1. Let's start by understanding what trigonometry is. Trigonometry is the branch of mathematics that studies the relationships between the angles and sides of triangles, especially right triangles. 2. The fundamental functions in trigonometry are sine ($\sin$), cosine ($\cos$), and tangent ($\tan$). These functions relate the angles of a right triangle to the ratios of its sides. 3. For a right triangle with an angle $\theta$, the sides are named as follows: the side opposite to $\theta$ is called the opposite side, the side adjacent to $\theta$ is called the adjacent side, and the longest side opposite the right angle is the hypotenuse. 4. The definitions of the primary trigonometric functions are: $$\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}}$$ $$\cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}}$$ $$\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}$$ 5. Important rules include the Pythagorean identity: $$\sin^2(\theta) + \cos^2(\theta) = 1$$ which means the square of sine plus the square of cosine of the same angle equals 1. 6. To solve problems, you often use these ratios to find unknown sides or angles. For example, if you know an angle and one side, you can find the other sides using these functions. 7. Example: Suppose you have a right triangle where $\theta = 30^\circ$ and the hypotenuse is 10 units. To find the opposite side: $$\sin(30^\circ) = \frac{\text{opposite}}{10}$$ Since $\sin(30^\circ) = 0.5$, then: $$0.5 = \frac{\text{opposite}}{10} \Rightarrow \text{opposite} = 10 \times 0.5 = 5$$ 8. Similarly, to find the adjacent side: $$\cos(30^\circ) = \frac{\text{adjacent}}{10}$$ Since $\cos(30^\circ) = \frac{\sqrt{3}}{2} \approx 0.866$, then: $$0.866 = \frac{\text{adjacent}}{10} \Rightarrow \text{adjacent} = 10 \times 0.866 = 8.66$$ 9. These functions also extend beyond right triangles to the unit circle and periodic phenomena, but this is the foundational understanding. This concludes the basic introduction to trigonometry and how to use sine, cosine, and tangent to solve right triangle problems.