Planned Quiz 10 practice
This practice focuses on Section 5.3 — using reference angles to evaluate trig functions for any angle, identifying quadrants from sign conditions, and finding all angles that produce a given trig value. Quiz 10 is listed in Canvas for Nov 11 at 2:10 PM; confirm official coverage there or in class.
▸Finding and using reference angles (θ′)
▸Evaluating sin, cos, tan for non-acute angles
▸Determining the quadrant from sign conditions
▸Finding two angles (0°–360°) satisfying a trig equation
▸All six trig functions from a point (x, y)
▸Sketching angles in standard position
The 3-Step Process — use this on every trig value problem
1
Find θ′
Locate the quadrant, then use its formula:
Q I: θ′ = θ
Q II: θ′ = 180° − θ
Q III: θ′ = θ − 180°
Q IV: θ′ = 360° − θ
Q I: θ′ = θ
Q II: θ′ = 180° − θ
Q III: θ′ = θ − 180°
Q IV: θ′ = 360° − θ
2
Evaluate at θ′
Look up the trig value for the reference angle using a special triangle. This step always gives a positive number — the magnitude of the answer.
3
Apply the Sign
Use the ASTC chart to decide whether the answer is positive or negative based on which quadrant the original angle θ is in.
📖 Worked Example — click to expand and see the 3 steps in action
▼
Problem: For θ = 210°, find the reference angle θ′ and then find sin(210°) exactly.
Step 1
Which quadrant? 210° is between 180° and 270° → Quadrant III.
Reference angle formula for Q III: θ′ = θ − 180°
θ′ = 210° − 180° = 30°
Reference angle formula for Q III: θ′ = θ − 180°
θ′ = 210° − 180° = 30°
Step 2
Look up sin at the reference angle:
From the 30-60-90 triangle: sin(30°) = 1/2
The magnitude of the answer is 1/2.
From the 30-60-90 triangle: sin(30°) = 1/2
The magnitude of the answer is 1/2.
Step 3
Apply the sign using ASTC:
In Quadrant III, only Tangent is positive. Sine is negative.
Final answer: sin(210°) = −1/2
In Quadrant III, only Tangent is positive. Sine is negative.
Final answer: sin(210°) = −1/2
Bonus
What if the angle is negative, like θ = −5π/6?
Negative angles rotate clockwise. First add 2π to get a positive co-terminal angle:
−5π/6 + 2π = 7π/6 (which is 210° — same situation as above!)
Then follow the same 3 steps. sin(−5π/6) = −1/2
Negative angles rotate clockwise. First add 2π to get a positive co-terminal angle:
−5π/6 + 2π = 7π/6 (which is 210° — same situation as above!)
Then follow the same 3 steps. sin(−5π/6) = −1/2
ASTC Sign Chart — which functions are positive?
Q II
S positive
sin, csc ✓
cos & tan negative
cos & tan negative
Q I
All positive
sin, cos,
tan all ✓
tan all ✓
Q III
T positive
tan, cot ✓
sin & cos negative
sin & cos negative
Q IV
C positive
cos, sec ✓
sin & tan negative
sin & tan negative
Memory: "All Students Take Calculus"
Special Triangle Values
30–60–90
sin 30° = 1/2
cos 30° = √3/2
sin 60° = √3/2
cos 60° = 1/2
cos 30° = √3/2
sin 60° = √3/2
cos 60° = 1/2
45–45–90
sin 45° = √2/2
cos 45° = √2/2
tan 45° = 1
cos 45° = √2/2
tan 45° = 1
📍 Worked Example — Six Trig Functions from a Point
▼
Problem: The point (−3, −4) lies on the terminal side of angle θ. Find r and all six trig functions of θ.
Step 1
Identify x and y, then compute r:
x = −3, y = −4
r = √(x² + y²) = √(9 + 16) = √25 = 5
Note: r is always positive — it is a distance.
x = −3, y = −4
r = √(x² + y²) = √(9 + 16) = √25 = 5
Note: r is always positive — it is a distance.
Step 2
Apply the definitions directly — signs come from x and y, not from a quadrant rule:
sin θ = y/r = −4/5
cos θ = x/r = −3/5
tan θ = y/x = (−4)/(−3) = 4/3 ← both negative, so tan is positive (Q III ✓)
sin θ = y/r = −4/5
cos θ = x/r = −3/5
tan θ = y/x = (−4)/(−3) = 4/3 ← both negative, so tan is positive (Q III ✓)
Step 3
Find the three reciprocal functions by flipping:
csc θ = 1/sin θ = r/y = 5/(−4) = −5/4
sec θ = 1/cos θ = r/x = 5/(−3) = −5/3
cot θ = 1/tan θ = x/y = (−3)/(−4) = 3/4
csc θ = 1/sin θ = r/y = 5/(−4) = −5/4
sec θ = 1/cos θ = r/x = 5/(−3) = −5/3
cot θ = 1/tan θ = x/y = (−3)/(−4) = 3/4
Check
Sanity-check with ASTC: (−3, −4) is in Quadrant III — only tangent and cotangent should be positive. ✓
sin and cos are negative ✓ · tan and cot are positive ✓ · csc and sec are negative ✓
sin and cos are negative ✓ · tan and cot are positive ✓ · csc and sec are negative ✓
Section A — Reference Angles
1
What is the reference angle for θ = 150°?
150° falls between 90° and 180° — that's Quadrant II. For Q II, the formula is θ′ = 180° − θ.
2
Find the reference angle for θ = −5π/6.
Negative angles rotate clockwise. Add 2π to get a positive co-terminal angle first: −5π/6 + 2π = 7π/6. Now, 7π/6 is between π and 3π/2 — which quadrant is that? Then apply the right formula.
3
What is the reference angle for θ = 300°?
300° is between 270° and 360°, placing it in Quadrant IV. For Q IV: θ′ = 360° − θ.
Section B — Exact Trig Values
4
Find the exact value of cos(240°).
Step 1: 240° is in Q III → θ′ = 240° − 180° = 60°.
Step 2: From your special triangle: cos(60°) = 1/2.
Step 3: Check your ASTC chart — is cosine positive or negative in Q III?
Step 2: From your special triangle: cos(60°) = 1/2.
Step 3: Check your ASTC chart — is cosine positive or negative in Q III?
5
Find the exact value of tan(135°).
135° is in Q II. Reference angle: 180° − 135° = 45°. From the 45-45-90 triangle: tan(45°) = 1. Now check: is tangent positive or negative in Q II?
Section C — Quadrant Identification
6
The terminal side of θ satisfies cos θ > 0 and sin θ < 0. In which quadrant does θ lie?
Translate into coordinates: since r > 0 always, cos θ = x/r > 0 means x > 0 (right side) and sin θ = y/r < 0 means y < 0 (below x-axis). Which quadrant is right and below?
7
Identify the quadrant where csc θ > 0 and cos θ < 0.
First, translate the reciprocal: csc θ = 1/sin θ, so csc > 0 means sin > 0 as well (same sign). Now you have: sin > 0 and cos < 0. Which two quadrants have sin > 0? Of those, which also has cos < 0?
Section D — Finding Angles from a Trig Value
8
Find two angles between 0° and 360° satisfying sin θ = −√3/2.
1) Ignore the sign first — what reference angle gives sin θ′ = √3/2? (Look at your 30-60-90 triangle.)
2) The value is negative, so θ must be in a quadrant where sine is negative. That's Q III and Q IV.
3) Build the two angles: Q III → 180° + θ′, and Q IV → 360° − θ′.
2) The value is negative, so θ must be in a quadrant where sine is negative. That's Q III and Q IV.
3) Build the two angles: Q III → 180° + θ′, and Q IV → 360° − θ′.
Section E — Six Trig Functions from a Point
9
The point (−4, 3) lies on the terminal side of θ. Find r, then compute sin θ, cos θ, and tan θ.
With x = −4 and y = 3:
r = √(x² + y²) — always positive.
sin θ = y/r · cos θ = x/r · tan θ = y/x
Pay close attention to signs — x is negative, so cos θ will be negative too.
r = √(x² + y²) — always positive.
sin θ = y/r · cos θ = x/r · tan θ = y/x
Pay close attention to signs — x is negative, so cos θ will be negative too.
10
Solve the right triangle. Given the labeled triangle below, find the missing side and both unknown angles. Round angles to the nearest tenth of a degree.
Missing side: Use the Pythagorean theorem — a² + b² = c². You know two sides.
Angle β: You know the side opposite β and the hypotenuse, so use sin β = opp/hyp → β = sin⁻¹(opp/hyp).
Angle α: Either use inverse trig on another ratio, or use the fact that the three angles of a triangle sum to 180°.
Angle β: You know the side opposite β and the hypotenuse, so use sin β = opp/hyp → β = sin⁻¹(opp/hyp).
Angle α: Either use inverse trig on another ratio, or use the fact that the three angles of a triangle sum to 180°.
Fill in all missing measurements:
| Element | Value |
|---|---|
| Side BC | 21 |
| Side AC (b) | |
| Hypotenuse AB | 29 |
| Angle β (at B) | |
| Angle α (at A) | |
| Angle C | 90° |