Trigonometric Identities

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Answer Key & Explanations

Trigonometric Identities · Grade 12 · Worksheet 1

1. A civil engineer is designing a suspension bridge where the main cable forms a parabolic curve described by the function f(x) = ax² + bx + c. The cable passes through points (-100, 50), (0, 10), and (100, 50) relative to the bridge's center. The engineer needs to verify that the cable's shape satisfies the trigonometric identity sin²θ + cos²θ = 1 at the point where the cable makes a 30° angle with the horizontal. Find the value of sin²θ + cos²θ at this point to confirm the identity holds. Answer: 1 Solution: (-100, 50), (0, 10), (100, 50) Equation: f(x) = a x^2 + b x + c From (0, 10): f(0) = a*0 + b*0 + c = 10 So c = 10.
   Full step-by-step solution

   Let's go step by step. --- **Step 1: Find the parabolic equation** We have points: (-100, 50), (0, 10), (100, 50) Equation: f(x) = a x^2 + b x + c From (0, 10): f(0) = a*0 + b*0 + c = 10 So c = 10. From (-100, 50): a*(10000) + b*(-100) + 10 = 50 10000a - 100b = 40 … (1) From (100, 50): a*(10000) + b*(100) + 10 = 50 10000a + 100b = 40 … (2) Add (1) and (2): 20000a = 80 a = 80 / 20000 = 8 / 2000 = 4 / 1000 = 0.004 From (2): 10000*(0.004) + 100b = 40 40 + 100b = 40 100b = 0 b = 0 So f(x) = 0.004 x^2 + 10 --- **Step 2: Find slope at a point where the cable makes 30° with horizontal** Slope = f'(x) = 0.008 x Let slope = tan θ = tan 30° = 1 / sqrt(3) ≈ 0.57735 So 0.008 x = 1 / sqrt(3) x = 1 / (0.008 * sqrt(3)) = 125 / sqrt(3) --- **Step 3: Find sin²θ + cos²θ** We know the trigonometric identity: sin²θ + cos²θ = 1 for any θ. So even without knowing the exact point on the cable, the identity holds. --- **Step 4: Conclusion** The problem says: "verify that the cable's shape satisfies the trigonometric identity sin²θ + cos²θ = 1 at the point where the cable makes a 30° angle with the horizontal." Since sin²θ + cos²θ = 1 is always true for any angle θ, the answer is 1. --- **Final answer:** 1

2. 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 measured to be 12 meters. Using calculus, determine the rate at which the area of the oil spill is increasing at that moment. Answer: 12π m²/s Solution: Write down what we know. A = \pi r^2 - Radius \( r \) increases at \( \frac{dr}{dt} = 0.5 \) m/s. - At the moment we care about, \( r = 12 \) m.
   Full step-by-step solution

   Let's go step-by-step. --- **Step 1: Write down what we know.** The area \( A \) of a circle is: \[ A = \pi r^2 \] We are told: - Radius \( r \) increases at \( \frac{dr}{dt} = 0.5 \) m/s. - At the moment we care about, \( r = 12 \) m. - We want \( \frac{dA}{dt} \) at that moment. --- **Step 2: Differentiate with respect to time \( t \).** Using the chain rule: \[ \frac{dA}{dt} = \frac{d}{dt} \left( \pi r^2 \right) = \pi \cdot 2r \cdot \frac{dr}{dt} \] So: \[ \frac{dA}{dt} = 2\pi r \cdot \frac{dr}{dt} \] --- **Step 3: Substitute the given values.** \[ r = 12, \quad \frac{dr}{dt} = 0.5 \] \[ \frac{dA}{dt} = 2\pi (12) \cdot (0.5) \] --- **Step 4: Simplify the calculation.** \[ 2 \times 12 = 24 \] \[ 24 \times 0.5 = 12 \] So: \[ \frac{dA}{dt} = 12\pi \] --- **Step 5: Include units.** Since \( A \) is in m² and \( t \) is in seconds, \( \frac{dA}{dt} \) is in m²/s. Thus: \[ \frac{dA}{dt} = 12\pi \ \text{m²/s} \] --- **Final Answer:** 12π m²/s

3. An engineer is designing a roller coaster that follows the path of a trigonometric function. The track's height above ground is modeled by h(t) = 3sin(t) + 4cos(t), where t is time in seconds and h is height in meters. To ensure safety, the engineer needs to verify that this function can be rewritten in the form h(t) = Rsin(t + α), where R > 0 and α is a phase shift. Find the values of R and α that satisfy this identity. Answer: R = 5, α = arctan(4/3) Solution: When combining sine and cosine functions into a single trigonometric function, we use the identity Asin(x) + Bcos(x) = Rsin(x + α), where R represents the amplitude and α represents the phase shift.
   Full step-by-step solution

   When combining sine and cosine functions into a single trigonometric function, we use the identity Asin(x) + Bcos(x) = Rsin(x + α), where R represents the amplitude and α represents the phase shift. This transformation is useful for analyzing the maximum and minimum values of periodic functions in real-world applications like wave motion or oscillating systems.

4. Verify: (csc⁶θ - cot⁶θ) / (csc⁴θ + csc²θ cot²θ + cot⁴θ) = 1. Answer: Verified identity Solution: Write the left side: (csc⁶θ - cot⁶θ) / (csc⁴θ + csc²θ cot²θ + cot⁴θ). Recognize the numerator as a difference of cubes: a³ - b³ = (a - b)(a² + ab + b²). Here a = csc²θ, b = cot²θ.
   Full step-by-step solution

   Step 1: Write the left side: (csc⁶θ - cot⁶θ) / (csc⁴θ + csc²θ cot²θ + cot⁴θ). Step 2: Recognize the numerator as a difference of cubes: a³ - b³ = (a - b)(a² + ab + b²). Here a = csc²θ, b = cot²θ. So csc⁶θ - cot⁶θ = (csc²θ - cot²θ)(csc⁴θ + csc²θ cot²θ + cot⁴θ). Step 3: Substitute into the left side: [(csc²θ - cot²θ)(csc⁴θ + csc²θ cot²θ + cot⁴θ)] / (csc⁴θ + csc²θ cot²θ + cot⁴θ). Step 4: Cancel the common factor (csc⁴θ + csc²θ cot²θ + cot⁴θ), provided it is not zero. Step 5: This leaves csc²θ - cot²θ. Step 6: Use the Pythagorean identity: csc²θ - cot²θ = 1. Step 7: Therefore, the left side simplifies to 1, which equals the right side. The identity is verified.

5. 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.

6. A civil engineer is designing a suspension bridge where the main cable forms a parabolic curve. The cable's shape is modeled by the function f(x) = ax² + bx + c. At the left support (x = 0), the cable is 50 meters high. At the right support (x = 200), the cable is also 50 meters high. The lowest point of the cable occurs at x = 100 and is 10 meters high. Determine the values of a, b, and c that define the cable's parabolic equation. Answer: f(x) = 0.004x² - 0.8x + 50 Solution: f(x) = a x^2 + b x + c At x = 0, f(0) = 50. f(0) = a(0)^2 + b(0) + c = c So c = 50. At x = 200, f(200) = 50.
   Full step-by-step solution

   We are given the parabolic function: f(x) = a x^2 + b x + c --- **Step 1: Use the left support condition** At x = 0, f(0) = 50. f(0) = a(0)^2 + b(0) + c = c So c = 50. --- **Step 2: Use the right support condition** At x = 200, f(200) = 50. f(200) = a(200)^2 + b(200) + 50 = 50 So: 40000 a + 200 b + 50 = 50 40000 a + 200 b = 0 Divide by 200: 200 a + b = 0 … (Equation 1) --- **Step 3: Use the lowest point condition** The lowest point is at x = 100, height 10. So f(100) = 10: a(100)^2 + b(100) + 50 = 10 10000 a + 100 b + 50 = 10 10000 a + 100 b = -40 Divide by 20: 500 a + 5 b = -2 … (Equation 2) --- **Step 4: Solve the system of equations** From Equation 1: b = -200 a Substitute into Equation 2: 500 a + 5(-200 a) = -2 500 a - 1000 a = -2 -500 a = -2 a = (-2)/(-500) = 2/500 = 1/250 = 0.004 Then b = -200 a = -200 * (1/250) = -200/250 = -4/5 = -0.8 --- **Step 5: Write the final equation** a = 0.004, b = -0.8, c = 50 So f(x) = 0.004 x^2 - 0.8 x + 50 --- **Final Answer:** f(x) = 0.004x^2 - 0.8x + 50

7. 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.