From a black box to a function

So far we've looked at trig functions as ratios of sides, and where those are known, that's easy to calculate. The other way we've used them is as "black boxes" — we put a number into a calculator and we get one back out. Here's a triangle for which we know the lengths of all sides:

$$ \begin{align} sin(\theta) &= \frac{o}{h} = \frac{5}{9.434} = 0.530 \\[5pt] cos(\theta) &= \frac{a}{h} = \frac{8}{9.434} = 0.056 \\[5pt] tan(\theta) &= \frac{o}{a} = \frac{5}{8} = 0.625 \end{align}$$

Later we'll learn how to recover angles when the side lengths are shown, but let's move on for now.

The unit circle – the fundamental basis for trigonometry

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Unit circle learning tool

The special triangles

Two common triangles you should memorize.

These two essential right triangles, their angle measures and the ratios of their side lengths will pop up time and again throughout your studies and work in math and science. Knowing how to jot them down and use them, and from where the measurements came, will be invaluable to you. Study them carefully and memorize their proportions.

The 45-45-90 triangle

The 45-45-90 triangle is just what it sounds like, a right triangle that has two 45˚ angles, and is thus an isosceles triangle. We'll work with just symbols for the lengths of the sides, like this:

Now the Pythagorean theorem has to hold for this right triangle, so

$$^2 + x^2 = r^2$$

Now combine the like terms on the left,

$$2x^2 = r^2$$

and let's solve for $x$ (you'll see why in a minute), by first dividing both sides by 2.

$$x^2 = \frac{r^2}{2}$$

Now to solve for $x$, take the square root of both sides:

$$x = \sqrt{\frac{r^2}{2}}$$

The 30-60-90 triangle

The 30-60-90 (degrees) right triangle is another frequently-encountered triangle that's worth knowing a lot about.

It isn't quite as easy to find an expression for the side lengths x and y, unless we combine two of these triangles to form an equilateral triangle like this:

Notice here that I've replaced x from the original figure with r/2. We know that the gray vertical line both bisects the top 60˚ angle (by construction) and bisects the base of the triangle. For a proof the the angle bisector of an isosceles triangle also bisects the opposite side, click here.

Now we can set up the Pythagorean theorem for one of the two 30-60-90 triangles shown:

$$r^2 = y^2 + \left( \frac{r}{2} \right)^2$$

Trig ratios in the special triangles

Trigonometric functions

Now we want to use the unit circle to create graphs of the three major trigonometric functions,

$$ \begin{align} f(x) &= sin(x) \\[5pt] f(x) &= cos(x) \\[5pt] f(x) &= tan(x) \end{align}$$

We'll do that by drawing a series of triangles, both 45-45-90 and 30-60-90, inside of the unit circle, and calculating the trig function values.

Let's start at 30˚ = θ/6. Notice that we can use that triangle to calculate all three trig functions. We'll set those aside as we work around the circle with triangles, and look at all of the values together at the end of the trip.

Now we really should work our way around the circle in increments of 30˚, but I'm going to skip around a bit and hope you'll get the idea. Next is a 135˚ angle, or 45˚ past 90˚, working counterclockwise (as usual) around the unit circle.

One quirk of the unit circle is that the length of the radius (the hypotenuses of all of our triangles) is always 1, but the other triangle sides can have negative or positive values that match with the x,y axis directions. That's how certain sine, cosine or tangent values end up negative.

Trig functions are periodic

Here's a table of trig function values at key angles in one round trip of the unit circle.

The trigonometric functions are repetitive, or cyclic, and we usually refer to them as periodic, meaning that they repeat the same basic pattern predictably.

Trig function graphs

Graphs of the sine (black) and cosine (magenta) functions are drawn below as $f(\theta)$ vs.$\theta$, for $\theta$ between $0$ and a littl more than $4\pi$ - twice around the unit circle plus a little more. Notice that these are periodic functions. Every $2\pi$ radians, they repeat, and that goes on infinitely in both positive and negative directions. The negative direction just means traversing the unit circle in the opposite (clockwise) direction.

All of the points in the table above are plotted on this graph. Be sure to think about these graphs for a while.

Transformations

In order to model periodic behavior using sines and cosines, we need, as usual, some transformations to translate and scale our function at will. In our typical way (see functions section), the transformations of $f(x) = sin(x)$ are shown on the right.

When we use trig. functions to model real data later, we'll make a slight modification of this prototype equation in order to keep the units straight.

The transformations of the other trig. functions are analogous.

Horizontal stretching

Vertical stretching

Move the slider to see the effect of the vertical-stretching parameter, $A$, on our sine function.

If this function represented a wave like a sound or light wave, the vertical distance between $f(x) = 0$ and the maximum (or minimum) value of the sine function would be the amplitude, which would be proportional to volume (loudness) of sound or brightness (intensity) of light.

Notice that when $A \lt 0$, the function is reflected across the x-axis, as we would expect for any function.

Vertical and horizontal translation

Vertical translation of a periodic function is performed just like on any other function: Simply add a constant number (k) to the value of the function for each point in the domain. Vertical translation is used in modeling to move the average value of the function up or down. For example, to model a tide that fluctuates by ± 3 feet from an average depth of 10 feet, we'd want a sine or cosine function (turns out it doesn't matter which), that oscillates between 7 and 13 feet. The function would read f(some stuff) + 10.

The tangent function

Now let's think about a class of trig functions called the reciprocal trig functions. The tangent is the most prominent of these, but there are three more. We can look at the tangent using SOH-CAH-TOA; we know that

$$sin(\theta) = \frac{o}{h}, \phantom{00} cos(\theta) = \frac{a}{h}, \phantom{00} tan(\theta) = \frac{o}{a}.$$

Now notice that

$$\frac{sin(\theta)}{cos(\theta)} = \frac{\frac{o}{h}}{\frac{a}{h}} = \frac{o}{\cancel{h}} \frac{\cancel{h}}{a} = \frac{o}{a}$$

This means that the tangent function is the ratio of the sine and cosine functions:

$$tan(\theta) = \frac{sin(\theta)}{cos(\theta)}$$

Now because $cos(\theta)$ — the denominator of the tangent expression — can be zero, the graph of the tangent function will have vertical asymptotes, as shown in the graph. The tangent of odd multiples of $\pi/2$ is infinite because $cos(\pi/2)$ (and odd multiples of $\pi/2$) is zero.

All six trigonometric functions

Finally, let's define the three other trig. functions. The reciprocal of the sine function (not to be confused with the inverse) is called the cosecant (csc) function. The reciprocal of the cosine is (paradoxically, I know) the secant (sec) and the reciprocal of the tangent is the cotangent (cot). csc(θ), sec(θ) and cot(θ) are defined below.

Graphs of all six trig functions

Several cycles of each of the six trig functions are plotted below. Note that the sin(x) and cos(x) graphs are vertically expanded by a factor of ten compared to the other graphs. This means that the sine graph would fit in between the U-shaped graphs (they are not parabolas!) in the csc(x) graph, same for the cos(x) and sec(x) graphs. Except for sine and cosine,

Example 1

Practice problems

For problems 1 and 2, calculate the measure of any missing angles and the lengths of any missing sides of the triangle.

1.

3. The cosine of an angle of a right triangle is $\frac{3}{11}$. Calculate the values of the other five trigonometric functions of this angle, and sketch the triangle.

   Solution

   
   

   Here's a diagram of the triangle, including the length of the missing side:

   

   The missing side is $o = \sqrt{11^2 - 3^2} = 10.583$.

   
   

   Now the values of the six trig functions are:

   $$ \begin{align} sin(x) &= \frac{10.583}{11} = 0.962 \\[5pt] cos(x) &= \frac{3}{11} = 0.273 \\[5pt] tan(x) &= \frac{10.583}{3} = 3.528 \\[5pt] csc(x) &= \frac{11}{10.583} = 1.039 \\[5pt] sec(x) &= \frac{11}{3} = 3.667 \\[5pt] cot(x) &= \frac{3}{10.583} = 0.283 \end{align}$$




4. The tangent of an angle of a right triangle is $\frac{7}{5}$. Calculate the values of the other five trigonometric functions of this angle and sketch the triangle.

   Solution

   
   

   Here's a diagram of the triangle, including the length of the missing side:

   

   The missing side is $o = \sqrt{11^2 - 3^2} = 10.583$.

   
   

   Now the values of the six trig functions are:

   $$ \begin{align} sin(x) &= \frac{10.583}{11} = 0.962 \\[5pt] cos(x) &= \frac{3}{11} = 0.273 \\[5pt] tan(x) &= \frac{10.583}{3} = 3.528 \\[5pt] csc(x) &= \frac{11}{10.583} = 1.039 \\[5pt] sec(x) &= \frac{11}{3} = 3.667 \\[5pt] cot(x) &= \frac{3}{10.583} = 0.283 \end{align}$$




5. A certain point on the unit circle is $\left(\frac{7}{25}, \; y \right)$ with $y \gt 0$. Determine the exact value of $y$.

   Solution

   
   

   Here's a diagram of the unit circle with our triangle roughly placed in it:

   

   
   

   Now it's clear that our task is just to find the missing side of the triangle, the $y$ coordinate of the point on the circle. That is

   $$ \begin{align} y &= \sqrt{1^2 - \left( \frac{7}{25} \right)^2} \\[5pt] &= \sqrt{\frac{625-49}{625}} \\[5pt] &= \sqrt{\frac{576}{625}} \\[5pt] &= \frac{24}{25} \end{align}$$




6. A certain point on the unit circle is $\left(\frac{15}{17}, \; y \right)$ with $y \lt 0$. Determine the exact value of $y$.

   Solution

   
   

   Here's a diagram of the unit circle with our triangle roughly placed in it:

   

   
   

   Now it's clear that our task is just to find the missing side of the triangle, the $y$ coordinate of the point on the circle. That is

   $$ \begin{align} y &= -\sqrt{1^2 - \left( \frac{15}{17} \right)^2} \\[5pt] &= -\sqrt{\frac{289-225}{289}} \\[5pt] &= -\sqrt{\frac{64}{289}} \\[5pt] &= -\frac{8}{17} \end{align}$$




7. Sketch a graph of $f(x) = sin(x) + cos(x)$ over the domain $[0, \, 2\pi]$.

   Solution

   
   

   Let's make a quick table of sums of sin(x) and cos(x) at a few select points:

   $\theta$	$sin(\theta)$	$cos(\theta)$	$s(\theta) + c(\theta)$
   $0^{\circ}$	$0$	$1$	$1$
   $45^{\circ}$	$\frac{\sqrt{2}}{2}$	$\frac{\sqrt{2}}{2}$	$\sqrt{2}$
   $90^{\circ}$	$1$	$0$	$1$
   $135^{\circ}$	$\frac{\sqrt{2}}{2}$	$-\frac{\sqrt{2}}{2}$	$0$
   $180^{\circ}$	$0$	$-1$	$-1$
   $225^{\circ}$	$-\frac{\sqrt{2}}{2}$	$-\frac{\sqrt{2}}{2}$	$-\sqrt{2}$
   $270^{\circ}$	$\frac{\sqrt{2}}{2}$	$-\frac{\sqrt{2}}{2}$	$0$
   $305^{\circ}$	$-\frac{\sqrt{2}}{2}$	$-\frac{\sqrt{2}}{2}$	$-\sqrt{2}$

   
   

   Here is what the graph looks like:

   

   It's still a periodic function with a period of $2 \pi$ radians or 360˚, but it has been shifted by -45˚.




8. Sketch a graph of $g(x) = sin^2(x) + cos^2(x)$. Note that $sin^(x)$ is just a commonly-used shorthand notation for $[sin(x)]^2$, just the square of the result of $sin(x)$.

   Solution

   
   

   Let's make a quick table of sums of sin²(x) and cos²(x) at a few select points:

   $\theta$	$sin^2(\theta)$	$cos^(\theta)$	$s^2(\theta) + c^2(\theta)$
   $0^{\circ}$	$0$	$1$	$1$
   $45^{\circ}$	$\frac{1}{2}$	$\frac{1}{2}$	$1$
   $90^{\circ}$	$1$	$0$	$1$
   $135^{\circ}$	$\frac{1}{2}$	$\frac{1}{2}$	$1$
   $180^{\circ}$	$0$	$1$	$1$
   $225^{\circ}$	$\frac{1}{2}$	$\frac{1}{2}$	$1$
   $270^{\circ}$	$\frac{1}{2}$	$\frac{1}{2}$	$1$
   $305^{\circ}$	$\frac{1}{2}$	$\frac{1}{2}$	$1$

   
   

   Notice that the sum $sin^2(\theta) + cos^2(\theta) = 1$ for all values of $\theta$. We haven't calculated any values in between, but it's true. This is a relationship that we'll use a lot going into analytic trigonometry and it's called the Pythagorean Identity:

   $$sin^2(\theta) + cos^2(\theta) = 1$$

   For any angle, $\theta$.

Video examples

1. Counting around the unit circle

You should know how to label the most-frequently-used angles around the unit circle (circle of radius r = 1), both in degrees and radians. It's not that difficult if you just think of it as counting around the circle in different increments.

Minutes of your life: 3:25