Trigonometric Equations

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Answer Key & Explanations

Trigonometric Equations · Grade 11 · Worksheet 3

1. Matiu is a marine biologist studying the rhythmic flashing patterns of a rare species of jellyfish. The light intensity emitted by a jellyfish is modeled by the equation 4 sin²θ - 2 sinθ - 2 = 0, where θ is the phase angle of the jellyfish's internal biological clock, measured in radians from 0 to 2π. Find all possible phase angles θ in the interval [0, 2π) that cause the light intensity to drop to zero. Answer: θ = π/2, 7π/6, 11π/6 Solution: Start with the equation 4 sin²θ - 2 sinθ - 2 = 0. Factor out the common factor of 2: 2(2 sin²θ - sinθ - 1) = 0. Divide both sides by 2: 2 sin²θ - sinθ - 1 = 0.
   Full step-by-step solution

   Step 1: Start with the equation 4 sin²θ - 2 sinθ - 2 = 0. Step 2: Factor out the common factor of 2: 2(2 sin²θ - sinθ - 1) = 0. Step 3: Divide both sides by 2: 2 sin²θ - sinθ - 1 = 0. Step 4: Factor the quadratic in sinθ. We look for two numbers that multiply to 2 * (-1) = -2 and add to -1. These numbers are -2 and 1. Step 5: Rewrite: 2 sin²θ - 2 sinθ + sinθ - 1 = 0. Step 6: Factor by grouping: 2 sinθ(sinθ - 1) + 1(sinθ - 1) = 0. Step 7: Factor out (sinθ - 1): (sinθ - 1)(2 sinθ + 1) = 0. Step 8: Set each factor to zero: Case 1: sinθ - 1 = 0 → sinθ = 1. Case 2: 2 sinθ + 1 = 0 → sinθ = -1/2. Step 9: Solve Case 1: sinθ = 1. In [0, 2π), the only solution is θ = π/2. Step 10: Solve Case 2: sinθ = -1/2. The reference angle is π/6. Since sine is negative in quadrants III and IV: θ = π + π/6 = 7π/6 and θ = 2π - π/6 = 11π/6. Step 11: All solutions in [0, 2π) are θ = π/2, 7π/6, and 11π/6. The answer is θ = π/2, 7π/6, 11π/6.

2. 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.

3. A right triangle is drawn on a coordinate plane with vertices at (0,0), (4,0), and (4,3). A circle is circumscribed around this triangle such that all three vertices lie on the circle. What is the exact length of the radius of this circumscribed circle? Answer: 2.5 Solution: We have a right triangle with vertices at (0,0), (4,0), and (4,3). The circle circumscribed around the triangle passes through all three vertices. We need the radius of this circumcircle.
   Full step-by-step solution

   Step 1: Understand the problem We have a right triangle with vertices at (0,0), (4,0), and (4,3). The circle circumscribed around the triangle passes through all three vertices. We need the radius of this circumcircle. Step 2: Identify the right angle Plotting the points: (0,0) to (4,0) is horizontal. (4,0) to (4,3) is vertical. So the angle at (4,0) is 90 degrees. Thus, this is a right triangle with legs of length 4 and 3. Step 3: Recall the property of a circumscribed circle around a right triangle For a right triangle, the hypotenuse is the diameter of the circumscribed circle. So the center of the circle is the midpoint of the hypotenuse, and the radius is half the length of the hypotenuse. Step 4: Find the hypotenuse The hypotenuse is between (0,0) and (4,3). Length = sqrt( (4-0)^2 + (3-0)^2 ) = sqrt(16 + 9) = sqrt(25) = 5. Step 5: Find the radius Since the hypotenuse is the diameter, Radius = (1/2) * Hypotenuse = (1/2) * 5 = 2.5. Step 6: Conclusion The exact radius of the circumscribed circle is 2.5.

4. Aroha is an astronomer analyzing the light curve of a variable star. The brightness of the star varies over time and is modeled by the function B(t) = 7 sin(3t) + 3 cos(3t), where t is time in hours. To calibrate her telescope, Aroha needs to find all times in the interval 0 ≤ t < 2π when the brightness is exactly 5 units. Solve the resulting trigonometric equation to determine these times. Answer: t = (π - arcsin(4/√58) + arcsin(3/√58))/3, (2π - arcsin(4/√58) + arcsin(3/√58))/3 Solution: Start with 7 sin(3t) + 3 cos(3t) = 5. Rewrite as R sin(3t + φ) where R = sqrt(7² + 3²) = sqrt(49 + 9) = sqrt(58). We have R sin(3t + φ) = 5, so sin(3t + φ) = 5/√58.
   Full step-by-step solution

   Step 1: Start with 7 sin(3t) + 3 cos(3t) = 5. Step 2: Rewrite as R sin(3t + φ) where R = sqrt(7² + 3²) = sqrt(49 + 9) = sqrt(58). Step 3: We have R sin(3t + φ) = 5, so sin(3t + φ) = 5/√58. Step 4: Find φ such that sin φ = 3/√58 and cos φ = 7/√58 (since 7 sin(3t) + 3 cos(3t) = R sin(3t)cos φ + R cos(3t) sin φ). Thus φ = arcsin(3/√58). Step 5: The equation becomes sin(3t + arcsin(3/√58)) = 5/√58. Step 6: Let θ = 3t + arcsin(3/√58). Then sin θ = 5/√58. Step 7: The reference angle is α = arcsin(5/√58). Since sine is positive in quadrants I and II: θ = α + 2πk or θ = π - α + 2πk, for integer k. Step 8: So 3t + arcsin(3/√58) = arcsin(5/√58) + 2πk or 3t + arcsin(3/√58) = π - arcsin(5/√58) + 2πk. Step 9: Solve for t: t = (arcsin(5/√58) - arcsin(3/√58) + 2πk)/3 or t = (π - arcsin(5/√58) - arcsin(3/√58) + 2πk)/3. Step 10: For k = 0: t₁ = (arcsin(5/√58) - arcsin(3/√58))/3, t₂ = (π - arcsin(5/√58) - arcsin(3/√58))/3. Step 11: For k = 1: t₃ = (arcsin(5/√58) - arcsin(3/√58) + 2π)/3, t₄ = (π - arcsin(5/√58) - arcsin(3/√58) + 2π)/3. Step 12: For k = 2: t = (arcsin(5/√58) - arcsin(3/√58) + 4π)/3, which exceeds 2π, so stop. Step 13: The solutions in 0 ≤ t < 2π are the four values above. The answer is t = (arcsin(5/√58) - arcsin(3/√58))/3, (π - arcsin(5/√58) - arcsin(3/√58))/3, (arcsin(5/√58) - arcsin(3/√58) + 2π)/3, (π - arcsin(5/√58) - arcsin(3/√58) + 2π)/3.

5. Mason is a civil engineer designing a wave barrier for a harbor. The height of the water surface near the barrier is modeled by the equation 9 cos²θ - 7 = 0, where θ represents the angle of wave approach in radians, measured from 0 to 2π. Find all angles θ in the interval [0, 2π) that satisfy this equation, indicating the directions where the water surface height reaches a specific equilibrium point. Answer: θ = arccos(√7/3), 2π - arccos(√7/3), arccos(-√7/3), 2π - arccos(-√7/3) Solution: Start with the equation 9 cos²θ - 7 = 0. Add 7 to both sides: 9 cos²θ = 7. Divide both sides by 9: cos²θ = 7/9.
   Full step-by-step solution

   Step 1: Start with the equation 9 cos²θ - 7 = 0. Step 2: Add 7 to both sides: 9 cos²θ = 7. Step 3: Divide both sides by 9: cos²θ = 7/9. Step 4: Take the square root of both sides: cosθ = ±√(7/9) = ±√7/3. Step 5: Solve cosθ = √7/3. Since √7/3 ≈ 0.8819, the reference angle is arccos(√7/3). Cosine is positive in quadrants I and IV. So: θ = arccos(√7/3) and θ = 2π - arccos(√7/3). Step 6: Solve cosθ = -√7/3. Since -√7/3 ≈ -0.8819, the reference angle is arccos(√7/3). Cosine is negative in quadrants II and III. So: θ = π - arccos(√7/3) and θ = π + arccos(√7/3). But note that arccos(-√7/3) directly gives the angle in quadrant II: arccos(-√7/3) = π - arccos(√7/3). The quadrant III solution is 2π - arccos(-√7/3) = π + arccos(√7/3). Step 7: All solutions in [0, 2π) are: θ = arccos(√7/3), 2π - arccos(√7/3), arccos(-√7/3), 2π - arccos(-√7/3). The answer is θ = arccos(√7/3), 2π - arccos(√7/3), arccos(-√7/3), 2π - arccos(-√7/3).

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