Trigonometric Equations

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Trigonometric Equations · Grade 11 · Worksheet 3

1. A right triangle is drawn on a coordinate plane with vertices at (0,0), (4,0), and (4,3). The angle θ is located at the origin, between the positive x-axis and the hypotenuse. Find the exact value of sin(2θ) using trigonometric identities. Answer: 24/25 Solution: A = (0,0) B = (4,0) C = (4,3) This is a right triangle with the right angle at B = (4,0). So AB is along the x-axis from (0,0) to (4,0) → length 4. BC is vertical from (4,0) to (4,3) → length 3.
   Full step-by-step solution

   Let's go step-by-step. --- **Step 1: Understand the triangle and θ** Vertices: A = (0,0) B = (4,0) C = (4,3) This is a right triangle with the right angle at B = (4,0). So AB is along the x-axis from (0,0) to (4,0) → length 4. BC is vertical from (4,0) to (4,3) → length 3. AC is the hypotenuse from (0,0) to (4,3) → length 5 (by Pythagoras: sqrt(4^2 + 3^2) = sqrt(25) = 5). Angle θ is at vertex A = (0,0), between the positive x-axis (side AB) and the hypotenuse AC. --- **Step 2: Find sin θ and cos θ** From the triangle: Opposite side to θ = vertical difference from A to C = 3. Adjacent side to θ = horizontal difference from A to C = 4. Hypotenuse = 5. So: sin θ = opposite/hypotenuse = 3/5 cos θ = adjacent/hypotenuse = 4/5 --- **Step 3: Use the double-angle identity for sine** The identity is: sin(2θ) = 2 sin θ cos θ Substitute: sin(2θ) = 2 * (3/5) * (4/5) = 2 * 12/25 = 24/25 --- **Step 4: Conclusion** The exact value of sin(2θ) is 24/25. --- **Final answer:** 24/25

2. Sophia is an acoustic engineer designing noise-canceling headphones. The interference pattern between the ambient noise and the anti-noise wave can be modeled by the equation 8cos²θ - 14cosθ + 5 = 0, where θ represents the phase angle between the two waves in radians, measured from 0 to 2π. Find all possible phase angles θ in the interval [0, 2π) that result in perfect destructive interference (zero net sound). Answer: θ = π/3, 5π/3, arccos(5/8), 2π - arccos(5/8) Solution: Start with the equation 8cos²θ - 14cosθ + 5 = 0. Treat cosθ as the variable. Factor the quadratic.
   Full step-by-step solution

   Step 1: Start with the equation 8cos²θ - 14cosθ + 5 = 0. Step 2: Treat cosθ as the variable. Factor the quadratic. We look for two numbers that multiply to 8*5 = 40 and add to -14. These numbers are -10 and -4. Step 3: Rewrite: 8cos²θ - 10cosθ - 4cosθ + 5 = 0. Step 4: Factor by grouping: 2cosθ(4cosθ - 5) - 1(4cosθ - 5) = 0. Step 5: Factor out (4cosθ - 5): (4cosθ - 5)(2cosθ - 1) = 0. Step 6: Set each factor to zero: Case 1: 4cosθ - 5 = 0 → cosθ = 5/4 = 1.25. Case 2: 2cosθ - 1 = 0 → cosθ = 1/2. Step 7: Check Case 1: cosθ = 1.25 is outside the range of cosine (which is [-1, 1]), so there are no solutions from this case. Step 8: Solve Case 2: cosθ = 1/2. In [0, 2π), cosine is positive in quadrants I and IV. The reference angle is arccos(1/2) = π/3. Quadrant I: θ = π/3. Quadrant IV: θ = 2π - π/3 = 5π/3. Step 9: All solutions in [0, 2π) are θ = π/3 and θ = 5π/3. The answer is θ = π/3, 5π/3.

3. 2sin²(x) - sin(x) - 1 = 0 for 0 ≤ x ≤ 2π Answer: x = π/2, 7π/6, 11π/6 Solution: 2 sin²(x) - sin(x) - 1 = 0, for 0 ≤ x ≤ 2π. Substitute to make it look like a quadratic equation. Let u = sin(x).
   Full step-by-step solution

   Let's solve the equation step by step. We have: 2 sin²(x) - sin(x) - 1 = 0, for 0 ≤ x ≤ 2π. --- **Step 1: Substitute to make it look like a quadratic equation.** Let u = sin(x). Then the equation becomes: 2u² - u - 1 = 0. --- **Step 2: Solve the quadratic equation for u.** Use the quadratic formula: u = [ -b ± sqrt(b² - 4ac) ] / (2a) Here a = 2, b = -1, c = -1. Discriminant: D = (-1)² - 4(2)(-1) = 1 + 8 = 9. So: u = [ 1 ± sqrt(9) ] / (4) u = [ 1 ± 3 ] / 4. Thus: u = (1 + 3)/4 = 4/4 = 1 or u = (1 - 3)/4 = -2/4 = -1/2. So sin(x) = 1 or sin(x) = -1/2. --- **Step 3: Solve sin(x) = 1 in [0, 2π].** sin(x) = 1 at x = π/2 only in the given interval. So one solution: x = π/2. --- **Step 4: Solve sin(x) = -1/2 in [0, 2π].** We know sin is negative in Quadrants III and IV. Reference angle: sin(π/6) = 1/2, so reference angle = π/6. In QIII: x = π + π/6 = 7π/6. In QIV: x = 2π - π/6 = 12π/6 - π/6 = 11π/6. So solutions: x = 7π/6, 11π/6. --- **Step 5: Combine all solutions in [0, 2π].** From sin(x) = 1: x = π/2. From sin(x) = -1/2: x = 7π/6, 11π/6. Thus final answer: x = π/2, 7π/6, 11π/6. --- **Final Answer:** x = π/2, 7π/6, 11π/6

4. Mason is a civil engineer modeling the vibrations of a suspension bridge caused by wind. The horizontal displacement (in meters) of a point on the bridge deck over time is given by the equation: 2 sin²(t) + 7 sin(t) - 4 = 0, where t is time in seconds measured from 0 to 2π. Find all times t in the interval [0, 2π) when the displacement is exactly zero. Answer: t = π/6, 5π/6 Solution: Start with the equation 2 sin²(t) + 7 sin(t) - 4 = 0. Factor the quadratic. We look for two numbers that multiply to 2 * (-4) = -8 and add to 7.
   Full step-by-step solution

   Step 1: Start with the equation 2 sin²(t) + 7 sin(t) - 4 = 0. Step 2: Factor the quadratic. We look for two numbers that multiply to 2 * (-4) = -8 and add to 7. These numbers are 8 and -1. Step 3: Rewrite: 2 sin²(t) + 8 sin(t) - sin(t) - 4 = 0. Step 4: Factor by grouping: 2 sin(t)(sin(t) + 4) - 1(sin(t) + 4) = 0. Step 5: Factor out (sin(t) + 4): (sin(t) + 4)(2 sin(t) - 1) = 0. Step 6: Set each factor to zero: Case 1: sin(t) + 4 = 0 → sin(t) = -4. This has no solution because sine is always between -1 and 1. Case 2: 2 sin(t) - 1 = 0 → sin(t) = 1/2. Step 7: Solve sin(t) = 1/2. In [0, 2π), sine is positive in quadrants I and II. The reference angle is π/6. Step 8: The solutions are t = π/6 and t = π - π/6 = 5π/6. The answer is t = π/6, 5π/6.

5. sin(2x) = cos(x) for x ∈ [0, 2π] Answer: π/6, π/2, 5π/6, 3π/2 Solution: Use the double-angle identity: sin(2x) = 2sin(x)cos(x) Substitute into the equation: 2sin(x)cos(x) = cos(x) Rearrange: 2sin(x)cos(x) - cos(x) = 0 Factor: cos(x)(2sin(x) - 1) = 0 Case 1: cos(x) = 0 Case 2: 2sin(x) - 1 = 0 Solve Case 1: cos(x) = 0 in [0, 2π] gives x = π/2, 3π/2 Solve Case 2:…
   Full step-by-step solution

   Step 1: Use the double-angle identity: sin(2x) = 2sin(x)cos(x) Step 2: Substitute into the equation: 2sin(x)cos(x) = cos(x) Step 3: Rearrange: 2sin(x)cos(x) - cos(x) = 0 Step 4: Factor: cos(x)(2sin(x) - 1) = 0 Step 5: Set each factor equal to zero: Case 1: cos(x) = 0 Case 2: 2sin(x) - 1 = 0 Step 6: Solve Case 1: cos(x) = 0 in [0, 2π] gives x = π/2, 3π/2 Step 7: Solve Case 2: 2sin(x) - 1 = 0 → sin(x) = 1/2 in [0, 2π] gives x = π/6, 5π/6 Step 8: Combine all solutions: x = π/6, π/2, 5π/6, 3π/2

6. Liam is designing a suspension bridge where the main cable forms a parabolic shape. The height of the cable above the roadway can be modeled by the function h(x) = 50 - 0.02x², where x is the horizontal distance from the center of the bridge in meters. Support cables hang vertically from the main cable to the roadway every 10 meters. At what horizontal distances from the center are the support cables exactly 32 meters long? Answer: x = ±30 meters Solution: h(x) = 50 - 0.02x² where x is the horizontal distance from the center. The roadway is at height 0.
   Full step-by-step solution

   Let's go step-by-step. --- **Step 1: Understand the problem** The main cable height is given by h(x) = 50 - 0.02x² where x is the horizontal distance from the center. The roadway is at height 0. The support cables hang vertically from the main cable to the roadway, so their length is equal to the height of the main cable at that x. We are told: support cables are 32 meters long. That means h(x) = 32. --- **Step 2: Set up the equation** h(x) = 32 50 - 0.02x² = 32 --- **Step 3: Solve for x²** 50 - 32 = 0.02x² 18 = 0.02x² Multiply both sides by 100 to avoid decimals: 1800 = 2x² Divide by 2: 900 = x² --- **Step 4: Solve for x** x² = 900 x = ±√900 x = ±30 --- **Step 5: Interpret the result** The support cables are 32 meters long at 30 meters to the right of the center and 30 meters to the left of the center. --- **Final answer:** x = ±30 meters

7. A right triangle is inscribed in a unit circle such that the hypotenuse is the diameter. If one acute angle measures θ, and the point where the terminal side intersects the circle has coordinates (x, y), express the area of the triangle in terms of trigonometric functions of θ. Answer: \frac{1}{2}|\sin(2θ)| Solution: In a unit circle, the coordinates of a point on the circumference correspond to cosine and sine of the angle. The area of a right triangle can be found using the legs as base and height.
   Full step-by-step solution

   In a unit circle, the coordinates of a point on the circumference correspond to cosine and sine of the angle. The area of a right triangle can be found using the legs as base and height. For a triangle inscribed in a circle with the hypotenuse as diameter, the legs are related to the sine and cosine functions. This demonstrates the connection between trigonometric functions and geometric properties.