What are the coordinates of the point at angle π/6 on the unit circle? — (√3/2, 1/2).At π/6 (30°): cos(π/6) = √3/2 (x-coordinate), sin(π/6) = 1/2 (y-coordinate). Point = (√3/2, 1/2).
Using "All Students Take Calculus," in which quadrant is only TANGENT positive? — Q3.ASTC: All (Q1), Sine (Q2), Tangent (Q3), Cosine (Q4). In Q3, only tangent (and cotangent) are positive.
What is the arc length subtended by a central angle of π/3 on a unit circle? — π/3.Arc length = radius × angle (in radians) = 1 × π/3 = π/3. On a unit circle, arc length equals the radian measure.
What is sin(−π/2)? — −1.−π/2 = −90°. The terminal point is (0, −1). sin(−π/2) = −1. Or use the odd function property: sin(−θ) = −sin(θ), so sin(−π/2) = −sin(π/2) = −1.
What is sin(3π/4)? — √2/2.3π/4 = 135° is in Q2. Reference angle = π − 3π/4 = π/4. sin(π/4) = √2/2. In Q2, sine is positive: sin(3π/4) = √2/2.
For angle θ in standard position, what does the "reference angle" mean? — The positive acute angle between the terminal side and the x-axis.The reference angle is the positive acute angle formed between the terminal side of θ and the nearest part of the x-axis. It helps find trig values in any quadrant.
Two angles are coterminal if they: — Differ by a multiple of 2π.Coterminal angles have the same terminal side. They differ by multiples of 2π (full revolutions): α and α + 2nπ are coterminal for any integer n.
What is the value of sec(π/3)? — 2.sec(π/3) = 1/cos(π/3) = 1/(1/2) = 2.
What is sin(π/6) + cos(π/3)? — 1.sin(π/6) = 1/2 and cos(π/3) = 1/2. Sum = 1/2 + 1/2 = 1. Note: sin(π/6) = cos(π/3) is not a coincidence — they are complementary angles.
For y = 2sin(3x − π) + 4: what is the amplitude? — 2.A = 2. Amplitude = |A| = 2.
For y = 2sin(3(x − π/3)) + 4: what is the period? — 2π/3.Period = 2π/|B| = 2π/3.
For y = 2sin(3(x − π/3)) + 4: what is the midline? — y = 4.Midline is y = D = 4.
What is the maximum value of y = 3 cos(x) + 1? — 4.Maximum = amplitude + vertical shift = 3 + 1 = 4. (When cos(x) = 1.)
What is the minimum value of y = 5 sin(x) − 2? — −7.Minimum = −amplitude + vertical shift = −5 + (−2) = −7.
The period of y = tan(x) is: — π.tan(x) has period π (not 2π like sine and cosine). It repeats every π units.
The graph of y = sin(x + π/4) is y = sin(x) shifted: — Left π/4.y = sin(x + π/4): the + inside means LEFT shift by π/4. Horizontal shifts are counterintuitive.
From the same function (max 7, min −3): what is the midline? — y = 2.Midline = (max + min)/2 = (7 + (−3))/2 = 4/2 = 2. Midline: y = 2.
What is the period of y = tan(2x)? — π/2.Period of tan(Bx) = π/|B| = π/2.
y = A sin(Bx − C) + D. Which parameter controls horizontal shift? — C.C (divided by B) gives the horizontal phase shift. The shift is C/B to the right.
What is the range of y = 4 sin(x) + 3? — [−1, 7].Amplitude 4, midline y = 3. Range: [3 − 4, 3 + 4] = [−1, 7].
cos(2θ) in terms of cosine only: — 2cos²θ − 1.cos(2θ) = 2cos²θ − 1. (Substitute sin²θ = 1 − cos²θ into cos²θ − sin²θ.)
cos(2θ) in terms of sine only: — 1 − 2sin²θ.cos(2θ) = 1 − 2sin²θ. (Substitute cos²θ = 1 − sin²θ into cos²θ − sin²θ.)
Simplify: (1 − cos²θ)/sin θ — sin θ.1 − cos²θ = sin²θ. So sin²θ/sin θ = sin θ.
Which identity follows from dividing sin²θ + cos²θ = 1 by sin²θ? — 1 + cot²θ = csc²θ.Dividing by sin²θ: 1 + cot²θ = csc²θ.
Express sin²θ using the double-angle formula for cos(2θ): — (1 − cos2θ)/2.cos(2θ) = 1 − 2sin²θ → sin²θ = (1 − cos2θ)/2. This is the "power-reducing" formula.
Express cos²θ using the double-angle formula for cos(2θ): — (1 + cos2θ)/2.cos(2θ) = 2cos²θ − 1 → cos²θ = (1 + cos2θ)/2. This is the power-reducing formula for cosine.
Simplify: sec²θ − tan²θ — 1.From tan²θ + 1 = sec²θ, we get sec²θ − tan²θ = 1.
Which is a valid form of cos(2θ)? — cos²θ − sin²θ.cos(2θ) = cos²θ − sin²θ is a standard double-angle form.
When proving a trig identity, you should: — Work on one side only and transform it into the other.Prove identities by working on one side (usually the more complex) and transforming it into the other. Never "cross the equals sign."
What is the exact value of sin(75°)? — (√6 + √2)/4.sin(75°) = sin(45° + 30°) = sin45°cos30° + cos45°sin30° = (√2/2)(√3/2) + (√2/2)(1/2) = (√6 + √2)/4.
What is the exact value of cos(75°)? — (√6 − √2)/4.cos(75°) = cos(45° + 30°) = cos45°cos30° − sin45°sin30° = (√2/2)(√3/2) − (√2/2)(1/2) = √6/4 − √2/4 = (√6 − √2)/4.
sin(15°) can be written as sin(45° − 30°). Its exact value is: — (√6 − √2)/4.sin(15°) = sin45°cos30° − cos45°sin30° = (√2/2)(√3/2) − (√2/2)(1/2) = (√6 − √2)/4.
Using the sum identity, derive sin(2A) = ? — 2sinA cosA.sin(2A) = sin(A + A) = sinA cosA + cosA sinA = 2sinA cosA.
Using the sum identity, derive cos(2A) = ? — cos²A − sin²A.cos(2A) = cos(A + A) = cosA cosA − sinA sinA = cos²A − sin²A.
What is sin(90° + θ)? — cos θ.sin(90° + θ) = sin90°cosθ + cos90°sinθ = (1)cosθ + (0)sinθ = cosθ.
What is cos(90° − θ)? — sin θ.cos(90° − θ) = cos90°cosθ + sin90°sinθ = (0)cosθ + (1)sinθ = sinθ. (Complementary angle identity.)
What is arcsin(−√3/2)? — −π/3.sin(−π/3) = −√3/2, and −π/3 ∈ [−π/2, π/2]. So arcsin(−√3/2) = −π/3.
What is arccos(−1)? — π.cos(π) = −1, and π ∈ [0, π]. So arccos(−1) = π.
The range of arctan(x) is: — (−π/2, π/2).arctan has an open interval range (−π/2, π/2) because tan is undefined at ±π/2.
What does arcsin(sin θ) = θ require? — θ ∈ [−π/2, π/2].arcsin(sin θ) = θ only when θ is in the principal range [−π/2, π/2].
The inverse trig functions require restricted domains because: — Trig functions are periodic, so not one-to-one.Periodic functions repeat values, so they are many-to-one. We must restrict the domain to make them invertible.
Which is NOT true about y = arcsin(x)? — Its range is [0, π].The range of arcsin is [−π/2, π/2], not [0, π]. The range [0, π] belongs to arccos.
Solve for θ: 2 sin θ = √3, θ ∈ [0, 2π) — θ = π/3 and 2π/3.sin θ = √3/2. arcsin(√3/2) = π/3 (Q1). Also θ = π − π/3 = 2π/3 (Q2). Both in [0, 2π).