Trigonometric Identities

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Answer Key & Explanations

Trigonometric Identities · Grade 12 · Worksheet 1

1. Verify: (csc²θ - cot²θ) / (sec²θ - tan²θ) = 1. Answer: 1 Solution: Recall the Pythagorean identities: csc²θ = 1 + cot²θ and sec²θ = 1 + tan²θ. Substitute into the numerator: csc²θ - cot²θ = (1 + cot²θ) - cot²θ = 1.
   Full step-by-step solution

   Step 1: Recall the Pythagorean identities: csc²θ = 1 + cot²θ and sec²θ = 1 + tan²θ. Step 2: Substitute into the numerator: csc²θ - cot²θ = (1 + cot²θ) - cot²θ = 1. Step 3: Substitute into the denominator: sec²θ - tan²θ = (1 + tan²θ) - tan²θ = 1. Step 4: The expression becomes 1 / 1 = 1. Therefore, the identity is verified: (csc²θ - cot²θ) / (sec²θ - tan²θ) = 1.

2. Liam is designing a suspension bridge where the main cable follows the curve y = 15sin(x/10) + 20, where y is the height in meters above the water and x is the horizontal distance in meters from the left tower. Engineers need to know the steepest slope of the cable to design proper connections. What is the maximum slope of the cable? Answer: 1.5 Solution: y = 15 sin(x/10) + 20 We want the maximum slope of the cable. Slope is the derivative dy/dx.
   Full step-by-step solution

   Step 1: Understand the problem The cable's curve is given by y = 15 sin(x/10) + 20 We want the maximum slope of the cable. Slope is the derivative dy/dx. Step 2: Find the derivative y = 15 sin(x/10) + 20 dy/dx = 15 * cos(x/10) * (1/10) dy/dx = (15/10) cos(x/10) dy/dx = 1.5 cos(x/10) Step 3: Interpret the slope expression Slope m(x) = 1.5 cos(x/10) The maximum possible slope occurs when cos(x/10) is maximum. The maximum value of cos(anything) is 1. Step 4: Find the maximum slope Maximum slope = 1.5 * 1 = 1.5 Step 5: Conclusion The steepest slope of the cable is 1.5 meters rise per meter horizontally. This occurs where cos(x/10) = 1, i.e., where x/10 = 2nπ for integer n. Final answer: 1.5

3. A circular oil spill is expanding such that its radius increases at a constant rate of 0.5 meters per second. At a particular moment, the radius is 12 meters. Using calculus, determine the rate at which the area of the oil spill is increasing at that moment. (Use π in your calculation) Answer: 12π Solution: Write down what we know. We have a circular oil spill. A = π × r² dr/dt = 0.5 m/s (radius increasing at constant rate) At the moment we care about: r = 12 m We want: dA/dt at that moment.
   Full step-by-step solution

   Let's go step-by-step. --- **Step 1: Write down what we know.** We have a circular oil spill. Area of a circle: A = π × r² We are told: dr/dt = 0.5 m/s (radius increasing at constant rate) At the moment we care about: r = 12 m We want: dA/dt at that moment. --- **Step 2: Differentiate the area formula with respect to time t.** A = π × r² Differentiate both sides with respect to t: dA/dt = π × (2 × r) × dr/dt (Here we used the chain rule: dA/dt = dA/dr × dr/dt, and dA/dr = 2πr.) So: dA/dt = 2 × π × r × dr/dt --- **Step 3: Substitute the given values.** r = 12 m dr/dt = 0.5 m/s dA/dt = 2 × π × 12 × 0.5 --- **Step 4: Simplify the calculation.** 2 × 12 = 24 24 × 0.5 = 12 So: dA/dt = 12 × π --- **Step 5: Interpret the result.** At the moment when the radius is 12 m, the area is increasing at a rate of 12π square meters per second. --- **Final Answer:** 12π

4. Verify: (sec⁶θ - tan⁶θ) / (sec²θ - tan²θ) = 1 + 3 sec²θ tan²θ. Answer: Verified identity Solution: Start with the left side: (sec⁶θ - tan⁶θ) / (sec²θ - tan²θ). Recognize sec⁶θ - tan⁶θ as a difference of cubes: a³ - b³ = (a - b)(a² + ab + b²), where a = sec²θ and b = tan²θ.
   Full step-by-step solution

   Step 1: Start with the left side: (sec⁶θ - tan⁶θ) / (sec²θ - tan²θ). Step 2: Recognize sec⁶θ - tan⁶θ as a difference of cubes: a³ - b³ = (a - b)(a² + ab + b²), where a = sec²θ and b = tan²θ. Step 3: Factor: sec⁶θ - tan⁶θ = (sec²θ - tan²θ)(sec⁴θ + sec²θ tan²θ + tan⁴θ). Step 4: Substitute into the left side: [(sec²θ - tan²θ)(sec⁴θ + sec²θ tan²θ + tan⁴θ)] / (sec²θ - tan²θ). Step 5: Cancel (sec²θ - tan²θ) (provided sec²θ ≠ tan²θ, i.e., θ not a multiple of π/2). Step 6: We now have: sec⁴θ + sec²θ tan²θ + tan⁴θ. Step 7: Use the identity sec²θ = 1 + tan²θ, so sec⁴θ = (1 + tan²θ)² = 1 + 2 tan²θ + tan⁴θ. Step 8: Substitute: (1 + 2 tan²θ + tan⁴θ) + sec²θ tan²θ + tan⁴θ = 1 + 2 tan²θ + 2 tan⁴θ + sec²θ tan²θ. Step 9: Replace sec²θ with 1 + tan²θ in the term sec²θ tan²θ: sec²θ tan²θ = (1 + tan²θ) tan²θ = tan²θ + tan⁴θ. Step 10: So the expression becomes: 1 + 2 tan²θ + 2 tan⁴θ + tan²θ + tan⁴θ = 1 + 3 tan²θ + 3 tan⁴θ. Step 11: Factor: 1 + 3 tan²θ + 3 tan⁴θ = 1 + 3 tan²θ (1 + tan²θ) = 1 + 3 tan²θ sec²θ. Step 12: This equals the right side: 1 + 3 sec²θ tan²θ. The identity is verified.

5. A right circular cone has a height of 15 cm and a base radius of 8 cm. A plane parallel to the base cuts the cone at a height of 5 cm from the vertex, creating a smaller cone at the top and a frustum below. What is the volume of the frustum? (Use π in your calculation) Answer: 2480π/3 Solution: Find the volume of the entire cone. Volume of large cone = (1/3)πr²h = (1/3)π(8)²(15) = (1/3)π(64)(15) = (1/3)π(960) = 320π cm³ Use similar triangles to find the radius of the smaller cone.
   Full step-by-step solution

   Step 1: Find the volume of the entire cone. Volume of large cone = (1/3)πr²h = (1/3)π(8)²(15) = (1/3)π(64)(15) = (1/3)π(960) = 320π cm³ Step 2: Use similar triangles to find the radius of the smaller cone. The cross-section creates a smaller cone with height 5 cm. By similar triangles: radius_small / height_small = radius_large / height_large radius_small / 5 = 8 / 15 radius_small = (8 × 5) / 15 = 40/15 = 8/3 cm Step 3: Find the volume of the smaller cone. Volume of small cone = (1/3)π(r_small)²h_small = (1/3)π(8/3)²(5) = (1/3)π(64/9)(5) = (1/3)π(320/9) = 320π/27 cm³ Step 4: Subtract to find the volume of the frustum. Volume of frustum = Volume_large_cone - Volume_small_cone = 320π - 320π/27 = (8640π/27 - 320π/27) = 8320π/27 = 2480π/3 cm³ The answer is 2480π/3.