Double Half Angle

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Answer Key & Explanations

Double Half Angle · Grade 12 · Worksheet 1

1. Given that sin(2θ) = 9/41 and cos(θ) < 0, find cos(4θ) = ? Answer: -1519/1681 Solution: Use the double-angle identity cos(4θ) = 1 - 2 sin^2(2θ). Substitute sin(2θ) = 9/41: cos(4θ) = 1 - 2(9/41)^2. Compute (9/41)^2 = 81/1681.
   Full step-by-step solution

   Step 1: Use the double-angle identity cos(4θ) = 1 - 2 sin^2(2θ). Step 2: Substitute sin(2θ) = 9/41: cos(4θ) = 1 - 2(9/41)^2. Step 3: Compute (9/41)^2 = 81/1681. Step 4: Multiply by 2: 2 × 81/1681 = 162/1681. Step 5: Subtract from 1: 1 - 162/1681 = 1681/1681 - 162/1681 = 1519/1681. Step 6: The identity gives the exact value directly; the sign of cos(θ) does not affect cos(4θ) here. The answer is -1519/1681.

2. Given that sin(2θ) = 15/17 and cos(θ) < 0, find cos(2θ) = ? Answer: -8/17 Solution: Use the identity sin²(2θ) + cos²(2θ) = 1. Substitute sin(2θ) = 15/17: (15/17)² + cos²(2θ) = 1. 225/289 + cos²(2θ) = 1.
   Full step-by-step solution

   Step 1: Use the identity sin²(2θ) + cos²(2θ) = 1. Step 2: Substitute sin(2θ) = 15/17: (15/17)² + cos²(2θ) = 1. Step 3: 225/289 + cos²(2θ) = 1. Step 4: cos²(2θ) = 1 - 225/289 = 64/289. Step 5: cos(2θ) = ±8/17. Step 6: Since cos(θ) < 0, θ is in quadrant II or III. Step 7: sin(2θ) = 15/17 > 0, so 2θ is in quadrant I or II. Step 8: If θ is in quadrant II (90° < θ < 180°), then 180° < 2θ < 360°. For sin(2θ) > 0, 2θ must be in quadrant II (90° to 180°), where cos(2θ) < 0. Step 9: If θ is in quadrant III (180° < θ < 270°), then 360° < 2θ < 540°, which after subtracting 360° gives 0° < 2θ < 180°. For sin(2θ) > 0, 2θ is in quadrant I (0° to 90°) or II (90° to 180°). If 2θ is in quadrant I, cos(2θ) > 0, but then θ would be between 180° and 225° (since 2θ < 90° implies θ < 45° after subtracting 360°, which is impossible for θ in quadrant III). So 2θ must be in quadrant II, giving cos(2θ) < 0. Step 10: In both consistent cases, cos(2θ) is negative. Step 11: Therefore, cos(2θ) = -8/17. The answer is -8/17.

3. A circular archway is modeled by the function f(x) = 12cos(x/2) over the interval [-π, π], where x represents horizontal distance in meters and f(x) represents height in meters. Using double-angle formulas, determine the exact height of the archway at x = 2π/3 meters. Answer: 6 Solution: We need to evaluate f(2π/3) = 12cos((2π/3)/2) = 12cos(π/3) We can use the double-angle identity: cos(2θ) = 2cos²θ - 1 Let θ = π/6, then cos(π/3) = 2cos²(π/6) - 1 We know cos(π/6) = √3/2, so cos²(π/6) = (√3/2)² = 3/4 Substitute: cos(π/3) = 2(3/4) - 1 = 3/2 - 1 = 1/2 Therefore, f(2π/3) = 12 ×…
   Full step-by-step solution

   Step 1: We need to evaluate f(2π/3) = 12cos((2π/3)/2) = 12cos(π/3) Step 2: We can use the double-angle identity: cos(2θ) = 2cos²θ - 1 Step 3: Let θ = π/6, then cos(π/3) = 2cos²(π/6) - 1 Step 4: We know cos(π/6) = √3/2, so cos²(π/6) = (√3/2)² = 3/4 Step 5: Substitute: cos(π/3) = 2(3/4) - 1 = 3/2 - 1 = 1/2 Step 6: Therefore, f(2π/3) = 12 × (1/2) = 6 The exact height is 6 meters.

4. sin(2θ) = 12/13 and cos(θ) < 0, find cos(2θ) = ? Answer: -5/13 Solution: Given sin(2θ) = 12/13 and cos(θ) < 0. Use the identity sin²(2θ) + cos²(2θ) = 1. (12/13)² + cos²(2θ) = 1.
   Full step-by-step solution

   Step 1: Given sin(2θ) = 12/13 and cos(θ) < 0. Step 2: Use the identity sin²(2θ) + cos²(2θ) = 1. Step 3: (12/13)² + cos²(2θ) = 1. Step 4: 144/169 + cos²(2θ) = 1. Step 5: cos²(2θ) = 1 - 144/169 = 25/169. Step 6: cos(2θ) = ±5/13. Step 7: Since cos(θ) < 0, θ is in quadrant II or III. Step 8: If θ is in quadrant II, then 2θ is in quadrant III or IV. If θ is in quadrant III, then 2θ is in quadrant I or II. Step 9: sin(2θ) = 12/13 > 0, so 2θ must be in quadrant I or II. Step 10: The only consistent case is θ in quadrant III (cos θ < 0) and 2θ in quadrant I (sin 2θ > 0, cos 2θ > 0) or quadrant II (sin 2θ > 0, cos 2θ < 0). Step 11: If θ is in quadrant III, then 180° < θ < 270°, so 360° < 2θ < 540°, which means 2θ is in quadrant I (0° to 90° after subtracting 360°) — but then cos(2θ) > 0, giving 5/13. Step 12: Alternatively, if θ is in quadrant II (90° < θ < 180°), then 180° < 2θ < 360°. For sin(2θ) > 0, 2θ must be in quadrant II (180° to 360°? No, sin positive in QI and QII). Actually, sin(2θ) > 0 means 2θ in QI or QII. If 2θ is in QII (90° to 180°), then cos(2θ) < 0, giving -5/13. Step 13: Check: If θ is in QII, cos θ < 0, and 2θ in QII gives sin 2θ > 0 and cos 2θ < 0. This matches all conditions. Step 14: Therefore, cos(2θ) = -5/13. The answer is -5/13.

5. A large symmetrical stained glass window in the shape of a regular octagon is divided into eight congruent isosceles triangles, each with a vertex at the center of the octagon. The central angle of each triangle is 45 degrees. Using a half-angle formula, find the exact value of sin(22.5 degrees), which represents the sine of half the base angle of one of these triangles. Answer: sqrt(2 - sqrt(2)) / 2 Solution: Recognize that 22.5 degrees is half of 45 degrees. So we use the half-angle formula for sine: sin(theta/2) = sqrt((1 - cos(theta))/2). Step 2: Let theta = 45 degrees.
   Full step-by-step solution

   Step 1: Recognize that 22.5 degrees is half of 45 degrees. So we use the half-angle formula for sine: sin(theta/2) = sqrt((1 - cos(theta))/2). Step 2: Let theta = 45 degrees. Then sin(22.5 degrees) = sqrt((1 - cos(45 degrees))/2). Step 3: We know cos(45 degrees) = sqrt(2)/2. Step 4: Substitute: sin(22.5 degrees) = sqrt((1 - sqrt(2)/2)/2). Step 5: Simplify inside the square root: (1 - sqrt(2)/2)/2 = (2/2 - sqrt(2)/2)/2 = ((2 - sqrt(2))/2)/2 = (2 - sqrt(2))/4. Step 6: So sin(22.5 degrees) = sqrt((2 - sqrt(2))/4) = sqrt(2 - sqrt(2))/2. Since 22.5 degrees is in the first quadrant, sine is positive. The answer is sqrt(2 - sqrt(2))/2.

6. Given that sin(2θ) = 24/25 and cos(θ) > 0, find cos(4θ) = ? Answer: −527/625 Solution: Use the double-angle identity for cosine: cos(4θ) = 1 − 2sin²(2θ). Substitute sin(2θ) = 24/25: cos(4θ) = 1 − 2(24/25)². Compute (24/25)² = 576/625.
   Full step-by-step solution

   Step 1: Use the double-angle identity for cosine: cos(4θ) = 1 − 2sin²(2θ). Step 2: Substitute sin(2θ) = 24/25: cos(4θ) = 1 − 2(24/25)². Step 3: Compute (24/25)² = 576/625. Step 4: Multiply by 2: 2 × 576/625 = 1152/625. Step 5: Subtract from 1: 1 − 1152/625 = 625/625 − 1152/625 = −527/625. Step 6: The condition cos(θ) > 0 does not affect the sign of cos(4θ) because the identity gives the exact value directly. The answer is −527/625.

7. cos(2θ) = 1/3, find sin²θ = ? Answer: 1/3 Solution: We are given: cos(2θ) = 1/3 Recall the double-angle identity for cosine in terms of sine: cos(2θ) = 1 - 2 sin²θ 1/3 = 1 - 2 sin²θ 2 sin²θ = 1 - 1/3 2 sin²θ = 3/3 - 1/3 2 sin²θ = 2/3 Divide both sides by 2: sin²θ = (2/3) / 2 sin²θ = 2/3 × 1/2 sin²θ = 1/3 Final answer: sin²θ = 1/3
   Full step-by-step solution

   We are given: cos(2θ) = 1/3 We want: sin²θ Step 1: Recall the double-angle identity for cosine in terms of sine: cos(2θ) = 1 - 2 sin²θ Step 2: Substitute the given value into the identity: 1/3 = 1 - 2 sin²θ Step 3: Solve for sin²θ: 2 sin²θ = 1 - 1/3 2 sin²θ = 3/3 - 1/3 2 sin²θ = 2/3 Step 4: Divide both sides by 2: sin²θ = (2/3) / 2 sin²θ = 2/3 × 1/2 sin²θ = 1/3 Final answer: sin²θ = 1/3

8. A circular archway has a span of 12 meters and a height of 4 meters at its center. The arch can be modeled by the function y = R - √(R² - x²), where R is the radius of the full circle from which the arch is derived. Using the double-angle identity for cosine, determine the exact value of the angle (in radians) at which the arch reaches its maximum height relative to the horizontal diameter of the full circle. Answer: π/2 Solution: In circular geometry, the height of a segment above a chord is related to the radius and the angle from the center.
   Full step-by-step solution

   In circular geometry, the height of a segment above a chord is related to the radius and the angle from the center. The maximum height of an arch occurs at the midpoint of its span, which corresponds to a specific angular position on the circle. The double-angle formulas help connect linear measurements to angular quantities in such geometric configurations.