1.7 Trigonometry: Right Triangle Ratios, Unit Circle & Basic Identities
Key Takeaways
- 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 $\theta$:
\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}$$ --- ## 2. Special Angle Values & The Unit Circle The unit circle $x^2 + y^2 = 1$ defines trigonometric values for any angle $\theta$ where $x = \cos \theta$ and $y = \sin \theta$. ### Exact Values Table | Angle $\theta$ (Deg) | Angle $\theta$ (Rad) | $\sin \theta$ | $\cos \theta$ | $\tan \theta$ | | :---: | :---: | :---: | :---: | :---: | | **$0^\circ$** | $0$ | $0$ | $1$ | $0$ | | **$30^\circ$** | $\pi/6$ | $1/2$ | $\sqrt{3}/2$ | $\sqrt{3}/3$ | | **$45^\circ$** | $\pi/4$ | $\sqrt{2}/2$ | $\sqrt{2}/2$ | $1$ | | **$60^\circ$** | $\pi/3$ | $\sqrt{3}/2$ | $1/2$ | $\sqrt{3}$ | | **$90^\circ$** | $\pi/2$ | $1$ | $0$ | Undefined | | **$180^\circ$** | $\pi$ | $0$ | $-1$ | $0$ | | **$270^\circ$** | $3\pi/2$ | $-1$ | $0$ | Undefined | ### Quadrant Signs: The ASTC Rule - **Quadrant I ($0^\circ$ to $90^\circ$):** **A**ll functions are POSITIVE. - **Quadrant II ($90^\circ$ to $180^\circ$):** **S**ine ($\& \csc$) is POSITIVE. - **Quadrant III ($180^\circ$ to $270^\circ$):** **T**angent ($\& \cot$) is POSITIVE. - **Quadrant IV ($270^\circ$ to $360^\circ$):** **C**osine ($\& \sec$) is POSITIVE. --- ## 3. Fundamental Trigonometric Identities ### Pythagorean Identities $$\sin^2 \theta + \cos^2 \theta = 1$$ $$1 + \tan^2 \theta = \sec^2 \theta$$ $$1 + \cot^2 \theta = \csc^2 \theta$$ ### Co-Function Identities $$\sin(90^\circ - \theta) = \cos \theta \quad \text{and} \quad \cos(90^\circ - \theta) = \sin \theta$$ $$\tan(90^\circ - \theta) = \cot \theta \quad \text{and} \quad \sec(90^\circ - \theta) = \csc \theta$$ ### Double-Angle & Sum Formulas $$\sin(2\theta) = 2 \sin \theta \cos \theta$$ $$\cos(2\theta) = \cos^2 \theta - \sin^2 \theta = 2\cos^2 \theta - 1 = 1 - 2\sin^2 \theta$$ $$\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$$ --- ## 4. Oblique Triangles: Law of Sines & Law of Cosines For any triangle $\triangle ABC$ with side lengths $a, b, c$ opposite angles $A, B, C$: $$\mathbf{Law of Sines:} \quad \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$$ $$\mathbf{Law of Cosines:} \quad c^2 = a^2 + b^2 - 2 a b \cos C$$ - Use **Law of Sines** when given ASA, AAS, or SSA (ambiguous case). - Use **Law of Cosines** when given SAS or SSS. --- ## Worked Step-by-Step ACET Exam Problems ### Problem 1: Simplifying Trigonometric Expressions *Simplify the expression $E = \frac{\tan \theta \cdot \csc \theta}{\sec^2 \theta - 1}$ into a single trigonometric ratio.* **Step 1: Convert terms to sine and cosine equivalents.** $$\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$$ **Step 2: Simplify the denominator using a Pythagorean identity.** $$\sec^2 \theta - 1 = \tan^2 \theta$$ **Step 3: Combine numerator and denominator.** $$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}$$ **Step 4: Express as reciprocal product.** $$E = \left(\frac{\cos \theta}{\sin \theta}\right) \cdot \left(\frac{1}{\sin \theta}\right) = \cot \theta \csc \theta$$ --- ### Problem 2: Oblique Triangle Distance via Law of Cosines *Two ships leave a port simultaneously. Ship A sails on a bearing of $030^\circ$ at 15 km/h, while Ship B sails on a bearing of $090^\circ$ at 20 km/h. How far apart are the two ships after 2 hours?* **Step 1: Determine distances traveled.** - Distance for Ship A: $d_A = 15 \text{ km/h} \times 2 \text{ h} = 30\text{ km}$. - Distance for Ship B: $d_B = 20 \text{ km/h} \times 2 \text{ h} = 40\text{ km}$. **Step 2: Find included angle $C$.** $$\text{Included angle } C = 090^\circ - 030^\circ = 60^\circ$$ **Step 3: Apply Law of Cosines.** $$c^2 = a^2 + b^2 - 2ab \cos C$$ $$c^2 = (30)^2 + (40)^2 - 2(30)(40) \cos 60^\circ$$ $$c^2 = 900 + 1600 - 2400 \left(\frac{1}{2}\right) = 2500 - 1200 = 1300$$ $$c = \sqrt{1300} = 10\sqrt{13} \approx 36.06\text{ 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 $\sin 30^\circ$ and $\cos 30^\circ$ | Remember $\sin 30^\circ = 1/2$; cosine gets $\sqrt{3}/2$ | | **Double Angle** | Writing $\sin(2\theta) = 2\sin\theta$ | Remember double angle adds cosine factor: $2\sin\theta\cos\theta$ | | **Law of Cosines** | Forgetting to multiply by $\cos C$ at the end | Keep $-2ab\cos C$ grouped together |Test Your Knowledge
If sin(theta) = 3/5 and theta is in Quadrant II, what is the exact value of tan(theta)?
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Test Your Knowledge
Which of the following expressions is equivalent to (1 - cos(2theta)) / sin(2theta)?
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Test Your Knowledge
A 10-meter ladder leans against a vertical wall, making a 60-degree angle with the ground. How high up the wall does the ladder reach?
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