1. Reciprocal Relations
Some trigonometric ratios naturally pair up as reciprocals of each other. This comes directly from their definitions in a right triangle.
Here are the reciprocal identities:
\( \sin \theta = \dfrac{1}{\csc \theta}, \qquad \csc \theta = \dfrac{1}{\sin \theta} \)
\( \cos \theta = \dfrac{1}{\sec \theta}, \qquad \sec \theta = \dfrac{1}{\cos \theta} \)
\( \tan \theta = \dfrac{1}{\cot \theta}, \qquad \cot \theta = \dfrac{1}{\tan \theta} \)
1.1. Example
If \( \sin \theta = \dfrac{4}{5} \), then:
\( \csc \theta = \dfrac{1}{\sin \theta} = \dfrac{5}{4} \)
2. Quotient Relations
Quotient relations show how tan and cot can be expressed using sin and cos. These relations are extremely helpful when simplifying trigonometric expressions.
\( \tan \theta = \dfrac{\sin \theta}{\cos \theta} \)
\( \cot \theta = \dfrac{\cos \theta}{\sin \theta} \)
2.1. Example
If \( \sin \theta = \dfrac{3}{5} \) and \( \cos \theta = \dfrac{4}{5} \), then:
\( \tan \theta = \dfrac{3/5}{4/5} = \dfrac{3}{4} \)
3. Pythagorean Relations
These relations come from the Pythagorean theorem: \( \text{(Opp)}^2 + \text{(Adj)}^2 = \text{(Hyp)}^2 \). When we divide every term by appropriate sides, we get identities linking different trigonometric ratios.
Here are the three key Pythagorean identities:
\( \sin^2 \theta + \cos^2 \theta = 1 \)
\( 1 + \tan^2 \theta = \sec^2 \theta \)
\( 1 + \cot^2 \theta = \csc^2 \theta \)
3.1. Example
If \( \tan \theta = \dfrac{3}{4} \), then:
\( \sec^2 \theta = 1 + \tan^2 \theta = 1 + \dfrac{9}{16} = \dfrac{25}{16} \)
4. Rewriting Ratios Using Relationships
Using reciprocal, quotient and Pythagorean identities, any trigonometric ratio can be rewritten in terms of others. This becomes very useful in algebraic simplification and solving equations.
4.1. Example
If \( \sin \theta = \dfrac{5}{13} \) and the angle lies in the first quadrant, then:
- \( \cos \theta = \sqrt{1 - \sin^2 \theta} = \sqrt{1 - 25/169} = \dfrac{12}{13} \)
- \( \tan \theta = \dfrac{\sin \theta}{\cos \theta} = \dfrac{5}{12} \)
- \( \sec \theta = \dfrac{1}{\cos \theta} = \dfrac{13}{12} \)
- \( \csc \theta = \dfrac{1}{\sin \theta} = \dfrac{13}{5} \)