Multiple Angle Trigonometry

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Multiple Angle Trigonometry · Grade 12 · Worksheet 3

1. An oceanographer is studying tidal patterns in a coastal bay. The water depth D(t) in meters is modeled by the function D(t) = 3.5 + 2.8cos(πt/6) + 1.2sin(πt/6), where t is time in hours after midnight. To determine when boats with a 4.2 meter draft can safely enter the harbor, she needs to find all times between 6:00 AM and 6:00 PM when the water depth is exactly 5.1 meters. Solve the trigonometric equation to find these times. Answer: 9:00 AM and 3:00 PM Solution: In trigonometric modeling of periodic phenomena like tides, equations often contain both sine and cosine terms. These can be combined using the identity R cos(x - α) = R cos x cos α + R sin x sin α, where R = sqrt(A² + B²) and α is determined by cos α = A/R and sin α = B/R.
   Full step-by-step solution

   In trigonometric modeling of periodic phenomena like tides, equations often contain both sine and cosine terms. These can be combined using the identity R cos(x - α) = R cos x cos α + R sin x sin α, where R = sqrt(A² + B²) and α is determined by cos α = A/R and sin α = B/R. This transformation simplifies solving the equation by reducing it to a basic cosine equation. The phase shift α represents how much the combined function is shifted from a standard cosine wave.

2. 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

3. Emma is observing a laser beam reflecting off a rotating mirror. The angle of the reflected beam, measured in degrees from a fixed reference line, is given by θ(t) = 15 sin(5t + 10°) + 25, where t is time in seconds. At what times t in the interval [0, 360) seconds does the reflected beam make an angle of exactly 10° with the reference line? Answer: t = 10, 50, 82, 122, 154, 194, 226, 266, 298, 338 seconds Solution: Set the equation: 15 sin(5t + 10°) + 25 = 10 Subtract 25 from both sides: 15 sin(5t + 10°) = -15 Divide by 15: sin(5t + 10°) = -1 The sine function equals -1 at angles: 270° + 360°k, where k is any integer.
   Full step-by-step solution

   Step 1: Set the equation: 15 sin(5t + 10°) + 25 = 10 Step 2: Subtract 25 from both sides: 15 sin(5t + 10°) = -15 Step 3: Divide by 15: sin(5t + 10°) = -1 Step 4: The sine function equals -1 at angles: 270° + 360°k, where k is any integer. So: 5t + 10° = 270° + 360°k Step 5: Subtract 10°: 5t = 260° + 360°k Step 6: Divide by 5: t = 52° + 72°k Step 7: Find all values of t in [0, 360) seconds by letting k = 0, 1, 2, 3, 4: - k = 0: t = 52° - k = 1: t = 52° + 72° = 124° - k = 2: t = 52° + 144° = 196° - k = 3: t = 52° + 216° = 268° - k = 4: t = 52° + 288° = 340° All five values are within [0, 360). The answer is t = 52, 124, 196, 268, 340 seconds.

4. An engineer is designing a roller coaster track where the height above ground follows the path h(t) = 12sin(2t) + 5cos(2t) meters, where t is time in seconds after the ride begins. To ensure proper safety clearance, she needs to find all times during the first 5 seconds when the track reaches exactly 8 meters above ground. Solve the trigonometric equation 12sin(2t) + 5cos(2t) = 8 for t in the interval [0, 5]. Answer: t = π/4, 3π/4, 5π/4, 7π/4, 9π/4 seconds Solution: Trigonometric equations with multiple angle terms often require combining them into a single trigonometric function using the method of auxiliary angles.
   Full step-by-step solution

   Trigonometric equations with multiple angle terms often require combining them into a single trigonometric function using the method of auxiliary angles. This technique involves finding an amplitude and phase shift that transforms the sum of sine and cosine terms into a single sine or cosine function. The resulting equation can then be solved using inverse trigonometric functions and considering the periodic nature of trigonometric functions within the given interval.

5. Charlotte is analyzing a wave pattern on an oscilloscope screen. The wave is generated by the function y = 8 sin(2θ), where θ is the angle in radians. A horizontal reference line is drawn across the screen at y = 4√2. Visualizing the sine curve oscillating between -8 and 8, find all angles θ in the interval [0, 2π) where the wave crosses this reference line. Answer: θ = π/8, 3π/8, 9π/8, 11π/8 Solution: Set up the equation: 8 sin(2θ) = 4√2. Divide both sides by 8: sin(2θ) = (4√2)/8 = √2/2. The reference angle for sin = √2/2 is π/4, since sin(π/4) = √2/2.
   Full step-by-step solution

   Step 1: Set up the equation: 8 sin(2θ) = 4√2. Step 2: Divide both sides by 8: sin(2θ) = (4√2)/8 = √2/2. Step 3: The reference angle for sin = √2/2 is π/4, since sin(π/4) = √2/2. Sine is positive in quadrants I and II. Step 4: General solutions for 2θ: 2θ = π/4 + 2πk or 2θ = π - π/4 + 2πk = 3π/4 + 2πk, where k is an integer. Step 5: Solve for θ: θ = π/8 + πk or θ = 3π/8 + πk. Step 6: Find solutions in [0, 2π) by testing integer k: For k = 0: θ = π/8 (valid), θ = 3π/8 (valid). For k = 1: θ = π/8 + π = 9π/8 (valid), θ = 3π/8 + π = 11π/8 (valid). For k = 2: θ = π/8 + 2π = 17π/8 (greater than 2π = 16π/8, not valid), θ = 3π/8 + 2π = 19π/8 (not valid). Step 7: The solutions in [0, 2π) are θ = π/8, 3π/8, 9π/8, 11π/8. The answer is θ = π/8, 3π/8, 9π/8, 11π/8.

6. 4sin(5x) - 2√3 = 0 for x ∈ [0, π] Answer: π/15, 2π/15, 7π/15, 8π/15, 13π/15, 14π/15 Solution: Isolate sin(5x): 4sin(5x) - 2√3 = 0 → 4sin(5x) = 2√3 → sin(5x) = (2√3)/4 = √3/2. Find general solutions for θ = 5x: sin(θ) = √3/2. The reference angle is π/3.
   Full step-by-step solution

   Step 1: Isolate sin(5x): 4sin(5x) - 2√3 = 0 → 4sin(5x) = 2√3 → sin(5x) = (2√3)/4 = √3/2. Step 2: Find general solutions for θ = 5x: sin(θ) = √3/2. The reference angle is π/3. Sine is positive in quadrants I and II. So θ = π/3 + 2πk or θ = 2π/3 + 2πk, where k is an integer. Step 3: Substitute back 5x = θ: 5x = π/3 + 2πk or 5x = 2π/3 + 2πk. Step 4: Solve for x: x = π/15 + 2πk/5 or x = 2π/15 + 2πk/5. Step 5: Find solutions in [0, π]. For the first family: k=0 → π/15; k=1 → π/15 + 2π/5 = π/15 + 6π/15 = 7π/15; k=2 → π/15 + 4π/5 = π/15 + 12π/15 = 13π/15; k=3 → π/15 + 6π/5 = π/15 + 18π/15 = 19π/15 > π (stop). For the second family: k=0 → 2π/15; k=1 → 2π/15 + 2π/5 = 2π/15 + 6π/15 = 8π/15; k=2 → 2π/15 + 4π/5 = 2π/15 + 12π/15 = 14π/15; k=3 → 2π/15 + 6π/5 = 2π/15 + 18π/15 = 20π/15 > π (stop). Step 6: List all solutions in [0, π]: π/15, 2π/15, 7π/15, 8π/15, 13π/15, 14π/15. Final answer: π/15, 2π/15, 7π/15, 8π/15, 13π/15, 14π/15.

7. 3tan(5x) - 3√3 = 0 for x ∈ [0, π] Answer: π/15, 4π/15, 7π/15, 2π/3, 13π/15 Solution: Isolate tan(5x). 3tan(5x) - 3√3 = 0 3tan(5x) = 3√3 tan(5x) = √3 Find general solutions for 5x. tan(θ) = √3 when θ = π/3 + πk, where k is any integer.
   Full step-by-step solution

   Step 1: Isolate tan(5x). 3tan(5x) - 3√3 = 0 3tan(5x) = 3√3 tan(5x) = √3 Step 2: Find general solutions for 5x. tan(θ) = √3 when θ = π/3 + πk, where k is any integer. So 5x = π/3 + πk. Step 3: Solve for x. x = π/15 + πk/5 Step 4: Find solutions in [0, π]. For k = 0: x = π/15 For k = 1: x = π/15 + π/5 = π/15 + 3π/15 = 4π/15 For k = 2: x = π/15 + 2π/5 = π/15 + 6π/15 = 7π/15 For k = 3: x = π/15 + 3π/5 = π/15 + 9π/15 = 10π/15 = 2π/3 For k = 4: x = π/15 + 4π/5 = π/15 + 12π/15 = 13π/15 For k = 5: x = π/15 + 5π/5 = π/15 + π = 16π/15, which is greater than π. Step 5: Verify all solutions are in [0, π]. π/15, 4π/15, 7π/15, 2π/3, 13π/15 are all within [0, π]. Final answer: π/15, 4π/15, 7π/15, 2π/3, 13π/15