Trigonometry: Key Concepts and Practice Problems (Quick Notes)

Origins of Trigonometry

Trigonometry began in the Greek world, with Hipparchus (c. 190–120 BCE) as the first to construct a table of values for a trigonometric function and is often called the father of trigonometry. The deeper origins trace to ancient Egyptian pyramids and Babylonian astronomy dating to about 3000 BCE.

Etymology and Scope

Trigonometry is the branch of mathematics that studies relationships between the sides and angles of triangles. The term derives from the Greek words trigono (triangle) and metron (measure), via a 16th century Latin derivative.

Early Tables: The Table of Chords

According to Theon of Alexandria, Hipparchus compiled a table of chords in a circle, effectively a trigonometric table, in 12 books.

Real-World Applications

Heights and distances: The practice uses angle measures such as the angle of elevation (the angle between the horizontal line and the line of sight above the observer's eye) and the angle of depression (downward). Use the relationships SOH CAH TOA: sinθ=oppositehypotenuse,cosθ=adjacenthypotenuse,tanθ=oppositeadjacent\sin\theta=\frac{opposite}{hypotenuse},\quad\cos\theta=\frac{adjacent}{hypotenuse},\quad\tan\theta=\frac{opposite}{adjacent}.

Cartesian Plane and Quadrants

Plot points as ordered pairs $(x,y)$. The Cartesian plane has four quadrants: I $(+,+)$, II $(−,+)$, III $(−,−)$, IV $(+,-)$. The positive direction is upward and to the right.

Distance Between Points

Distance formula: d=(x<em>2x</em>1)2+(y<em>2y</em>1)2d=\sqrt{(x<em>2-x</em>1)^2+(y<em>2-y</em>1)^2}. Example: distance between A(-4,-1) and B(1,2) is d=(1(4))2+(2(1))2=52+32=345.83d=\sqrt{(1-(-4))^2+(2-(-1))^2}=\sqrt{5^2+3^2}=\sqrt{34}\approx 5.83.

Worked Examples: Problems 1–3

Problem 1: P1(-2,-3), P2(2,3), P3(-3,2). Distances: AB=52=213,  BC=26,  AC=26AB=\sqrt{52}=2\sqrt{13},\; BC=\sqrt{26},\; AC=\sqrt{26}. Area: A=13A=13.

Problem 2: P1(-2,8), P2(3,6), P3(-2,-6). Distances: AB=29,  BC=13,  AC=14AB=\sqrt{29},\; BC=13,\; AC=14. Area: A=35A=35.

Problem 3: P1(8,4), P2(4,-6), P3(2,-1). Distances: AB=116=229,  BC=29,  AC=61AB=\sqrt{116}=2\sqrt{29},\; BC=\sqrt{29},\; AC=\sqrt{61}. Area: A=20A=20.