, 135°, 215°, 255°, 335°.

3. 2cos²(3x) - 1 = 0 for x ∈ [0, π] Answer: π/6, π/3, π/2, 2π/3, 5π/6 Solution: Step 1: Recognize that 2cos²(3x) - 1 = cos(6x) using the double-angle identity cos(2θ) = 2cos²θ - 1 Step 2: The equation becomes cos(6x) = 0 Step 3: Solve cos(6x) = 0: 6x = π/2 + kπ where k is an integer Step 4: Divide by 6: x = π/12 + kπ/6 Step 5: Find all solutions in [0, π]: When k = 0: x =…
   Full step-by-step solution

   Step 1: Recognize that 2cos²(3x) - 1 = cos(6x) using the double-angle identity cos(2θ) = 2cos²θ - 1 Step 2: The equation becomes cos(6x) = 0 Step 3: Solve cos(6x) = 0: 6x = π/2 + kπ where k is an integer Step 4: Divide by 6: x = π/12 + kπ/6 Step 5: Find all solutions in [0, π]: When k = 0: x = π/12 When k = 1: x = π/12 + π/6 = π/4 When k = 2: x = π/12 + 2π/6 = π/12 + π/3 = 5π/12 When k = 3: x = π/12 + 3π/6 = π/12 + π/2 = 7π/12 When k = 4: x = π/12 + 4π/6 = π/12 + 2π/3 = 9π/12 = 3π/4 When k = 5: x = π/12 + 5π/6 = π/12 + 10π/12 = 11π/12 Step 6: Verify all solutions are in [0, π]: π/12, π/4, 5π/12, 7π/12, 3π/4, 11π/12 Step 7: The solutions are π/12, π/4, 5π/12, 7π/12, 3π/4, 11π/12

4. 2cos(3x) + √2 = 0 for x ∈ [0, 2π] Answer: π/4, 5π/12, 11π/12, 5π/4, 19π/12, 23π/12 Solution: Step 1: Isolate cos(3x) 2cos(3x) + √2 = 0 2cos(3x) = -√2 cos(3x) = -√2/2 Step 2: Find reference angles cos(θ) = -√2/2 when θ = 3π/4 and θ = 5π/4 in [0, 2π] Step 3: Account for periodicity 3x = 3π/4 + 2πk or 3x = 5π/4 + 2πk, where k is an integer Step 4: Solve for x x = π/4 + 2πk/3 or x = 5π/12 +…
   Full step-by-step solution

   Step 1: Isolate cos(3x) 2cos(3x) + √2 = 0 2cos(3x) = -√2 cos(3x) = -√2/2 Step 2: Find reference angles cos(θ) = -√2/2 when θ = 3π/4 and θ = 5π/4 in [0, 2π] Step 3: Account for periodicity 3x = 3π/4 + 2πk or 3x = 5π/4 + 2πk, where k is an integer Step 4: Solve for x x = π/4 + 2πk/3 or x = 5π/12 + 2πk/3 Step 5: Find solutions in [0, 2π] For k = 0: x = π/4, 5π/12 For k = 1: x = π/4 + 2π/3 = 11π/12, 5π/12 + 2π/3 = 13π/12 For k = 2: x = π/4 + 4π/3 = 19π/12, 5π/12 + 4π/3 = 7π/4 Step 6: Verify all solutions are in [0, 2π] Final solutions: π/4, 5π/12, 11π/12, 13π/12, 19π/12, 7π/4 Step 7: Check for duplicates and order π/4 = 3π/12, 5π/12, 11π/12, 13π/12, 19π/12, 7π/4 = 21π/12 All values are between 0 and 2π and distinct.

5. A lighthouse beacon rotates at a constant speed, projecting a beam of light that sweeps across the ocean. The horizontal distance (in kilometers) from the lighthouse to the point where the beam hits a straight shoreline is modeled by the function d(t) = 12 tan(π t / 9), where t is the time in seconds after the beam passes the point directly perpendicular to the shoreline. At what times t in the interval 0 < t < 9 seconds is the beam exactly 12√3 kilometers from the lighthouse along the shoreline? Answer: t = 3 seconds Solution: Set the distance equal to 12√3: 12 tan(π t / 9) = 12√3. Divide both sides by 12: tan(π t / 9) = √3. Recall that tan θ = √3 when θ = π/3 + nπ, where n is any integer (since tangent has period π).
   Full step-by-step solution

   Step 1: Set the distance equal to 12√3: 12 tan(π t / 9) = 12√3. Step 2: Divide both sides by 12: tan(π t / 9) = √3. Step 3: Recall that tan θ = √3 when θ = π/3 + nπ, where n is any integer (since tangent has period π). Step 4: So, π t / 9 = π/3 + nπ. Step 5: Multiply both sides by 9/π: t = 3 + 9n. Step 6: We need t in the interval 0 < t < 9. The only integer n that gives a t in this interval is n = 0, giving t = 3. Step 7: For n = 1, t = 12, which is outside the interval. For n = -1, t = -6, also outside. Step 8: Therefore, the only time in the given interval is t = 3 seconds. The answer is t = 3 seconds.

6. 4sin(3x) - 2 = 0 for x ∈ [0, 2π]. Answer: π/18, 5π/18, 13π/18, 17π/18, 25π/18, 29π/18 Solution: Isolate sin(3x): 4sin(3x) - 2 = 0 → 4sin(3x) = 2 → sin(3x) = 1/2. Find the general solutions for 3x. sin(θ) = 1/2 when θ = π/6 + 2πk or θ = 5π/6 + 2πk, where k is an integer.
   Full step-by-step solution

   Step 1: Isolate sin(3x): 4sin(3x) - 2 = 0 → 4sin(3x) = 2 → sin(3x) = 1/2. Step 2: Find the general solutions for 3x. sin(θ) = 1/2 when θ = π/6 + 2πk or θ = 5π/6 + 2πk, where k is an integer. Step 3: Substitute back: 3x = π/6 + 2πk or 3x = 5π/6 + 2πk. Step 4: Solve for x: x = π/18 + 2πk/3 or x = 5π/18 + 2πk/3. Step 5: Find all solutions in [0, 2π] by trying integer values of k. For x = π/18 + 2πk/3: k=0 → π/18; k=1 → π/18 + 2π/3 = 13π/18; k=2 → π/18 + 4π/3 = 25π/18; k=3 → π/18 + 2π = 37π/18 > 2π (stop). For x = 5π/18 + 2πk/3: k=0 → 5π/18; k=1 → 5π/18 + 2π/3 = 17π/18; k=2 → 5π/18 + 4π/3 = 29π/18; k=3 → 5π/18 + 2π = 41π/18 > 2π (stop). Step 6: List all solutions in [0, 2π]: π/18, 5π/18, 13π/18, 17π/18, 25π/18, 29π/18. Final answer: π/18, 5π/18, 13π/18, 17π/18, 25π/18, 29π/18.

7. An engineer is designing a roller coaster with a track section that follows the path y = 4sin(2θ) + 3cos(2θ), where y represents the height in meters and θ is the angle parameter in radians. To ensure proper banking, she needs to find all angles between 0 and 2π where the track reaches exactly 2 meters above the reference level. Solve the trigonometric equation to determine these critical angles. Answer: π/4, 3π/4, 5π/4, 7π/4 Solution: Trigonometric equations with both sine and cosine terms can be solved by expressing them as a single sine or cosine function using amplitude-phase form.
   Full step-by-step solution

   Trigonometric equations with both sine and cosine terms can be solved by expressing them as a single sine or cosine function using amplitude-phase form. The key insight is recognizing that A sin(x) + B cos(x) can be rewritten as R sin(x + φ) or R cos(x - θ), where R is the amplitude calculated from the coefficients. This transformation simplifies solving by reducing the equation to a basic trigonometric form where standard solution methods apply.