Right Triangle Trigonometry

7.4 The Other Trigonometric Functions

Trig on a General Circle

The unit circle is great for exact benchmark values, but trig functions are not limited to radius 1. We can define all six trig functions on any circle centered at the origin.

Definition: Trig Functions on Radius r

Let $P(x,y)$ lie on a circle centered at the origin with radius $$ r=\sqrt{x^2+y^2}>0 $$ for angle $\theta$ in standard position.

Then: $$ \sin\theta=\frac{y}{r},\quad \cos\theta=\frac{x}{r},\quad \tan\theta=\frac{y}{x} $$ $$ \csc\theta=\frac{r}{y},\quad \sec\theta=\frac{r}{x},\quad \cot\theta=\frac{x}{y} $$

Warning: Domain Restrictions

  • $\tan\theta$ and $\sec\theta$ are undefined when $x=0$.
  • $\cot\theta$ and $\csc\theta$ are undefined when $y=0$.

Example: Compute all six functions from a point

Find all six trig functions for angle $\theta$ through point $P(-3,4)$.

Show Solution

Step 1: Compute radius. $$ r=\sqrt{(-3)^2+4^2}=\sqrt{9+16}=\sqrt{25}=5 $$

Step 2: Use definitions. $$ \sin\theta=\frac{y}{r}=\frac{4}{5} $$ $$ \cos\theta=\frac{x}{r}=\frac{-3}{5} $$ $$ \tan\theta=\frac{y}{x}=\frac{4}{-3}=-\frac{4}{3} $$ $$ \csc\theta=\frac{r}{y}=\frac{5}{4} $$ $$ \sec\theta=\frac{r}{x}=\frac{5}{-3}=-\frac{5}{3} $$ $$ \cot\theta=\frac{x}{y}=\frac{-3}{4}=-\frac{3}{4} $$

Example: Compute all six functions from a fractional point

Find all six trig functions for point $$ P\left(-\frac{1}{5},\frac{2\sqrt{6}}{5}\right) $$

Show Solution

Step 1: Compute $r$. $$ r=\sqrt{\left(-\frac{1}{5}\right)^2+\left(\frac{2\sqrt{6}}{5}\right)^2} =\sqrt{\frac{1}{25}+\frac{24}{25}} =\sqrt{1}=1 $$

So this point is already on the unit circle.

Step 2: Function values. $$ \sin\theta=\frac{2\sqrt{6}}{5},\quad \cos\theta=-\frac{1}{5},\quad \tan\theta=\frac{\frac{2\sqrt{6}}{5}}{-\frac{1}{5}}=-2\sqrt{6} $$ $$ \csc\theta=\frac{1}{\sin\theta}=\frac{5}{2\sqrt{6}}=\frac{5\sqrt{6}}{12} $$ $$ \sec\theta=\frac{1}{\cos\theta}=-5 $$ $$ \cot\theta=\frac{1}{\tan\theta}=-\frac{1}{2\sqrt{6}}=-\frac{\sqrt{6}}{12} $$

MyOpenMath: Try your own!

Signs by Quadrant (ASTC)

As soon as angles leave Quadrant I, signs matter as much as magnitudes. Remember All Students Take Classes.

Example: Determine quadrant from sign clues

Name the quadrant if:

  1. $\cos\theta>0$ and $\tan\theta<0$
  2. $\sin\theta>0$ and $\cos\theta<0$
  3. $\cot\theta<0$ and $\cos\theta>0$
Show Solution
  1. $\cos\theta>0$ means Quadrant I or IV. $\tan\theta<0$ means Quadrant II or IV. Intersection is Quadrant IV.
  2. $\sin\theta>0$ means Quadrant I or II. $\cos\theta<0$ means Quadrant II or III. Intersection is Quadrant II.
  3. $\cot\theta<0$ means tangent negative, so Quadrant II or IV. $\cos\theta>0$ means Quadrant I or IV. Intersection is Quadrant IV.

Example: Mixed sign analysis

Find quadrant if $\csc\theta<0$ and $\sec\theta>0$.

Show Solution

$\csc\theta<0$ means $\sin\theta<0$ (Quadrant III or IV).

$\sec\theta>0$ means $\cos\theta>0$ (Quadrant I or IV).

Intersection is Quadrant IV.

MyOpenMath: Try your own!

Finding Remaining Functions from One Function Value

This is where algebra and quadrant logic combine. We recover the missing coordinate, assign sign correctly, then compute everything else.

Example: Given sin and quadrant

Given $$ \sin\theta=\frac{3}{5} $$ and $\theta$ is in Quadrant II, find all six trig functions.

Show Solution

Step 1: Interpret as triangle ratio. $$ \sin\theta=\frac{y}{r}=\frac{3}{5} $$ So $r=5$ and $|y|=3$. In Quadrant II, $y>0$, so $y=3$.

Step 2: Solve for $x$ using $x^2+y^2=r^2$. $$ x^2+3^2=5^2 $$ $$ x^2+9=25 $$ $$ x^2=16 $$ $$ x=\pm4 $$ In Quadrant II, $x<0$, so $x=-4$.

Step 3: Compute all functions. $$ \sin\theta=\frac{3}{5},\quad \cos\theta=\frac{-4}{5},\quad \tan\theta=\frac{3}{-4}=-\frac{3}{4} $$ $$ \csc\theta=\frac{5}{3},\quad \sec\theta=\frac{5}{-4}=-\frac{5}{4},\quad \cot\theta=\frac{-4}{3} $$

Example: Given sec and acute angle

Given $$ \sec\theta=\frac{7}{4} $$ and $\theta$ is acute, find the remaining five functions.

Show Solution

Step 1: Convert sec to cosine. $$ \cos\theta=\frac{1}{\sec\theta}=\frac{4}{7} $$

Since $\theta$ is acute, all principal right-triangle sides are positive.

Step 2: Let $x=4$, $r=7$, solve for $y$. $$ x^2+y^2=r^2 $$ $$ 4^2+y^2=7^2 $$ $$ 16+y^2=49 $$ $$ y^2=33 $$ $$ y=\sqrt{33} $$

Step 3: Compute remaining functions. $$ \sin\theta=\frac{y}{r}=\frac{\sqrt{33}}{7} $$ $$ \tan\theta=\frac{y}{x}=\frac{\sqrt{33}}{4} $$ $$ \csc\theta=\frac{1}{\sin\theta}=\frac{7}{\sqrt{33}}=\frac{7\sqrt{33}}{33} $$ $$ \cot\theta=\frac{1}{\tan\theta}=\frac{4}{\sqrt{33}}=\frac{4\sqrt{33}}{33} $$

MyOpenMath: Try your own!

Four Things You Need to Know

So, as I stole from my pre-calculus teacher and she stole from hers, I hope you will steal from me: there are really only 4 things you need to know to survive trig.

  1. $1,2,3$ and $3,2,1$
  2. Quadrant Angles $0^\circ$ means $(1,0)$, $90^\circ$ means $(0,1)$, $180^\circ$ means $(-1,0)$, $270^\circ$ means $(0,-1)$
  3. Reference Angles
  4. All Students Take Classes