1.7 Trigonometry: Right Triangle Ratios, Unit Circle & Basic Identities

Ateneo College Entrance Test ACET

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1 Chapter 1: Mathematics

1.1 Real Numbers, Ratios, Rates & Percentages 1.2 Algebraic Expressions, Polynomials & Linear Equations 1.3 Functions, Domain/Range, Graphs & Transformations 1.4 Quadratic, Polynomial, Exponential & Logarithmic Equations 1.5 Plane Geometry: Triangles, Circles, Polygons & Perimeter/Area 1.6 Coordinate Geometry: Lines, Distance, Midpoints & Conic Sections 1.7 Trigonometry: Right Triangle Ratios, Unit Circle & Basic Identities 1.8 Arithmetic & Geometric Sequences, Series & Word Patterns 1.9 Permutations, Combinations, Probability & Statistics

2 Chapter 2: English Proficiency

2.1 Subject-Verb Agreement, Verb Tenses & Aspect 2.2 Pronoun-Antecedent Agreement, Case, Misplaced & Dangling Modifiers 2.3 Sentence Structure, Run-Ons, Fragments & Parallelism 2.4 Contextual Vocabulary, Word Roots, Prefixes & Suffixes 2.5 Advanced Synonyms, Antonyms & Nuanced Vocabulary 2.6 Verbal Analogies: Relationship Types & Solving Strategies 2.7 Punctuation, Capitalization, Mechanics & Common Errors 2.8 Diction, Idiomatic Expressions, Prepositions & Frequently Confused Words

3 Chapter 3: Reading Comprehension

3.1 Identifying Main Ideas, Central Themes & Paragraph Summaries 3.2 Locating Explicit Details, Supporting Evidence & Data 3.3 Drawing Logical Inferences & Unstated Conclusions 3.4 Analyzing Author's Purpose, Point of View, Tone & Attitude 3.5 Interpreting Figurative Language, Tone Words & Context Clues 3.6 Text Structure, Rhetorical Devices & Evaluating Arguments

4 Chapter 4: Logical and Abstract Reasoning

4.1 Number & Letter Series Patterns & Formulas 4.2 Abstract Reasoning: Matrix Patterns & Figure Rotations 4.3 Formal Syllogisms, Categorical Logic & Deductive Proofs 4.4 Coding-Decoding, Cipher Patterns & Symbol Operations 4.5 Spatial Visualization, Paper Folding & Venn Diagram Logic

5 Chapter 5: General Knowledge

5.1 Philippine History, Government, Constitution & Civics 5.2 Integrated Science: Biology, Chemistry, Physics & Earth Science 5.3 Philippine & World Geography, Basic Economics & Cultural Facts 5.4 Current Events, World Landmarks, Literature & General Trivia

6 Chapter 6: ACET Essay Writing & Exam Strategies

6.1 ACET Essay: Thesis Development, Structuring & Argumentation 6.2 Time Allocation, Pacing Strategies & Subtest Tactics 6.3 Guessing Strategies (No Penalty), Test Anxiety & Exam Day Readiness

1.7 Trigonometry: Right Triangle Ratios, Unit Circle & Basic Identities

Key Takeaways Explain This Topic AI

Memorize exact trigonometric values for special reference angles (0, 30, 45, 60, and 90 degrees).

Use ASTC quadrant rules (All Students Take Calculus) to assign signs to trigonometric functions across all four quadrants.

Apply fundamental Pythagorean identities (sin^2 + cos^2 = 1, 1 + tan^2 = sec^2) to simplify complex trigonometric expressions.

Solve oblique triangles using Law of Sines and Law of Cosines appropriately.

Last updated: July 2026

Trigonometry: Right Triangle Ratios, Unit Circle & Basic Identities

Trigonometry on the ACET evaluates right triangle ratios, unit circle values, identity simplifications, and oblique triangle laws.

1. Right Triangle Trigonometric Ratios (SOH CAH TOA)

For a right triangle with acute angle θ \theta θ : sin ⁡ θ = Opposite Hypotenuse csc ⁡ θ = Hypotenuse Opposite = 1 sin ⁡ θ cos ⁡ θ = Adjacent Hypotenuse sec ⁡ θ = Hypotenuse Adjacent = 1 cos ⁡ θ tan ⁡ θ = Opposite Adjacent cot ⁡ θ = Adjacent Opposite = 1 tan ⁡ θ \begin{aligned}

\sin \theta &= \frac{\text{Opposite}}{\text{Hypotenuse}} & \csc \theta &= \frac{\text{Hypotenuse}}{\text{Opposite}} = \frac{1}{\sin \theta}

\\ \cos \theta &= \frac{\text{Adjacent}}{\text{Hypotenuse}} & \sec \theta &= \frac{\text{Hypotenuse}}{\text{Adjacent}} = \frac{1}{\cos \theta}

\\ \tan \theta &= \frac{\text{Opposite}}{\text{Adjacent}} & \cot \theta &= \frac{\text{Adjacent}}{\text{Opposite}} = \frac{1}{\tan \theta}

\end{aligned} sin θ cos θ tan θ ​ = Hypotenuse Opposite ​ = Hypotenuse Adjacent ​ = Adjacent Opposite ​ ​ csc θ sec θ cot θ ​ = Opposite Hypotenuse ​ = sin θ 1 ​ = Adjacent Hypotenuse ​ = cos θ 1 ​ = Opposite Adjacent ​ = tan θ 1 ​ ​

2. Special Angle Values & The Unit Circle

The unit circle x 2 + y 2 = 1 x^2 + y^2 = 1 x 2 + y 2 = 1 defines trigonometric values for any angle θ \theta θ where x = cos ⁡ θ x = \cos \theta x = cos θ and y = sin ⁡ θ y = \sin \theta y = sin θ .

Exact Values Table

Angle θ \theta θ (Deg) Angle θ \theta θ (Rad) sin ⁡ θ \sin \theta sin θ cos ⁡ θ \cos \theta cos θ tan ⁡ θ \tan \theta tan θ

• 0 ∘ 0^\circ 0 ∘ 0 0 0 0 0 0 1 1 1 0 0 0
• 30 ∘ 30^\circ 3 0 ∘ π / 6 \pi/6 π /6 1 / 2 1/2 1/2 3 / 2 \sqrt{3}/2 3 ​ /2 3 / 3 \sqrt{3}/3 3 ​ /3
• 45 ∘ 45^\circ 4 5 ∘ π / 4 \pi/4 π /4 2 / 2 \sqrt{2}/2 2 ​ /2 2 / 2 \sqrt{2}/2 2 ​ /2 1 1 1
• 60 ∘ 60^\circ 6 0 ∘ π / 3 \pi/3 π /3 3 / 2 \sqrt{3}/2 3 ​ /2 1 / 2 1/2 1/2 3 \sqrt{3} 3 ​
• 90 ∘ 90^\circ 9 0 ∘ π / 2 \pi/2 π /2 1 1 1 0 0 0 Undefined
• 180 ∘ 180^\circ 18 0 ∘ π \pi π 0 0 0 − 1 -1 − 1 0 0 0
• 270 ∘ 270^\circ 27 0 ∘ 3 π / 2 3\pi/2 3 π /2 − 1 -1 − 1 0 0 0 Undefined
• Quadrant Signs: The ASTC Rule

Quadrant I ( 0 ∘ 0^\circ 0 ∘ to 90 ∘ 90^\circ 9 0 ∘ ): A ll functions are POSITIVE.

Quadrant II ( 90 ∘ 90^\circ 9 0 ∘ to 180 ∘ 180^\circ 18 0 ∘ ): S ine ( & csc ⁡ \& \csc & csc ) is POSITIVE.

Quadrant III ( 180 ∘ 180^\circ 18 0 ∘ to 270 ∘ 270^\circ 27 0 ∘ ): T angent ( & cot ⁡ \& \cot & cot ) is POSITIVE.

Quadrant IV ( 270 ∘ 270^\circ 27 0 ∘ to 360 ∘ 360^\circ 36 0 ∘ ): C osine ( & sec ⁡ \& \sec & sec ) is POSITIVE.

3. Fundamental Trigonometric Identities

Pythagorean Identities sin ⁡ 2 θ + cos ⁡ 2 θ = 1 \sin^2 \theta + \cos^2 \theta = 1 sin 2 θ + cos 2 θ = 1 1 + tan ⁡ 2 θ = sec ⁡ 2 θ 1 + \tan^2 \theta = \sec^2 \theta 1 + tan 2 θ = sec 2 θ 1 + cot ⁡ 2 θ = csc ⁡ 2 θ 1 + \cot^2 \theta = \csc^2 \theta 1 + cot 2 θ = csc 2 θ

Co-Function Identities sin ⁡ ( 90 ∘ − θ ) = cos ⁡ θ and cos ⁡ ( 90 ∘ − θ ) = sin ⁡ θ \sin(90^\circ - \theta) = \cos \theta \quad \text{and} \quad \cos(90^\circ - \theta) = \sin \theta sin ( 9 0 ∘ − θ ) = cos θ and cos ( 9 0 ∘ − θ ) = sin θ tan ⁡ ( 90 ∘ − θ ) = cot ⁡ θ and sec ⁡ ( 90 ∘ − θ ) = csc ⁡ θ \tan(90^\circ - \theta) = \cot \theta \quad \text{and} \quad \sec(90^\circ - \theta) = \csc \theta tan ( 9 0 ∘ − θ ) = cot θ and sec ( 9 0 ∘ − θ ) = csc θ

Double-Angle & Sum Formulas sin ⁡ ( 2 θ ) = 2 sin ⁡ θ cos ⁡ θ \sin(2\theta) = 2 \sin \theta \cos \theta sin ( 2 θ ) = 2 sin θ cos θ cos ⁡ ( 2 θ ) = cos ⁡ 2 θ − sin ⁡ 2 θ = 2 cos ⁡ 2 θ − 1 = 1 − 2 sin ⁡ 2 θ \cos(2\theta) = \cos^2 \theta - \sin^2 \theta = 2\cos^2 \theta - 1 = 1 - 2\sin^2 \theta cos ( 2 θ ) = cos 2 θ − sin 2 θ = 2 cos 2 θ − 1 = 1 − 2 sin 2 θ sin ⁡ ( A ± B ) = sin ⁡ A cos ⁡ B ± cos ⁡ A sin ⁡ B \sin(A \pm B) = \sin A \cos B \pm \cos A \sin B sin ( A ± B ) = sin A cos B ± cos A sin B cos ⁡ ( A ± B ) = cos ⁡ A cos ⁡ B ∓ sin ⁡ A sin ⁡ B \cos(A \pm B) = \cos A \cos B \mp \sin A \sin B cos ( A ± B ) = cos A cos B ∓ sin A sin B

4. Oblique Triangles: Law of Sines & Law of Cosines

For any triangle △ A B C \triangle ABC △ A BC with side lengths a , b , c a, b, c a , b , c opposite angles A , B , C A, B, C A , B , C : Law of Sines: a sin ⁡ A = b sin ⁡ B = c sin ⁡ C \textbf{Law of Sines:} \quad \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} Law of Sines: sin A a ​ = sin B b ​ = sin C c ​ Law of Cosines: c 2 = a 2 + b 2 − 2 a b cos ⁡ C \textbf{Law of Cosines:} \quad c^2 = a^2 + b^2 - 2 a b \cos C Law of Cosines: c 2 = a 2 + b 2 − 2 ab cos C

Use Law of Sines when given ASA, AAS, or SSA (ambiguous case).

Use Law of Cosines when given SAS or SSS.

• Worked Step-by-Step ACET Exam Problems
• Problem 1: Simplifying Trigonometric Expressions

Simplify the expression E = tan ⁡ θ ⋅ csc ⁡ θ sec ⁡ 2 θ − 1 E = \frac{\tan \theta \cdot \csc \theta}{\sec^2 \theta - 1} E = s e c 2 θ − 1 t a n θ ⋅ c s c θ ​ into a single trigonometric ratio.

Step 1: Convert terms to sine and cosine equivalents. Numerator: tan ⁡ θ ⋅ csc ⁡ θ = ( sin ⁡ θ cos ⁡ θ ) ⋅ ( 1 sin ⁡ θ ) = 1 cos ⁡ θ = sec ⁡ θ \text{Numerator: } \tan \theta \cdot \csc \theta = \left(\frac{\sin \theta}{\cos \theta}\right) \cdot \left(\frac{1}{\sin \theta}\right) = \frac{1}{\cos \theta} = \sec \theta Numerator: tan θ ⋅ csc θ = ( cos θ sin θ ​ ) ⋅ ( sin θ 1 ​ ) = cos θ 1 ​ = sec θ

Step 2: Simplify the denominator using a Pythagorean identity. sec ⁡ 2 θ − 1 = tan ⁡ 2 θ \sec^2 \theta - 1 = \tan^2 \theta sec 2 θ − 1 = tan 2 θ

Step 3: Combine numerator and denominator. E = sec ⁡ θ tan ⁡ 2 θ = 1 / cos ⁡ θ sin ⁡ 2 θ / cos ⁡ 2 θ = 1 cos ⁡ θ ⋅ cos ⁡ 2 θ sin ⁡ 2 θ = cos ⁡ θ sin ⁡ 2 θ E = \frac{\sec \theta}{\tan^2 \theta} = \frac{1/\cos \theta}{\sin^2 \theta / \cos^2 \theta} = \frac{1}{\cos \theta} \cdot \frac{\cos^2 \theta}{\sin^2 \theta} = \frac{\cos \theta}{\sin^2 \theta} E = tan 2 θ sec θ ​ = sin 2 θ / cos 2 θ 1/ cos θ ​ = cos θ 1 ​ ⋅ sin 2 θ cos 2 θ ​ = sin 2 θ cos θ ​

Step 4: Express as reciprocal product. E = ( cos ⁡ θ sin ⁡ θ ) ⋅ ( 1 sin ⁡ θ ) = cot ⁡ θ csc ⁡ θ E = \left(\frac{\cos \theta}{\sin \theta}\right) \cdot \left(\frac{1}{\sin \theta}\right) = \cot \theta \csc \theta E = ( sin θ cos θ ​ ) ⋅ ( sin θ 1 ​ ) = cot θ csc θ

Problem 2: Oblique Triangle Distance via Law of Cosines

Two ships leave a port simultaneously. Ship A sails on a bearing of 030 ∘ 030^\circ 03 0 ∘ at 15 km/h, while Ship B sails on a bearing of 090 ∘ 090^\circ 09 0 ∘ at 20 km/h. How far apart are the two ships after 2 hours?

Step 1: Determine distances traveled.

Distance for Ship A: d A = 15 km/h × 2 h = 30 km d_A = 15 \text{ km/h} \times 2 \text{ h} = 30\text{ km} d A ​ = 15 km/h × 2 h = 30 km .

Distance for Ship B: d B = 20 km/h × 2 h = 40 km d_B = 20 \text{ km/h} \times 2 \text{ h} = 40\text{ km} d B ​ = 20 km/h × 2 h = 40 km .

Step 2: Find included angle C C C . Included angle C = 090 ∘ − 030 ∘ = 60 ∘ \text{Included angle } C = 090^\circ - 030^\circ = 60^\circ Included angle C = 09 0 ∘ − 03 0 ∘ = 6 0 ∘

Step 3: Apply Law of Cosines. c 2 = a 2 + b 2 − 2 a b cos ⁡ C c^2 = a^2 + b^2 - 2ab \cos C c 2 = a 2 + b 2 − 2 ab cos C c 2 = ( 30 ) 2 + ( 40 ) 2 − 2 ( 30 ) ( 40 ) cos ⁡ 60 ∘ c^2 = (30)^2 + (40)^2 - 2(30)(40) \cos 60^\circ c 2 = ( 30 ) 2 + ( 40 ) 2 − 2 ( 30 ) ( 40 ) cos 6 0 ∘ c 2 = 900 + 1600 − 2400 ( 1 2 ) = 2500 − 1200 = 1300 c^2 = 900 + 1600 - 2400 \left(\frac{1}{2}\right) = 2500 - 1200 = 1300 c 2 = 900 + 1600 − 2400 ( 2 1 ​ ) = 2500 − 1200 = 1300 c = 1300 = 10 13 ≈ 36.06 km c = \sqrt{1300} = 10\sqrt{13} \approx 36.06\text{ km} c = 1300 ​ = 10 13 ​ ≈ 36.06 km

ACET Speed Tactics & Traps Summary

• Problem Type Common Student Trap ACET Speed Tactic
• ASTC Signs Forgetting cosine is negative in Q2/Q3 Use ASTC acronym to verify function signs per quadrant

Special Angles Mixing up sin ⁡ 30 ∘ \sin 30^\circ sin 3 0 ∘ and cos ⁡ 30 ∘ \cos 30^\circ cos 3 0 ∘ Remember sin ⁡ 30 ∘ = 1 / 2 \sin 30^\circ = 1/2 sin 3 0 ∘ = 1/2 ; cosine gets 3 / 2 \sqrt{3}/2 3 ​ /2

Double Angle Writing sin ⁡ ( 2 θ ) = 2 sin ⁡ θ \sin(2\theta) = 2\sin\theta sin ( 2 θ ) = 2 sin θ Remember double angle adds cosine factor: 2 sin ⁡ θ cos ⁡ θ 2\sin\theta\cos\theta 2 sin θ cos θ

Law of Cosines Forgetting to multiply by cos ⁡ C \cos C cos C at the end Keep − 2 a b cos ⁡ C -2ab\cos C − 2 ab cos C grouped together