## Graph of the sine function

To easily draw a sine function, on x – axis we’ll put values from $ -2 \pi$ to $ 2 \pi$, and on y – axis real numbers.

First, codomain of the sine is [-1, 1], that means that your graphs highest point on y – axis will be 1, and lowest -1, it’s easier to draw lines parallel to x – axis through -1 and 1 on y axis to know where is your boundary.

Next, find the zeros. Zeros are the points where your graph intersects x – axis. Look at the unit line. Where is function sine equal to zero?

$ Sin(x) = 0$ where x – axis cuts the unit line. Why? You try to find your angles just in a way you did before. Set your value on y – axis, here it is right in the origin of the unit circle, and draw parallel lines to x – axis. This is exactly x – axis.

That means that the angles whose sine value is equal to 0 are $ 0, \pi, 2 \pi, 3 \pi, 4 \pi$ And those are your zeros, mark them on the x – axis.

This is what you should have now:

Now you need your maximum values and minimum values. **Maximum** is a point where your graph reaches its highest value, and **minimum** is a point where a graph reaches its lowest value on a certain area. Again, take a look at a unit line. The highest value is 1, and the angle in which the sine reaches that value is $\frac{\pi}{2}$, and the lowest is $ -1$ in $\frac{3 \pi}{2}$. This will also repeat so the highest points will be $\frac{\pi}{2}, \frac{5 \pi}{2}, \frac{9 \pi}{2}$ … ($\frac{\pi}{2}$ and every other angle you get when you get into that point in second lap, third and so on..), and lowest points $\frac{3 \pi}{2}, \frac{7 \pi}{2}, \frac{11 \pi}{2}$ …

So mark those points on the graph also.

And when you connect your points you should get something like this:

## Graph of the cosine function

Graph of cosine function is drawn just like the graph of sine value, the only difference are the zeros.

Take a look at a unit circle again. Where is the cosine value equal to zero? It is equal to zero where y-axis cuts the circle, that means in $ –\frac{\pi}{2}, \frac{\pi}{2}, \frac{3 \pi}{2}$ …

Just follow the same steps we used for sine function.

First, mark the zeros. Again, since the codomain of the cosine is [-1, 1] your graph will only have values in that area, so draw lines that go through -1, 1 and are parallel to x – axis.

Now you need points where your function reaches maximum, and points where it reaches minimum. Again, look at the unit circle. The greatest value cosine can have is 1, and it reaches it in $ 0, 2 \pi, 4 \pi$ …

The lowest value cosine can have is -1 and it reaches it in $\pi, 3 \pi, 5 \pi$ …

And when you connect your points you should get something like this:

From these graphs you can notice one very important property. These functions are periodic. For a function, to be periodical means that one point after a certain period will have the same value again, and after that same period will again have the same value.

A function is periodic if $ f(x) = f (x + p)$, where p is a certain period.

This is best seen from extremes. Take a look at maximums, they are always of value 1, and minimums of value -1, and that is constant. Their period is $2 \pi$.

**sin(x) = sin (x + 2 π)**

cos(x) = cos (x + 2 π)

cos(x) = cos (x + 2 π)

*odd or even*.

A function is even if $f(-x) = f(x)$

For example function $ f(x) = x^2$ is even because $ f(-x) = (-x)^2 = – x^2$, and function $ f( x )= x^3$ is odd because $ f(-x) = (-x)^3= – x^3$.

Now let’s get back to our trigonometry functions.

Function sine is an odd function. Why? This is easily seen from the unit circle. To find out whether the function is odd or even, we must compare its value in x and –x.

Draw any point, for example $\frac{\pi}{4}$, and then also draw $ – \frac{\pi}{4}$. For now we’ll mark that value with a b.

$ sin (\frac{\pi}{4}) = b \rightarrow sin (-\frac{\pi}{4}) = ?$

From the drawing we conclude that

$ Sin(-\frac{\pi}{4}) = – sin(\frac{\pi}{4})$. This means that the __function sine is odd function__.

$ Sin(-x) = – sin(x)$.

Now take a look at the cosine value: *they are equal*.

This means that the cosine is an even function.

$ Cos(-x) = cos(x)$.

## Tangent and cotangent

**Tangent** is a function derived from functions sine and cosine.

It is defined as

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

Since tangent is defined as a fraction, he has to have particular conditions about his domain. Denominator can not be zero. This means that the domain of tangent will be whole set of real numbers except the points where cosine reaches zero. Those points are $ (\frac{\pi}{2} + k \pi, k \epsilon Z)$. What does this notation mean? You have all whole numbers k. that means $ {…-3, -2, -1, 0, 1, 2, 3…}$ when you multiply that with $\pi$ and add $\frac{\pi}{2}$ you’ll always have a zero for cosine. It means that on your first π/2 you add or subtract as many π as you want, you’ll always get zero.

That means that tangent is a function whose domain is $ \Re / ( \frac{\pi}{2} + k \pi, k \epsilon Z)$, and codomain whole $\Re$.

Again, conclude how the graph should look from the unit circle.

Which value will function tangent reach in points $\frac{\pi}{2}$ and $ -\frac{\pi}{2}$? In these points the line that tells us where the value of tangent is, is now parallel to the other line. That means that in those points their value will be $\pm \infty$.

The zeros of tangent will be the same as for sine (we simply equalize the tangent with zero and solve simple equation).

The lines where the function tangent goes to infinity are called asymptotes. Asymptotes are lines to who functions infinitely approaching but never touch.

For function tangent asymptotes will be lines vertical on x-axis that go through the $ (\frac{\pi}{2} + k \pi, k \epsilon Z)$.

This is how it should look now:

Just watch where your function goes to $ +\infty$ and where to $ -\infty$. Remember to help yourself with the unit circle. All you need for the graphs of trigonometric functions are there; you just need to practice concluding from it.

And the final look of the function tangent:

**Cotangent** is a function derived from functions sine and cosine.

Cotangent is defined as

$cot (x) = \frac{cos(x)}{sin(x)}$

Cotangent is also function defined as a fraction, which means that the domain of cotangent will be whole set of real numbers without the zeros of the sine function.

Domain = $ \Re / (k \pi : k \epsilon Z)$, codomain $\Re$.

The zeros are the points where value of cosine is equal to 0, and the value where the graph will go into the infinity are in the zeros of sine. Also again draw the asymptotes, the zeros, and watch where your graph goes to the $\infty$ and $ -\infty$.

Tangent and cotangent are not even functions. Tangent and cotangent are also periodic functions with period π.

$ tan(-x) = – tan(x), tan(x + \pi) = tan(x)$

$cot(-x) = – cot(x), cot(x + \pi) = cot(x)$

Like any other functions, trigonometric functions can also be altered, they can be translated and be more “dense” or “rare”.

Let’s break down this function. Since you know how to draw a sine function, this will be easy.

1. $ f(x) = 2 sin(x)$

Examine this function. For every value of x, it’s sine value will be doubled, which means that even those end points will be doubled. This means that the codomain of this function, instead of [-1,1] will now be [-2,2]. Zeros and extremes will remain in the same points.

If you have any number instead of that 2, your function will be $ f(x) = a * sin(x)$. That implies that codomain of this function will be [-a, a].

Analog for $ f(x)= 3 cos(x)$

2. $ f(x) = sin(2x)$

The bigger the argument, the more “dense” your function will be. How to find zeros for this function? Easiest way is to use substitution.

$ f(x) = sin(2x)$, to find zeros you make an equation $ sin(2x) = 0$ and use substitution $ t = 2x$.

Now you have $ sin(t) = 0$, and this you know. $ Sin(t)$ will be zero when t is 0 or $ \pi (+2k \pi)$. If you put that back in your substitution you get that $ 0 = 2x$, and $\pi = 2x$, which means that one zero is 0, and other is $\frac{\pi}{2} (+2k \pi)$. Codomain remains the same because your “a” is equal to 1. Now your function looks like:

On the other hand, when your “b” is lesser than 1, your function will be more sparse.

For $ f(x) = sin(\frac{1x}{2})$

The same with the cosine:

$ f(x) = cos(2x)$

$ f(x) = cos(\frac{1x}{2})$

3. $ f(x) = sin(x + 2)$

When you have addition or subtraction in your argument, that number marks the distance which your graph makes to the left or right. If you have addition, whole graph will be translated to the left, and if you have subtraction to the right.

Again, zeros are found using the substitution: $ sin(x + 2) = 0$, $ t = x + 2$, $ sin(t) = 0$ and so on.

$ f(x)= sin(x – \pi)$

And the same goes with cosine.

4. $ f(x) = sin(x) + 2$

The number after the sine function represents the translation on the y – axis. If it’s positive, it will go up, and if negative down. Here, the codomain also changes, the domain is the domain of the sine ± that number.

For the function $ f(x) = sin(x) + 2$, the codomain of the sine is [-1, 1] if we add 2, the codomain of the function $ f(x) = sin(x) + 2$ is [1, 3]. If there are no variations with the argument of the sine function, the maximums and minimums will remain in the same points.

And of course, to make it more interesting you can combine all of these and get for example:

$ f(x) = 2 sin(7x + 4) + 2$

**Tangent and cotangent function variations**

Once you learn how to draw simple trigonometry functions you’ll learn them all. Everything that applies to sine and cosine also applies to tangent and cotangent.

Again, you find your zeros using the substitution exactly like for the sine and cosine.

1. $ f(x)= a * tan(x)$

If “a” is greater than 1, the function will be narrower, which means that it will “grow” much faster, and if it is less than 1 it will grow slower.

$ f(x) = 2 tan(x)$

$ f(x) = 0.5 tan(x)$

2. $ f(x) = tan(bx)$

In this function “b” will change the domain. Domain of regular tangent is $\Re / (\frac{\pi}{2} + k \pi, k \epsilon Z)$, domain of function $f(x) = tan(bx)$ will be $\Re / (\frac{\pi}{b_2} + k \pi, k \epsilon Z)$.

Which means that for $ f(x) = tan(2x)$ its domain is $ R / (π\frac{\pi}{4} + k \pi, k \epsilon Z)$

3. $ f(x) = tan(x + c)$. “c” in this function will change the domain and the location. You have two ways of drawing this function.

First one is to find zeros, and find where your graph goes to infinity-asymptotes.

This is done using substitution $ t = x + c$.

Second one is to draw the graph $ f(x) = tan(x)$ and then translate it. If c is positive, the graph will be translated to the left, and if it is negative to the right.

$ f(x) = tan(x + 2)$

4. $ f(x) = tan(x) + d$

As you can already guess, “d” will translate graph according to y – axis. If d is positive, whole graph will be translated upwards, and if it is negative downwards.

Zeros will change, but the domain will remain the same.

The same graph transformations will apply to cotangent.

## Graphs of inverse trigonometric functions

If we want to draw graph of some inverse function, we must make sure we can do that. We can’t lose some properties that are strictly connected to the function definition.

Simplified, you can’t find inverse function of function that any line parallel to the x- axis cuts in more than one point. Let’s take a look at our sine function first.

If we also draw line $ y = \frac{1}{2}$:

This line will cut our sine function in many points. We want to restrict our function so that any line we draw parallel to x – axis cuts sine in exactly one point.

The usual restriction is on $ [-\frac{\pi}{2}, \frac{\pi}{2}]$.

Inverse function will in a way replace the properties of domain and codomain. That means that the domain for Arc sin will be [-1, 1] and codomain $ [-\frac{\pi}{2}, \frac{\pi}{2}]$.

Since all values will be inverted, this graph will be symmetrical to sine considering line $ x = y$. The easiest way to draw this is to draw line $ x = y$, and draw endpoints, on y – axis those will be $\frac{\pi}{2}$ and $ -\frac{\pi}{2}$, and on y – axis.

And then you simply translate every point. Just be careful not to cross over from domain or codomain.

Note how graph will never cross -1 and 1 on the x axis.

$ Arccos(x)$

For this function the usual restriction is $ [0, \pi]$

Domain of arcus cosine is [-1, 1] and codomain $ [0, \pi]$.

You do everything the same as for the sine. Graph of arcus cosine is symmetrical to the graph of cosine considering line $ x = y$. first mark your endpoints, and line $ x = y$, and do the symmetrical mapping.

Graph of arcus tangent is gained the same.

Its domain will be $ <-\infty, +\infty>$ and codomain $ <-\frac{\pi}{2}, \frac{\pi}{2}>$. Lines $ y = -\frac{\pi}{2}$ and $y = \frac{\pi}{2}$ will now be horizontal asymptotes.

And for arcus cotangent, common restriction is domain $ <0, +\infty>$ and codomain $ <0, +\infty>$.

$ Arcctg(x)$ is defined as $ arctan (\frac{1}{x})$ and because of that he will have a graph similar to $\frac{1}{x}$, but here graph will intersect y- axis in $\frac{\pi}{2}$ and $ -\frac{\pi}{2}$.

## Graphs of trigonometric functions worksheets

**Graphing trigonometric functions** (889.6 KiB, 711 hits)