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NBT Mathematics: Trigonometry Flashcards

6 cards from real NBT practice questions. Tap to flip, then mark Knew It or Still Learning — missed cards come back until you master them.

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  1. If sin θ = 3/5 and θ is in the second quadrant, what is the value of tan(2θ)?

    Answer: -24/7

    In Q2, cos θ = -4/5 (negative). tan θ = sin θ / cos θ = (3/5)/(-4/5) = -3/4. Using the double angle formula: tan(2θ) = 2tan θ / (1 - tan²θ) = 2(-3/4) / (1 - 9/16) = (-3/2) / (7/16) = (-3/2)(16/7) = -24/7.

  2. Which of the following is equivalent to cos 4x expressed entirely in terms of cos x?

    Answer: 8cos⁴x - 8cos²x + 1

    Use double-angle identities iteratively. cos 2x = 2cos²x - 1. Then cos 4x = 2cos²(2x) - 1 = 2(2cos²x - 1)² - 1 = 2(4cos⁴x - 4cos²x + 1) - 1 = 8cos⁴x - 8cos²x + 2 - 1 = 8cos⁴x - 8cos²x + 1. Note that options A and D are identical expressions, so A is correct.

  3. The general solution of the equation sin(2x) = cos(x) over all real numbers is:

    Answer: x = 30° + 120°k or x = 90° + 360°k, k ∈ ℤ

    Rewrite sin(2x) = cos(x) as 2sin(x)cos(x) - cos(x) = 0, so cos(x)(2sin(x) - 1) = 0. This gives cos(x) = 0 → x = 90° + 180°k, or sin(x) = 1/2 → x = 30° + 360°k or x = 150° + 360°k. Combining: x = 30° + 120°k captures 30°, 150°, 270° (since cos = 0 at 90° and 270°, and 90° + 180°k). The correct general form capturing all families is x = 30° + 120°k or x = 90° + 360°k.

  4. In triangle ABC, side a = 7, side b = 8, and angle C = 120°. What is the length of side c?

    Answer: √(169) = 13

    By the cosine rule: c² = a² + b² - 2ab·cos C = 49 + 64 - 2(7)(8)cos(120°) = 113 - 112(-1/2) = 113 + 56 = 169. Therefore c = 13.

  5. If f(x) = 2sin(3x - 60°), what is the x-coordinate of the first maximum of f(x) for x > 0°?

    Answer: 50°

    A maximum of 2sin(u) occurs when u = 90°. Set 3x - 60° = 90°, giving 3x = 150°, so x = 50°. This is the first maximum for x > 0° since the next would be at 3x - 60° = 90° + 360° = 450°, giving x = 170°.

  6. Given that tan α = p (where p > 0 and α is acute), express (sin α - cos α)² in terms of p:

    Answer: (p - 1)² / (p² + 1)

    Since tan α = p, we have sin α = p/√(p²+1) and cos α = 1/√(p²+1). Then (sin α - cos α)² = sin²α - 2sin α cos α + cos²α = 1 - 2sin α cos α = 1 - 2(p/(p²+1)) = 1 - 2p/(p²+1) = (p²+1-2p)/(p²+1) = (p-1)²/(p²+1).