A common approach in trigonometry is to use the unitary circle to represent the trigonometric functions. To represent an unitary circle we use a x-y Cartesian coordinate systems and a circle with the radius of 1. The center of the circle is in the origin of the x-y system.
The unitary circle is split in 4 parts, named quadrants. The first quadrant (I) is made by the positive x and y axes. If we draw any radius of the unitary circle in the first quadrant, the angle with the positive x axis can have any value between 0 and 90°.
An angle can be measured in degrees (°) or radians. A full circle has an angle of 360° or 2π radians.
In the table below we sum up the equivalent between degrees and radians for the quadrants of the unitary circle.
Quadrant I | Quadrant II | Quadrant III | Quadrant IV | |
Start | 0° (0) | 90° (π/2) | 180° (π) | 270° (3π/2) |
End | 90° (π/2) | 180° (π) | 270° (3π/2) | 360° (2π) |
In the picture below we can see the unitary circle with angles multiple of 30° and 45°. Also, for each angle degree value, we have the value in radians and the x-y coordinates of the point of intersection between the angle segment and unitary circle.
The conversion between degrees and radians is explained in detail in the article How to transform from degrees to radians and from radians to degrees.Now, let’s take any angle φ in the unitary circle. The angle φ is formed between the segment AB (length 1) and the x-axis. The x-y coordinates of the point B will be the length of the segments ABx and ABy (see below).
If we apply the definition of the trigonometric functions for the angle φ, we’ll get:
\[ \begin{equation*} \begin{split}\text{sin}(\varphi)&=\frac{BB_x}{AB}=\frac{BB_x}{1}=BB_x&=AB_y\\
\text{cos}(\varphi)&=\frac{AB_x}{AB}=\frac{AB_x}{1}&=AB_x\\
\text{tg}(\varphi)&=\frac{BB_x}{AB_x}&=\frac{AB_x}{AB_y}\\
\text{cosec}(\varphi)&=\frac{AB}{BB_x}=\frac{1}{BB_x}&=\frac{1}{AB_y}\\
\text{sec}(\varphi)&=\frac{AB}{AB_x}&=\frac{1}{AB_x}\\
\text{ctg}(\varphi)&=\frac{AB_x}{BB_x}&=\frac{AB_x}{AB_y}
\end{split} \end{equation*} \]
As you can see all the trigonometric functions are expressed only function of the x-y coordinates of the point B, the length of the segments ABx and ABy.
This way we can find the values of the trigonometric functions for any give angle φ.
As an exercise we are going to calculate and plot the values of the trigonometric functions for each quadrant. We’ll take 30° (quadrant I), 120° (quadrant II), 210° (quadrant III) and 300° (quadrant IV).
Quadrant I | Quadrant II | Quadrant III | Quadrant IV | |
Angle (φ) | 30° | 120° | 210° | 300° |
Sine | 0.50 | 0.87 | -0.50 | -0.87 |
Cosine | 0.87 | -0.50 | -0.87 | 0.50 |
Tangent | 0.58 | -1.73 | 0.58 | -1.73 |
Cosecant | 2.00 | 1.15 | -2.00 | -1.15 |
Secant | 1.15 | -2.00 | -1.15 | 2.00 |
Cotangent | 1.73 | -0.58 | 1.73 | -0.58 |
By analyzing the values of the trigonometric function we can see that:
- the sine function is positive in quadrants I and II and negative in quadrants II and IV
- the cosine function is positive in quadrants I and IV and negative in quadrants II and III
- the tangent function is positive in quadrants I and III and negative in quadrants II and IV
- the cosecant function is positive in quadrants I and II and negative in quadrants II and IV
- the secant function is positive in quadrants I and IV and negative in quadrants II and III
- the cotangent function is positive in quadrants I and III and negative in quadrants II and IV
There is a graphical representation of all the trigonometric functions of a right triangle inside a unitary circle. We can also visualize the result of the trigonometric functions as segments connected to each other.
We can see that the length of the segments are in fact the values returned by the trigonometric functions. What we get in each quadrant is a right triangle which’s segments and height are the results of the trigonometric functions.
By knowing how to represent the triangle for the first quadrant, we can figure out the other three by vertical and horizontal flipping of the first quadrant.
For any questions or observations regarding this tutorial please use the comment form below.
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