Ratio of Trigonometric Functions
In addition to the basic trigonometric functions (sine and cosine), additional useful functions can be defined as their ratios. In particular, the tangent (tan), cosecant (csc), secant (sec), and cotangent (cot) functions are defined as
\[\tan \theta = \frac{\sin\theta}{\cos\theta}, ~~\csc\theta=\frac{1}{\sin\theta}, ~~\sec\theta=\frac{1}{\cos\theta}, ~~\cot\theta=\frac{\cos\theta}{\sin\theta}.\]
Of these, tangent is the most interesting, due to having a natural geometric interpretation and satisfying several additional properties.
Contents
Formal Definitions
Consider the following right triangle:
The sides with respect to angle \( \theta\) are
\[\begin{align} a &= \text{ adjacent}\\ b &= \text{ opposite}\\ c &= \text{ hypotenuse}. \end{align}\]
Then the basic trigonometric functions are defined by
\[\begin{align} \sin \theta &= \frac{(\text{opposite})}{(\text{hypotenuse})} = \frac{b}{c} \\ \cos \theta &= \frac{(\text{adjacent})}{(\text{hypotenuse})} = \frac{a}{c}\\ \tan \theta &= \frac{(\text{opposite})}{(\text{adjacent})} = \frac{b}{a}. \end{align}\]
We also have the following reciprocal functions
\[\begin{align} \csc \theta &= \frac{1}{\sin \theta} \\ \sec \theta &= \frac{1}{\cos \theta} \\ \cot \theta &= \frac{1}{\tan \theta}. \end{align}\]
For example, \( \cot \theta\) can be expressed as
\[ \cot \theta = \frac{1}{\tan \theta} = \frac{1}{\frac{b}{a}} = \frac{a}{b} = \frac{(\text{adjacent})}{(\text{opposite})}. \]
Example Problems
If the lengths of hypotenuse and opposite are \(c=5\) and \(b=3\), respectively, in the right triangle below, what is \( \cos \theta ?\)
Since the length of hypotenuse is 5 and that of opposite is 3, from the Pythagorean theorem, the length of adjacent is
\[ \begin{align} (\text{hypotenuse})^2 &= (\text{opposite})^2+(\text{adjacent})^2 \\ 5^2 & = 3^2 + (\text{adjacent})^2\\ \Rightarrow (\text{adjacent}) &= 4. \end{align} \]
Hence, \( \cos \theta = \dfrac{(\text{adjacent})}{(\text{hypotenuse})} = \frac{4}{5}. \ _ \square \)
If \( \sin \theta= \frac{1}{2} \) and \( \cos \theta = \frac{1}{\sqrt{3}} \) in a right triangle, then what is \( \tan \theta ?\)
Since \( \tan \theta =\frac{\sin \theta}{\cos \theta} ,\)
\[ \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\ \, \frac{1}{2}\ }{\frac{1}{\sqrt{3}}} = \frac{\sqrt{3}}{2}. \ _ \square \]
If \( \cot \theta=\frac{1}{3} \) in a right triangle, what is \( \tan \theta ?\)
Since \( \tan \theta \) and \( \cot \theta\) are reciprocals of each other,
\[ \tan \theta = \frac{1}{\cot \theta} = \frac{1}{\ \frac{1}{3}\ } = 3. \ _ \square\]
What is the value of \(\tan 180^\circ ?\)
If \( \tan \theta + \cot \theta=2, \) what is \( \sin \theta \cdot \cos \theta ?\)
We have
\[ \begin{align} \tan \theta + \cot \theta &= \frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta} \\ &= \frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cdot \cos \theta}. \qquad (1) \end{align} \]
Also, observe that the value of \( \sin^2 \theta + \cos^2 \theta \) can be obtained as follows:
\[ \begin{align} \sin^2 \theta + \cos^2 \theta &= \left ( \frac{\text{opposite}}{\text{hypotenuse}} \right )^2 + \left ( \frac{\text{adjacent}}{\text{hypotenuse}} \right )^2 \\ &= \frac{(\text{opposite})^2+(\text{adjacent})^2}{(\text{hypotenuse})^2} \\ &= \frac{(\text{hypotenuse})^2}{(\text{hypotenuse})^2} =1. \qquad (2) \end{align} \]
Then from \( (1)\) and \( (2) ,\) \( \sin \theta \cdot \cos \theta \) can be obtained as follows:
\[ \begin{align} \tan \theta + \cot \theta &= \frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cdot \cos \theta} \\ 2 &= \frac{1}{\sin \theta \cdot \cos \theta} \\ \Rightarrow \sin \theta \cdot \cos \theta &= \frac{1}{2}. \ _ \square \end{align} \]