Grade 12
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Precalculus · Trigonometry

Master Trigonometry
From Zero to Hero

20 carefully curated problems — from core identities to real-world applications. Choose your answer and instantly see whether you're right.

20Problems
7Units
∞Retries
All Units Unit Circle Ratios & SOH-CAH-TOA Identities Graphs Inverse Trig Laws Word Problems

0 of 20 answered

Unit 1 · Unit Circle & Angle Measure
1
Unit Circle Easy High-Frequency Mistake
Convert \( 270° \) to radians.
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Memory Key
DEGREES × (π / 180) = RADIANS  ·  "Divide by 180, multiply by π"
Worked Example
Convert \(60°\) to radians:
\(60° \times \dfrac{\pi}{180°} = \dfrac{60\pi}{180} = \dfrac{\pi}{3}\)
2
Unit Circle Easy Tricky Sign
What is the exact value of \(\cos\!\left(\dfrac{5\pi}{6}\right)\)?
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Memory Key
ALL STUDENTS TAKE CALCULUS — Quadrant signs for sin/cos/tan
Q1: All +, Q2: Sin +, Q3: Tan +, Q4: Cos +
Worked Example
\(\dfrac{5\pi}{6}\) is in Quadrant II. Reference angle \(= \pi - \dfrac{5\pi}{6} = \dfrac{\pi}{6}\).
\(\cos\!\left(\dfrac{\pi}{6}\right) = \dfrac{\sqrt{3}}{2}\), but cosine is negative in Q2.
\(\therefore\;\cos\!\left(\dfrac{5\pi}{6}\right) = -\dfrac{\sqrt{3}}{2}\)
Unit 2 · Right Triangle Ratios
3
SOH-CAH-TOA Easy
In a right triangle, the side opposite angle \(\theta\) is 5 and the hypotenuse is 13. What is \(\sin\theta\)?
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Memory Key
SOH: Sin = Opposite/Hypotenuse  ·  CAH: Cos = Adjacent/Hypotenuse  ·  TOA: Tan = Opposite/Adjacent
4
SOH-CAH-TOA Medium Pythagorean Setup
If \(\cos\theta = \dfrac{8}{17}\) and \(\theta\) is in Quadrant I, find \(\tan\theta\).
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Memory Key
PYTHAGOREAN SETUP: opp² + adj² = hyp² → find missing side first, then TOA
Worked Example
\(\cos\theta = \frac{8}{17}\) → adjacent \(= 8\), hypotenuse \(= 17\)
\(\text{opposite} = \sqrt{17^2 - 8^2} = \sqrt{289 - 64} = \sqrt{225} = 15\)
\(\tan\theta = \dfrac{\text{opp}}{\text{adj}} = \dfrac{15}{8}\)
Unit 3 · Trigonometric Identities
5
Pythagorean Identity Easy Most-Tested
If \(\sin\theta = \dfrac{3}{5}\), what is \(\cos^2\!\theta\)?
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Memory Key
BIG THREE: sin²θ + cos²θ = 1  |  1 + tan²θ = sec²θ  |  1 + cot²θ = csc²θ
6
Simplify Identity Medium Tricky Cancel
Simplify: \(\dfrac{\sin^2\!\theta + \cos^2\!\theta}{\cos\theta}\)
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Memory Key
SIMPLIFY TRICK: Replace numerator with 1 first (Pythagorean), then deal with denominator
7
Double Angle Medium High-Frequency Mistake
If \(\sin\theta = \dfrac{1}{2}\), find \(\sin(2\theta)\) given \(\theta\) is in Quadrant I.
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Memory Key
DOUBLE ANGLE: sin(2θ) = 2·sinθ·cosθ  ·  cos(2θ) = cos²θ − sin²θ
Worked Example
\(\sin\theta = \tfrac{1}{2}\) → \(\cos\theta = \tfrac{\sqrt{3}}{2}\) (Q1, use Pythagorean identity)
\(\sin(2\theta) = 2\cdot\tfrac{1}{2}\cdot\tfrac{\sqrt{3}}{2} = \dfrac{\sqrt{3}}{2}\)
8
Sum Formula Hard
Use the angle addition formula to find the exact value of \(\cos(75°)\).
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Memory Key
SUM SPLIT: 75° = 45° + 30° → cos(A+B) = cosA·cosB − sinA·sinB
Worked Example
\(\cos(75°) = \cos(45°+30°)\)
\(= \cos45°\cos30° - \sin45°\sin30°\)
\(= \dfrac{\sqrt{2}}{2}\cdot\dfrac{\sqrt{3}}{2} - \dfrac{\sqrt{2}}{2}\cdot\dfrac{1}{2} = \dfrac{\sqrt{6}-\sqrt{2}}{4}\)
Unit 4 · Graphs of Trig Functions
9
Amplitude & Period Easy Most-Tested
What is the period of \(f(x) = 3\sin(2x) + 1\)?
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Memory Key
f(x) = A·sin(Bx + C) + D  →  Amplitude = |A|, Period = 2π/B, Phase Shift = −C/B
10
Phase Shift Medium Tricky Sign
Find the phase shift of \(g(x) = \cos\!\left(3x - \dfrac{\pi}{2}\right)\).
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Memory Key
PHASE SHIFT = −C/B (from Bx + C). Minus sign flips direction! − inside = shift RIGHT
Unit 5 · Solving Trig Equations
11
Solve for θ Easy
Solve for \(\theta \in [0°, 360°)\):   \(2\sin\theta - \sqrt{2} = 0\)
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Memory Key
SOLVE STEPS: Isolate trig → find reference angle → apply quadrant rules → list ALL solutions in range
12
Quadratic Trig Hard Classic Trap
Solve \(2\cos^2\!\theta - \cos\theta - 1 = 0\) for \(\theta \in [0, 2\pi)\).
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Memory Key
QUADRATIC TRIG: Treat cosθ as variable "u", factor like 2u²−u−1=0, solve each factor
Worked Example
Let \(u = \cos\theta\): \(\;2u^2 - u - 1 = 0\)
Factor: \((2u + 1)(u - 1) = 0\)
\(u = -\tfrac{1}{2}\) → \(\theta = \tfrac{2\pi}{3},\;\tfrac{4\pi}{3}\)    \(u = 1\) → \(\theta = 0\)
Unit 6 · Inverse Trigonometric Functions
13
Inverse Trig Easy Range Awareness
Evaluate: \(\arcsin\!\left(\!-\dfrac{1}{2}\right)\)
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Memory Key
RANGE LOCK: arcsin → [−π/2, π/2]  |  arccos → [0, π]  |  arctan → (−π/2, π/2). Always pick angle in that window!
14
Composite Inverse Hard Mind-Bender
Find the exact value of \(\cos\!\left(\arctan\dfrac{3}{4}\right)\).
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Memory Key
DRAW A TRIANGLE: arctan(3/4) means opp=3, adj=4 → hyp = √(9+16) = 5 → read off cosine
Worked Example
Let \(\theta = \arctan\!\dfrac{3}{4}\). Draw right triangle: opposite \(= 3\), adjacent \(= 4\), hypotenuse \(= 5\).
\(\cos\theta = \dfrac{\text{adjacent}}{\text{hypotenuse}} = \dfrac{4}{5}\)
Unit 7 · Law of Sines & Cosines
15
Law of Sines Medium
In triangle \(ABC\), \(\angle A = 45°\), \(\angle B = 60°\), and side \(a = 10\). Find side \(b\).
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Memory Key
LAW OF SINES: a/sinA = b/sinB = c/sinC  ·  Use when: AAS, ASA, SSA (ambiguous!)
16
Law of Cosines Hard
In triangle \(ABC\), \(a = 7\), \(b = 5\), \(C = 60°\). Find side \(c\).
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Memory Key
LAW OF COSINES: c² = a² + b² − 2ab·cosC  ·  Use when: SAS or SSS. Like Pythagorean theorem with extra term!
Worked Example
\(c^2 = 7^2 + 5^2 - 2(7)(5)\cos60°\)
\(= 49 + 25 - 70\cdot\tfrac{1}{2} = 74 - 35 = 39\)
\(c = \sqrt{39}\)
Unit 8 · Real-World Applications
17
Elevation Angle Medium Word Problem
A person standing 100 ft from the base of a building looks up at an angle of elevation of \(60°\) to reach the top. How tall is the building?
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Memory Key
ELEVATION/DEPRESSION: Always measured from HORIZONTAL. Draw diagram → identify opp/adj/hyp → pick right ratio
Worked Example
Adjacent \(= 100\) ft (distance from building), angle \(= 60°\).
\(\tan 60° = \dfrac{\text{height}}{100}\) → height \(= 100\tan60° = 100\sqrt{3} \approx 173.2\) ft
18
Navigation Hard Word Problem
Two ships leave port at the same time. Ship A travels 30 km due north. Ship B travels 40 km due east. How far apart are the ships?
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Memory Key
PERPENDICULAR PATHS → 90° angle → use Pythagorean theorem (or Law of Cosines with C=90°)
19
Modeling Hard Word Problem
A Ferris wheel has a radius of 20 m and its center is 25 m above the ground. It completes one full revolution every 40 seconds. Which function models the rider's height \(h(t)\) in meters?
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Memory Key
FERRIS WHEEL MODEL: h(t) = Amplitude·sin(2π/Period · t) + Midline  ·  Midline = center height, Amplitude = radius
20
Area of Triangle Hard Word Problem · Final Boss
A triangular plot of land has sides \(a = 12\) m, \(b = 9\) m, and an included angle \(C = 30°\). What is the area of the plot?
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Memory Key
TRIG AREA FORMULA: Area = ½·a·b·sinC  ·  Use when you know TWO sides and the INCLUDED angle (SAS)
Worked Example
\(\text{Area} = \dfrac{1}{2} \cdot 12 \cdot 9 \cdot \sin 30°\)
\(= \dfrac{1}{2} \cdot 108 \cdot \dfrac{1}{2} = 27 \text{ m}^2\)
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