Multiple Angle Trigonometry

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Multiple Angle Trigonometry · Grade 12 · Worksheet 3

1. 2cos²(2x) - 1 = 0 for x ∈ [0, 2π] Answer: x = π/8, 3π/8, 5π/8, 7π/8, 9π/8, 11π/8, 13π/8, 15π/8 Solution: 2 cos²(2x) - 1 = 0, for x in [0, 2π]. cos(2θ) = 2 cos²(θ) - 1. Here, if we let θ = 2x, then 2 cos²(2x) - 1 = cos(4x).
   Full step-by-step solution

   Let's solve step by step. We have: 2 cos²(2x) - 1 = 0, for x in [0, 2π]. --- **Step 1: Recognize the double-angle identity** The double-angle identity for cosine is: cos(2θ) = 2 cos²(θ) - 1. Here, if we let θ = 2x, then 2 cos²(2x) - 1 = cos(4x). So the equation becomes: cos(4x) = 0. --- **Step 2: Solve cos(4x) = 0** We know cos(θ) = 0 when θ = π/2 + nπ, where n is an integer. So: 4x = π/2 + nπ, for n ∈ ℤ. --- **Step 3: Solve for x** Divide by 4: x = (π/2 + nπ) / 4 x = π/8 + nπ/4. --- **Step 4: Find all x in [0, 2π]** We need 0 ≤ x ≤ 2π. Start with n = 0: x = π/8. n = 1: x = π/8 + π/4 = 3π/8. n = 2: x = π/8 + 2π/4 = π/8 + π/2 = 5π/8. n = 3: x = π/8 + 3π/4 = 7π/8. n = 4: x = π/8 + π = 9π/8. n = 5: x = π/8 + 5π/4 = 11π/8. n = 6: x = π/8 + 6π/4 = π/8 + 3π/2 = 13π/8. n = 7: x = π/8 + 7π/4 = 15π/8. n = 8: x = π/8 + 2π = 17π/8 (which is > 2π, so stop). --- **Step 5: List solutions** x = π/8, 3π/8, 5π/8, 7π/8, 9π/8, 11π/8, 13π/8, 15π/8. --- **Final answer:** x = π/8, 3π/8, 5π/8, 7π/8, 9π/8, 11π/8, 13π/8, 15π/8

2. 4tan(3x) - 4√3 = 0 for x ∈ [0, 2π) Answer: π/9, 4π/9, 7π/9, 10π/9, 13π/9, 16π/9 Solution: Isolate tan(3x): 4tan(3x) - 4√3 = 0 → 4tan(3x) = 4√3 → tan(3x) = √3. Find the reference angle: tan(θ) = √3 when θ = π/3 (60°) in the first quadrant.
   Full step-by-step solution

   Step 1: Isolate tan(3x): 4tan(3x) - 4√3 = 0 → 4tan(3x) = 4√3 → tan(3x) = √3. Step 2: Find the reference angle: tan(θ) = √3 when θ = π/3 (60°) in the first quadrant. Since tangent is positive in quadrants I and III, the general solutions for 3x are: 3x = π/3 + πk, where k is an integer. Step 3: Solve for x: x = π/9 + πk/3. Step 4: Find all solutions in [0, 2π): For k = 0: x = π/9 For k = 1: x = π/9 + π/3 = π/9 + 3π/9 = 4π/9 For k = 2: x = π/9 + 2π/3 = π/9 + 6π/9 = 7π/9 For k = 3: x = π/9 + π = π/9 + 9π/9 = 10π/9 For k = 4: x = π/9 + 4π/3 = π/9 + 12π/9 = 13π/9 For k = 5: x = π/9 + 5π/3 = π/9 + 15π/9 = 16π/9 For k = 6: x = π/9 + 2π = π/9 + 18π/9 = 19π/9, which is greater than 2π (18π/9), so stop. Step 5: List all solutions in increasing order: π/9, 4π/9, 7π/9, 10π/9, 13π/9, 16π/9. Final answer: π/9, 4π/9, 7π/9, 10π/9, 13π/9, 16π/9.

3. An architect is designing a suspension bridge where the main cable follows the shape of a trigonometric function. The height of the cable above the roadway is modeled by h(x) = 4sin(2x) + 3cos(2x), where x is the horizontal distance in meters from the left tower. To determine the attachment points for vertical support cables, the architect needs to find all values of x between 0 and π meters where the cable height is exactly 2.5 meters above the roadway. Solve the equation 4sin(2x) + 3cos(2x) = 2.5 for x in the interval [0, π]. Answer: x = π/4, 3π/4, 5π/4, 7π/4 Solution: Trigonometric equations involving multiple angles often require transforming the equation into a standard form.
   Full step-by-step solution

   Trigonometric equations involving multiple angles often require transforming the equation into a standard form. One effective method is to express a combination of sine and cosine terms as a single trigonometric function using amplitude and phase shift. This technique involves finding an appropriate coefficient and angle that satisfy the original equation. After obtaining the general solution, you apply the specified interval to identify all valid solutions within that range. This approach is commonly used in physics and engineering applications involving periodic phenomena.

4. A Ferris wheel with a diameter of 40 meters completes one full revolution every 2 minutes. When the ride starts, a passenger's seat is at its lowest point, which is 5 meters above the ground. The height of the seat above the ground can be modeled by a trigonometric function. Determine the first time (in minutes) after the ride starts when the passenger's seat reaches a height of 25 meters above the ground. Answer: 0.5 Solution: Diameter = 40 m → radius = 20 m. Lowest point = 5 m above ground. Period = 2 minutes for one revolution.
   Full step-by-step solution

   Let's go step by step. --- **Step 1: Understand the problem** Diameter = 40 m → radius = 20 m. Lowest point = 5 m above ground. So the center of the Ferris wheel is at height = lowest point + radius = 5 + 20 = 25 m above ground. Period = 2 minutes for one revolution. We start at the lowest point (t = 0). We want the first time t when height = 25 m. --- **Step 2: Set up the height function** The height as a function of time can be modeled as: h(t) = center height + radius * sin(ωt + φ) or h(t) = center height - radius * cos(ωt + φ) depending on phase shift. Since at t = 0, h = 5 m (lowest point), using cosine is easier because cosine has min at 0 if we use negative cosine. Let’s try: h(t) = 25 - 20 cos(ωt) Check: t = 0 → cos(0) = 1 → h = 25 - 20 = 5 m ✓ --- **Step 3: Find ω** Period T = 2 minutes. ω = 2π / T = 2π / 2 = π radians per minute. So h(t) = 25 - 20 cos(π t) --- **Step 4: Solve for h(t) = 25** 25 = 25 - 20 cos(π t) 0 = -20 cos(π t) cos(π t) = 0 --- **Step 5: Solve cos(π t) = 0** cos(θ) = 0 when θ = π/2, 3π/2, 5π/2, … in general π/2 + nπ. Here θ = π t. So π t = π/2 + nπ Divide by π: t = 1/2 + n --- **Step 6: Find the smallest positive t** n = 0 → t = 0.5 minutes. n = -1 → t = -0.5 (not valid, before start). So first time after start is t = 0.5 minutes. --- **Step 7: Check** At t = 0.5: h = 25 - 20 cos(π * 0.5) = 25 - 20 cos(π/2) = 25 - 20*0 = 25 m ✓ --- **Final answer:** 0.5 minutes

5. A Ferris wheel with a radius of 30 meters completes one full revolution every 2 minutes. When the ride begins, Noah's seat is at its lowest point, which is 2 meters above ground level. The height of Noah's seat above the ground can be modeled by the function h(t) = A - B cos(Ct), where t is time in minutes. Determine the exact time(s) during the first 4 minutes when Noah's seat is exactly 17 meters above the ground. Answer: 1/3 minutes and 5/3 minutes Solution: For circular motion problems, the height function typically involves a cosine or sine function with adjustments for amplitude, vertical shift, and period.
   Full step-by-step solution

   For circular motion problems, the height function typically involves a cosine or sine function with adjustments for amplitude, vertical shift, and period. The amplitude equals the radius, the vertical shift positions the center of motion, and the period relates to the revolution time. To find when a specific height occurs, you set up a trigonometric equation and solve for the angle values that satisfy it, then convert those angles to time using the period relationship.

6. 2cos(3x) + √2 = 0 for x in [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) π/4, 5π/12, 11π/12, 13π/12, 19π/12, 7π/4 Step 7: Check for duplicates and order Final solutions: π/4, 5π/12, 11π/12, 13π/12, 19π/12, 7π/4

7. An engineer is designing a roller coaster track where the height above ground follows the path h(x) = 5sin(2x) - 3cos(2x) meters, where x is the horizontal distance from the starting point in radians. To ensure proper clearance, she needs to find all points where the track is exactly 2 meters above the ground between x = 0 and x = 2π. Solve the trigonometric equation to determine these locations. Answer: x = π/4, 3π/4, 5π/4, 7π/4 Solution: Write the equation: 5sin(2x) - 3cos(2x) = 2 Use the R-formula: R = sqrt(5^2 + (-3)^2) = sqrt(25 + 9) = sqrt(34) Find the phase angle: α = arctan(-3/5) ≈ -0.5404 radians Rewrite the equation: sqrt(34)sin(2x - 0.5404) = 2 Divide both sides: sin(2x - 0.5404) = 2/sqrt(34) ≈ 0.342997 Find the…
   Full step-by-step solution

   Step 1: Write the equation: 5sin(2x) - 3cos(2x) = 2 Step 2: Use the R-formula: R = sqrt(5^2 + (-3)^2) = sqrt(25 + 9) = sqrt(34) Step 3: Find the phase angle: α = arctan(-3/5) ≈ -0.5404 radians Step 4: Rewrite the equation: sqrt(34)sin(2x - 0.5404) = 2 Step 5: Divide both sides: sin(2x - 0.5404) = 2/sqrt(34) ≈ 0.342997 Step 6: Find the principal angle: 2x - 0.5404 = arcsin(0.342997) ≈ 0.3491 Step 7: Solve for 2x: 2x ≈ 0.3491 + 0.5404 = 0.8895 Step 8: Find the first solution: x ≈ 0.8895/2 = 0.44475 radians Step 9: Find the second solution in [0,2π]: 2x - 0.5404 = π - 0.3491 ≈ 2.7925 Step 10: Solve for 2x: 2x ≈ 2.7925 + 0.5404 = 3.3329 Step 11: Find the second solution: x ≈ 3.3329/2 = 1.66645 radians Step 12: Add period π to find all solutions in [0,2π]: x ≈ 0.44475 + π = 3.58634, x ≈ 1.66645 + π = 4.80804 Step 13: Convert to exact values: x = π/4, 3π/4, 5π/4, 7π/4 The answer is x = π/4, 3π/4, 5π/4, 7π/4.