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 θ \theta θ x = cos θ x = \cos \theta x = cos θ y = sin θ y = \sin \theta y = sin θ
Exact Values Table
Angle θ \theta θ Angle θ \theta θ 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 ∘ 90 ∘ 90^\circ 9 0 ∘ A ll functions are POSITIVE.Quadrant II (90 ∘ 90^\circ 9 0 ∘ 180 ∘ 180^\circ 18 0 ∘ S ine (& csc \& \csc & csc Quadrant III (180 ∘ 180^\circ 18 0 ∘ 270 ∘ 270^\circ 27 0 ∘ T angent (& cot \& \cot & cot Quadrant IV (270 ∘ 270^\circ 27 0 ∘ 360 ∘ 360^\circ 36 0 ∘ C osine (& sec \& \sec & sec
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 a , b , c a, b, c a , b , c 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
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 ∘ 090 ∘ 090^\circ 09 0 ∘
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 ∘ cos 30 ∘ \cos 30^\circ cos 3 0 ∘ Remember sin 30 ∘ = 1 / 2 \sin 30^\circ = 1/2 sin 3 0 ∘ = 1/2 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 Keep − 2 a b cos C -2ab\cos C − 2 ab cos C