2.2.11 Trigonometric Values

Trigonometry problems have been increasingly popular for writers of the number sense test.

Special Angles (Quadrant I)

Function30°45°60°90°
sin01/22/2\sqrt{2}/23/2\sqrt{3}/21
cos13/2\sqrt{3}/22/2\sqrt{2}/21/20
tan03/3\sqrt{3}/313\sqrt{3}Undef
cscUndef22\sqrt{2}23/32\sqrt{3}/31
sec123/32\sqrt{3}/32\sqrt{2}2Undef
cotUndef3\sqrt{3}13/3\sqrt{3}/30

Signs in Quadrants (ASTC)

  • Quadrant I (All): All positive
  • Quadrant II (Students): Sin/Csc positive
  • Quadrant III (Take): Tan/Cot positive
  • Quadrant IV (Calculus): Cos/Sec positive

Reference Angles

QuadrantReference Angle Formula
Iθ\theta
II180θ180^\circ - \theta
IIIθ180\theta - 180^\circ
IV360θ360^\circ - \theta

Example: sin(210)\sin(210^\circ) Q-III (sin is negative). Ref angle = 210180=30210^\circ - 180^\circ = 30^\circ. sin(210)=sin(30)=1/2\sin(210^\circ) = -\sin(30^\circ) = -1/2.

Problem Set 2.2.11

sin(-30°)
cos θ = .375 then sec θ =
sin(3π)
tan(225°)
sin(sin⁻¹(1/2))
sin θ = -.1 then csc θ =
sin(11π/6)
cos(-5π)
π/18 = ___ °
cos(sec⁻¹ 3)
5π/8 = ___ °
π/5 = ___ °
cos(sin⁻¹ 1)
tan(-45°)
sin(-π)
cos(-300°)
sin⁻¹(sin 1)
csc(-150°)
sec(120°)
tan(-225°)
3π/5 = ___ °
tan(-45°)
tan(315°)
If 0° < x < 90° and tan x = cot x, x =
280° = kπ then k =
tan(5π/4)
cos θ = .08333... then sec θ =
sin(5π) + cos(5π)
sec(60°)
12° = π/k, k =
cos θ = -.25 then sec θ =
tan² 60°
1.25π = ___ °
cot² 60°
sin(cos⁻¹(√2/2))
cos(-3π) - sin(-3π)
cos(-4π/3) + sin(-5π/6)
2 sin 120° cos 30°
cos(240°) - sin(150°)
sin(cos⁻¹(√3/2))
sin(cos⁻¹ 1)
If csc θ = -3, where 270° < θ < 300°, then sin θ =
sin(-7π/6) - cos(-2π/3)
sec θ = -3, θ is in QIII, then cos θ =
cos(5π/6) × sin(2π/3)
sin(3π/4) × cos(5π/4)
sin 30° + cos 60° = tan x, 0° ≤ x ≤ 90°, x =
cos(sin⁻¹(√3/2))
sin(-π/3) × sin(π/3)
cos(120°) × cos(120°)
216° = kπ, k =
cos(-2π/3) × cos(4π/3)
tan(30°) × cot(60°)
cos(-π/3) × cos(π/3)
sin(π/6) + cos(π/3) = tan(π/k) then k =
cos⁻¹(.8) + cos⁻¹(.6) = kπ then k =
sin(300°) × cos(330°)
sin(-π/6) × cos(π/3)
630° = kπ, k =