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Precalculus
Trigonometry

20 Exam-Style Multiple Choice Questions · All Core Topics

📐 Unit Circle 📊 Graphing Trig 🔁 Identities ⚡ Solving Equations 📐 Law of Cosines
TIME 30:00
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01   CONCEPT REVIEW
TOPIC 01 The Unit Circle & Exact Values

The unit circle has radius 1, centered at the origin. For any angle \(\theta\), the coordinates on the unit circle are \((\cos\theta,\, \sin\theta)\).

Key Exact Values to Memorize:

\(\sin 0 = 0,\quad \cos 0 = 1,\quad \tan 0 = 0\)

\(\sin\tfrac{\pi}{6} = \tfrac{1}{2},\quad \cos\tfrac{\pi}{6} = \tfrac{\sqrt{3}}{2},\quad \tan\tfrac{\pi}{6} = \tfrac{1}{\sqrt{3}}\)

\(\sin\tfrac{\pi}{4} = \tfrac{\sqrt{2}}{2},\quad \cos\tfrac{\pi}{4} = \tfrac{\sqrt{2}}{2},\quad \tan\tfrac{\pi}{4} = 1\)

\(\sin\tfrac{\pi}{3} = \tfrac{\sqrt{3}}{2},\quad \cos\tfrac{\pi}{3} = \tfrac{1}{2},\quad \tan\tfrac{\pi}{3} = \sqrt{3}\)

\(\sin\tfrac{\pi}{2} = 1,\quad \cos\tfrac{\pi}{2} = 0,\quad \tan\tfrac{\pi}{2} = \text{undefined}\)

Quadrant Signs (ASTC): All positive (Q1) → Sine positive (Q2) → Tangent positive (Q3) → Cosine positive (Q4)
TOPIC 02 Trigonometric Identities

Pythagorean Identities:

\(\sin^2\theta + \cos^2\theta = 1\)

\(\tan^2\theta + 1 = \sec^2\theta\)

\(1 + \cot^2\theta = \csc^2\theta\)

Sum & Difference:

\(\sin(A \pm B) = \sin A\cos B \pm \cos A\sin B\)

\(\cos(A \pm B) = \cos A\cos B \mp \sin A\sin B\)

Double Angle:

\(\sin 2\theta = 2\sin\theta\cos\theta\)

\(\cos 2\theta = \cos^2\theta - \sin^2\theta = 1-2\sin^2\theta = 2\cos^2\theta-1\)

\(\tan 2\theta = \dfrac{2\tan\theta}{1-\tan^2\theta}\)

Co-function & Even/Odd:

\(\cos(\pi - \theta) = -\cos\theta, \quad \sin(\pi - \theta) = \sin\theta\)

\(\sin(-\theta) = -\sin\theta\) (odd), \quad \cos(-\theta) = \cos\theta\) (even)

TOPIC 03 Graphing Trig Functions

For \(y = A\sin(Bx + C) + D\):

Amplitude \(= |A|\)

Period \(= \dfrac{2\pi}{|B|}\)

Phase Shift \(= -\dfrac{C}{B}\) (positive = right)

Vertical Shift \(= D\)

Remember: Period of \(\tan\) and \(\cot\) is \(\pi\), not \(2\pi\).
TOPIC 04 Solving Trig Equations & Reference Angles
  • Reference angle: the acute angle between the terminal side and the x-axis.
  • For \(\theta = 210°\): reference angle \(= 210° - 180° = 30°\).
  • Use ASTC to determine sign in each quadrant.
  • General solutions in \([0, 2\pi)\): find all quadrants where the function has the given sign.

If \(\sin\theta = k\), then solutions in \([0, 2\pi)\) are in Q1 and Q2 (if \(k>0\)), or Q3 and Q4 (if \(k<0\)).

TOPIC 05 Law of Cosines & Triangle Area

Law of Cosines: \(c^2 = a^2 + b^2 - 2ab\cos C\)

Triangle Area: \(\text{Area} = \tfrac{1}{2}ab\sin C\)

Radian Conversion: \(\text{radians} = \text{degrees} \times \dfrac{\pi}{180}\)

02   WORKED EXAMPLES
EXAMPLE 01 — Unit Circle
Find the exact value of \(\sin\!\left(\tfrac{\pi}{6}\right) + \cos\!\left(\tfrac{\pi}{3}\right)\).
  1. 1Recall: \(\sin\!\left(\tfrac{\pi}{6}\right) = \tfrac{1}{2}\) (30° on unit circle)
  2. 2Recall: \(\cos\!\left(\tfrac{\pi}{3}\right) = \tfrac{1}{2}\) (60° on unit circle)
  3. 3Sum: \(\tfrac{1}{2} + \tfrac{1}{2} = 1\)
✓ Answer: 1
EXAMPLE 02 — Double Angle
If \(\sin\theta = \tfrac{1}{3}\), find \(\cos 2\theta\).
  1. 1Use identity: \(\cos 2\theta = 1 - 2\sin^2\theta\)
  2. 2Substitute: \(1 - 2\left(\tfrac{1}{3}\right)^2 = 1 - \tfrac{2}{9} = \tfrac{7}{9}\)
✓ Answer: 7/9
EXAMPLE 03 — Law of Cosines
In triangle \(ABC\), \(a = 5\), \(b = 7\), and \(C = 60°\). Find side \(c\).
  1. 1Law of Cosines: \(c^2 = a^2 + b^2 - 2ab\cos C\)
  2. 2\(c^2 = 25 + 49 - 2(5)(7)\cos 60° = 74 - 70 \cdot \tfrac{1}{2} = 74 - 35 = 39\)
  3. 3\(c = \sqrt{39}\)
✓ Answer: √39
★ PRACTICE EXAM ★
03   EXAM QUESTIONS
Unit Circle
ASTC / Quadrants
Exact Values
Graphing
Identities
Double Angle
Sum & Difference
Solving Equations
Reference Angles
Law of Cosines
01 Unit Circle · Exact Values Easy
Find the exact value of \(\sin\!\left(\dfrac{\pi}{6}\right) + \cos\!\left(\dfrac{\pi}{3}\right)\).
02 ASTC · Quadrant II Easy
If \(\sin\theta = \dfrac{3}{5}\) and \(\theta\) is in Quadrant II, what is \(\cos\theta\)?
03 Exact Values Easy
Compute \(\tan\!\left(\dfrac{\pi}{4}\right) + \sin\!\left(\dfrac{\pi}{2}\right)\).
04 Graphing · Period Easy
What is the period of \(y = \sin(2x)\)?
05 Co-function Identity Medium
Which expression equals \(\cos(\pi - \theta)\)?
06 Graphing · Amplitude Easy
What is the amplitude of \(y = 3\sin(x) + 1\)?
07 Pythagorean Identity Easy
Which identity is correct?
08 Solving Trig Equations Medium
Solve \(\sin\theta = \dfrac{1}{2}\) for \(\theta \in [0,\, 2\pi)\). Which lists ALL solutions?
09 Sum & Difference Formula Medium
Using the sum formula, find the exact value of \(\sin\!\left(\dfrac{\pi}{3} + \dfrac{\pi}{4}\right)\).
10 Double Angle Formula Medium
If \(\sin\theta = \dfrac{1}{3}\), find the exact value of \(\cos 2\theta\).
11 Graphing · Vertical Shift Easy
What is the vertical shift of \(y = \sin(x) - 2\)?
12 Reference Angles Easy
What is the reference angle for \(210°\)?
13 Even / Odd Functions Easy
Which expression equals \(\sin(-\theta)\)?
14 Graphing · Phase Shift Medium
What is the phase shift of \(y = \sin\!\left(x - \dfrac{\pi}{3}\right)\)?
15 Law of Cosines Hard
In triangle \(ABC\), \(a = 5\), \(b = 7\), and \(C = 60°\). Find the exact length of side \(c\).
16 Radian / Degree Conversion Easy
Convert \(135°\) to radians.
17 ASTC · Quadrant III Medium
If \(\cos\theta = -\dfrac{\sqrt{3}}{2}\) and \(\theta\) is in Quadrant III, what is \(\sin\theta\)?
18 Double Angle · Tangent Hard
If \(\tan\theta = 2\), find \(\tan 2\theta\).
19 Triangle Area Formula Hard
Find the area of a triangle with sides \(a = 6\), \(b = 8\), and included angle \(C = 45°\).
20 Solving Trig Equations Hard
How many solutions does \(\sin^2\theta = \dfrac{1}{4}\) have in \([0,\, 2\pi)\)?
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★ FULL SOLUTIONS ★
04   ANSWER KEY & EXPLANATIONS
Q01 Unit Circle · Exact Values Correct: B
Find \(\sin\!\left(\tfrac{\pi}{6}\right) + \cos\!\left(\tfrac{\pi}{3}\right)\).
  1. 1From the unit circle, \(\sin\!\left(\tfrac{\pi}{6}\right) = \tfrac{1}{2}\) (this is 30°).
  2. 2From the unit circle, \(\cos\!\left(\tfrac{\pi}{3}\right) = \tfrac{1}{2}\) (this is 60°).
  3. 3\(\tfrac{1}{2} + \tfrac{1}{2} = 1\).
✓ Answer: B — 1
Q02 ASTC · Quadrant II Correct: A
If \(\sin\theta = \tfrac{3}{5}\), Quadrant II, find \(\cos\theta\).
  1. 1Pythagorean identity: \(\cos^2\theta = 1 - \sin^2\theta = 1 - \tfrac{9}{25} = \tfrac{16}{25}\).
  2. 2So \(\cos\theta = \pm\tfrac{4}{5}\). In Quadrant II, cosine is negative.
  3. 3\(\cos\theta = -\tfrac{4}{5}\).
✓ Answer: A — −4/5
Q03 Exact Values Correct: C
Compute \(\tan\!\left(\tfrac{\pi}{4}\right) + \sin\!\left(\tfrac{\pi}{2}\right)\).
  1. 1\(\tan\!\left(\tfrac{\pi}{4}\right) = 1\) (45° on unit circle: opposite = adjacent).
  2. 2\(\sin\!\left(\tfrac{\pi}{2}\right) = 1\) (top of unit circle).
  3. 3Sum: \(1 + 1 = 2\).
✓ Answer: C — 2
Q04 Graphing · Period Correct: B
Period of \(y = \sin(2x)\).
  1. 1Period formula: \(T = \tfrac{2\pi}{|B|}\) where \(B\) is the coefficient of \(x\).
  2. 2Here \(B = 2\), so \(T = \tfrac{2\pi}{2} = \pi\).
✓ Answer: B — π
Q05 Co-function Identity Correct: D
Simplify \(\cos(\pi - \theta)\).
  1. 1Apply the difference formula: \(\cos(\pi-\theta) = \cos\pi\cos\theta + \sin\pi\sin\theta\).
  2. 2\(\cos\pi = -1\) and \(\sin\pi = 0\), so the expression becomes \((-1)\cos\theta + 0 = -\cos\theta\).
✓ Answer: D — −cos θ
Q06 Graphing · Amplitude Correct: A
Amplitude of \(y = 3\sin(x) + 1\).
  1. 1In \(y = A\sin(Bx+C)+D\), the amplitude is \(|A|\).
  2. 2Here \(A = 3\), \(D = 1\) (vertical shift). Amplitude \(= |3| = 3\). The \(+1\) shifts the graph up but does not change amplitude.
✓ Answer: A — 3
Q07 Pythagorean Identity Correct: C
Which identity is correct?
  1. 1Start from \(\sin^2\theta + \cos^2\theta = 1\). Divide every term by \(\cos^2\theta\).
  2. 2\(\tan^2\theta + 1 = \sec^2\theta\). This is the standard Pythagorean identity for tangent/secant.
  3. 3Options A, B, D are all incorrect algebraically.
✓ Answer: C — tan²θ + 1 = sec²θ
Q08 Solving Trig Equations Correct: B
Solve \(\sin\theta = \tfrac{1}{2}\) for \(\theta \in [0, 2\pi)\).
  1. 1The reference angle where \(\sin = \tfrac{1}{2}\) is \(\tfrac{\pi}{6}\) (30°).
  2. 2Sine is positive in Quadrants I and II, so the two solutions are \(\theta = \tfrac{\pi}{6}\) (Q1) and \(\theta = \pi - \tfrac{\pi}{6} = \tfrac{5\pi}{6}\) (Q2).
✓ Answer: B — π/6 and 5π/6
Q09 Sum Formula Correct: D
Find \(\sin\!\left(\tfrac{\pi}{3} + \tfrac{\pi}{4}\right)\).
  1. 1\(\sin(A+B) = \sin A\cos B + \cos A\sin B\).
  2. 2\(= \sin\tfrac{\pi}{3}\cos\tfrac{\pi}{4} + \cos\tfrac{\pi}{3}\sin\tfrac{\pi}{4}\).
  3. 3\(= \tfrac{\sqrt{3}}{2}\cdot\tfrac{\sqrt{2}}{2} + \tfrac{1}{2}\cdot\tfrac{\sqrt{2}}{2} = \tfrac{\sqrt{6}}{4} + \tfrac{\sqrt{2}}{4} = \dfrac{\sqrt{6}+\sqrt{2}}{4}\).
✓ Answer: D — (√6 + √2)/4
Q10 Double Angle Formula Correct: C
If \(\sin\theta = \tfrac{1}{3}\), find \(\cos 2\theta\).
  1. 1Use the identity: \(\cos 2\theta = 1 - 2\sin^2\theta\).
  2. 2Substitute: \(1 - 2\left(\tfrac{1}{3}\right)^2 = 1 - 2\cdot\tfrac{1}{9} = 1 - \tfrac{2}{9} = \tfrac{7}{9}\).
✓ Answer: C — 7/9
Q11 Graphing · Vertical Shift Correct: B
Vertical shift of \(y = \sin(x) - 2\).
  1. 1In \(y = A\sin(Bx+C)+D\), the vertical shift is \(D\).
  2. 2Here \(D = -2\), meaning the graph shifts 2 units downward.
✓ Answer: B — 2 units down
Q12 Reference Angles Correct: A
Reference angle for 210°.
  1. 1210° lies in Quadrant III (between 180° and 270°).
  2. 2Reference angle \(= 210° - 180° = 30°\).
✓ Answer: A — 30°
Q13 Even / Odd Functions Correct: C
Simplify \(\sin(-\theta)\).
  1. 1Sine is an odd function: \(f(-x) = -f(x)\).
  2. 2Therefore \(\sin(-\theta) = -\sin\theta\). (Contrast: \(\cos(-\theta) = \cos\theta\), cosine is even.)
✓ Answer: C — −sin θ
Q14 Graphing · Phase Shift Correct: B
Phase shift of \(y = \sin\!\left(x - \tfrac{\pi}{3}\right)\).
  1. 1Phase shift formula: \(-\tfrac{C}{B}\) where \(y = \sin(Bx + C)\).
  2. 2Rewrite as \(\sin(1 \cdot x + (-\tfrac{\pi}{3}))\): so \(C = -\tfrac{\pi}{3}\), \(B = 1\).
  3. 3Phase shift \(= -\tfrac{-\pi/3}{1} = +\tfrac{\pi}{3}\). Positive means shift right by \(\tfrac{\pi}{3}\).
✓ Answer: B — π/3 to the right
Q15 Law of Cosines Correct: A
Find \(c\) given \(a=5, b=7, C=60°\).
  1. 1Law of Cosines: \(c^2 = a^2 + b^2 - 2ab\cos C\).
  2. 2\(c^2 = 25 + 49 - 2(5)(7)\cos 60° = 74 - 70 \cdot \tfrac{1}{2} = 74 - 35 = 39\).
  3. 3\(c = \sqrt{39}\).
✓ Answer: A — √39
Q16 Radian Conversion Correct: C
Convert 135° to radians.
  1. 1Multiply by \(\tfrac{\pi}{180}\): \(135 \times \tfrac{\pi}{180}\).
  2. 2Simplify: \(\tfrac{135}{180} = \tfrac{3}{4}\), so the result is \(\tfrac{3\pi}{4}\).
✓ Answer: C — 3π/4
Q17 ASTC · Quadrant III Correct: B
If \(\cos\theta = -\tfrac{\sqrt{3}}{2}\), Quadrant III, find \(\sin\theta\).
  1. 1Pythagorean identity: \(\sin^2\theta = 1 - \cos^2\theta = 1 - \tfrac{3}{4} = \tfrac{1}{4}\).
  2. 2So \(\sin\theta = \pm\tfrac{1}{2}\). In Quadrant III, sine is negative.
  3. 3\(\sin\theta = -\tfrac{1}{2}\).
✓ Answer: B — −1/2
Q18 Double Angle · Tangent Correct: D
If \(\tan\theta = 2\), find \(\tan 2\theta\).
  1. 1Double angle formula: \(\tan 2\theta = \dfrac{2\tan\theta}{1 - \tan^2\theta}\).
  2. 2Substitute \(\tan\theta = 2\): \(\dfrac{2(2)}{1 - 4} = \dfrac{4}{-3} = -\dfrac{4}{3}\).
✓ Answer: D — −4/3
Q19 Triangle Area Correct: A
Area of triangle with \(a=6\), \(b=8\), \(C=45°\).
  1. 1Area formula: \(\text{Area} = \tfrac{1}{2}ab\sin C\).
  2. 2\(\text{Area} = \tfrac{1}{2}(6)(8)\sin 45° = 24 \cdot \tfrac{\sqrt{2}}{2} = 12\sqrt{2}\).
✓ Answer: A — 12√2
Q20 Solving Trig Equations Correct: C
How many solutions does \(\sin^2\theta = \tfrac{1}{4}\) have in \([0, 2\pi)\)?
  1. 1Take the square root: \(\sin\theta = \pm\tfrac{1}{2}\).
  2. 2For \(\sin\theta = +\tfrac{1}{2}\): solutions are \(\tfrac{\pi}{6}\) (Q1) and \(\tfrac{5\pi}{6}\) (Q2).
  3. 3For \(\sin\theta = -\tfrac{1}{2}\): solutions are \(\tfrac{7\pi}{6}\) (Q3) and \(\tfrac{11\pi}{6}\) (Q4).
  4. 4Total solutions: 4.
✓ Answer: C — 4 solutions

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