Trigonometry problems tend to fall into a few repeating patterns once you've seen enough of them. At its core, trigonometry describes the relationship between the angles and sides of a right triangle using three basic ratios — sine, cosine, and tangent — and everything else in the topic (identities, equations, graphs) is built on top of these three. These 15 worked examples cover the patterns that show up most often in board exams, organised into basic ratios, identity verification, and solving trig equations.
Before jumping into the examples, it helps to fix the definitions in your head clearly: for an angle θ in a right triangle, sin θ = opposite/hypotenuse, cos θ = adjacent/hypotenuse, and tan θ = opposite/adjacent. A useful memory aid many students use is SOH-CAH-TOA — Sine = Opposite/Hypotenuse, Cosine = Adjacent/Hypotenuse, Tangent = Opposite/Adjacent.
Once angles go beyond 90°, you need to know which ratios are positive in which quadrant of the unit circle. The rule students memorise is "All Students Take Calculus" — read anticlockwise starting from Quadrant I:
This rule is exactly why example 10 below (sinθ = 0.5) has two answers in Quadrants I and II, while example 11 (cosθ = −0.5) has two answers in Quadrants II and III — the sign of the ratio tells you which quadrants to look in.
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°.
How do I remember which ratio is positive in which quadrant?
Use the ASTC rule explained above — "All Students Take Calculus" going anticlockwise from Quadrant I.
Why do trig equations usually have two answers instead of one?
Because sine, cosine, and tangent repeat their values at different angles within a full 360° rotation — for example, sin30° and sin150° are both 0.5. An equation like sinθ = 0.5 is really asking "which angles give this ratio," and there's more than one within a full circle.
What's the fastest way to verify a trig identity?
Convert everything to sine and cosine first using the definitions tanθ = sinθ/cosθ and the reciprocal identities, then simplify both sides until they match.
Use Calvo's scientific calculator in DEG mode to double-check angle values as you work through similar problems. If your answer doesn't match the calculator, the mode setting (degrees vs radians) is the first thing worth checking before you assume the working is wrong.