NBT Mathematics: Trigonometry Flashcards

Given that sin α = 3/5 with α ∈ (90°, 180°) and cos β = −5/13 with β ∈ (180°, 270°), what is the exact value of cos(α + β)?

Answer: 56/65

From the quadrant conditions: cos α = −4/5 (Q2, cosine negative) and sin β = −12/13 (Q3, sine negative). Applying the compound angle formula: cos(α + β) = cos α cos β − sin α sin β = (−4/5)(−5/13) − (3/5)(−12/13) = 20/65 + 36/65 = 56/65. The result is positive because both terms are positive after accounting for the quadrant signs.

Simplify the expression: (tan x − sin x)(tan x + sin x) ÷ tan²x

Answer: sin²x

The numerator is the difference of squares: tan²x − sin²x = sin²x/cos²x − sin²x = sin²x(1 − cos²x)/cos²x = sin⁴x/cos²x. Dividing by tan²x = sin²x/cos²x gives (sin⁴x/cos²x) × (cos²x/sin²x) = sin²x. Note that sec²x − 1 = tan²x, making option D a tempting but incorrect distractor.

Fully simplify: sin(180° + x) · cos(360° − x) · tan(90° + x)

Answer: cos²x

Applying reduction formulas: sin(180° + x) = −sin x; cos(360° − x) = cos x; tan(90° + x) = −cot x = −cos x/sin x. Multiplying: (−sin x)(cos x)(−cos x/sin x) = (−sin x)(−cos²x/sin x) = cos²x. The two negatives cancel and sin x cancels, leaving cos²x.

If tan θ = −2 and θ ∈ (90°, 180°), what is the exact value of sin 2θ?

Answer: −4/5

With tan θ = −2 in Q2: construct a reference triangle with opposite = 2, adjacent = 1, hypotenuse = √5. Since θ is in Q2, sin θ = 2/√5 (positive) and cos θ = −1/√5 (negative). Then sin 2θ = 2 sin θ cos θ = 2(2/√5)(−1/√5) = −4/5. The negative sign is critical and is the most common error.

Find the complete general solution of the equation: cos 2x + cos x = 0

Answer: x = ±60° + 360°n or x = 180° + 360°n

Substituting cos 2x = 2cos²x − 1: 2cos²x − 1 + cos x = 0, which factors as (2cos x − 1)(cos x + 1) = 0. So cos x = 1/2 → x = ±60° + 360°n, and cos x = −1 → x = 180° + 360°n. Option A omits the x = 300° + 360°n family (the negative 60°), which is a critical omission. Option D incorrectly uses 180° period for cosine = 1/2.

For the equation 2sin²x − sin x − 1 = 0, how many solutions exist in the interval x ∈ [0°, 360°]?

Answer: 3

Factoring: (2sin x + 1)(sin x − 1) = 0 gives sin x = −1/2 or sin x = 1. For sin x = −1/2 in [0°, 360°]: x = 210° and x = 330° (two solutions). For sin x = 1: x = 90° (one solution). Total = 3 solutions: {90°, 210°, 330°}. A common mistake is including x = 270° (sin 270° = −1, not −1/2) or miscounting the negative half solutions.