Trigonometry problems tend to fall into a few repeating patterns once you've seen enough of them. These 15 worked examples cover the ones that show up most often in board exams.
1. In a right triangle, the opposite side is 3 and the hypotenuse is 5. Find sin θ.
sin θ = opposite/hypotenuse = 3/5 = 0.6.
2. If cos θ = 4/5, find sin θ (using the Pythagorean identity).
sin²θ = 1 − cos²θ = 1 − 16/25 = 9/25 → sin θ = 3/5.
3. Find tan 45°.
At 45°, opposite = adjacent, so tan 45° = 1.
4. A ladder leans against a wall making a 60° angle with the ground, reaching 10 m up. Find the ladder's length.
sin 60° = 10/L → L = 10/sin60° = 10/0.866 ≈ 11.55 m.
5. Verify sin²θ + cos²θ = 1 for θ = 30°.
sin30°=0.5, cos30°≈0.866. 0.25 + 0.75 = 1 ✓
6. Simplify sinθ·cscθ.
cscθ = 1/sinθ, so sinθ · (1/sinθ) = 1.
7. Simplify (1 − cos²θ)/sinθ.
1−cos²θ = sin²θ, so sin²θ/sinθ = sinθ.
8. Prove tanθ = sinθ/cosθ using θ = 45°.
sin45°/cos45° = (√2/2)/(√2/2) = 1 = tan45° ✓
9. Simplify 1 + tan²θ.
Using the identity, 1 + tan²θ = sec²θ.
10. Solve sinθ = 0.5 for 0° ≤ θ ≤ 360°.
θ = 30° or θ = 150° (sine is positive in Quadrants I and II).
11. Solve cosθ = −0.5 for 0° ≤ θ ≤ 360°.
θ = 120° or θ = 240° (cosine is negative in Quadrants II and III).
12. Solve 2sinθ − 1 = 0.
sinθ = 1/2 → θ = 30° or 150°.
13. Solve tanθ = 1 for 0° ≤ θ ≤ 360°.
θ = 45° or θ = 225° (tangent is positive in Quadrants I and III).
14. Solve 2cos²θ − 1 = 0.
cos²θ = 1/2 → cosθ = ±(√2/2) → θ = 45°, 135°, 225°, or 315°.
15. Solve sin²θ − sinθ = 0.
Factor: sinθ(sinθ − 1) = 0 → sinθ = 0 or sinθ = 1 → θ = 0°, 90°, 180°, or 360°.
Use Calvo's scientific calculator in DEG mode to double-check angle values as you work through similar problems.