The Unit Circle

The unit circle is a circle with a centre corresponding to the origin of the Cartesian plane |(0,0)| and a radius measuring |1| unit:

The equation of the circle with Cartesian coordinates is |x^2+y^2=1|.

• Each point is called a trigonometric point |P(\theta)| where |\theta| is the measure of an angle in radians. In short, a trigonometric point is a point on the unit circle.

• The coordinates |(x,y)| of each of the points represent |\cos \theta| and |\sin \theta|, respectively. ||P(\theta)=(x,y)= (\cos \theta,\sin \theta)||

• Therefore, the unit circle’s equation |x^2+y^2=1| becomes: ||\cos^2 \theta + \sin^2 \theta=1||

We generally use radians rather than degrees when working with the unit circle.

When the Cartesian coordinates of the point are known, check whether the equation |x^2+y^2=1| is respected. If not, the given point is not on the unit circle.

Is the point |(0.6,0.8)| on the unit circle?

Replace |x| and |y| in |x^2+y^2=1| by the values of the point. ||0.6^2+0.8^2=0.36+0.64=1||

Therefore, the point |(0.6,0.8)| is on the unit circle.

Is the point |\left(\dfrac{1}{4}, \dfrac{2}{3}\right)| on the unit circle?

Replace |x| and |y| in |x^2+y^2=1| by the values of the point. ||\left(\dfrac{1}{4}\right)^2 + \left(\dfrac{2}{3}\right)^2 = \dfrac{1}{16} + \dfrac{4}{9} = \dfrac{73}{144} \neq 1||

Therefore, the point |\left(\dfrac{1}{4}, \dfrac{2}{3}\right)| is not on the unit circle.

A cyclical function associates a point |P(\theta)| on the unit circle centred in |(0,0)| with any real number |\theta| on the real number line. It is a function |f| such that |\begin{eqnarray*}f: \mathbb{R} &\to& [-1,1] \times [-1,1] \\ \theta &\mapsto & (\cos \theta, \sin \theta) \end{eqnarray*}.|

A period is the smallest positive real number |p|, such as |P(\theta + p) = P(\theta).|

In terms of trigonometric points, the period is |2\pi| radians.

Different formulas can be established. Here are two of them:

|\bullet| |\cos(\theta + 2\pi n) = \cos \theta|

|\bullet| |\sin(\theta + 2\pi n) = \sin \theta|

The two formulas are valid regardless of the value of |\theta|. In addition, the values of |n| must be integers.

For example, take the trigonometric point |P(\theta) = \left(\dfrac{5}{13},\dfrac{12}{13}\right).|

We want to know the value of the point when the angle is |\theta + 8 \pi| radians.

Note that |8\pi| is a multiple of |2\pi|. Indeed, |4 \times 2\pi = 8\pi.| Thus, |n=4.|

Apply the two formulas shown above:

• |\cos(\theta + 8\pi) = \cos \theta = \dfrac{5}{13}|

• |\sin(\theta + 8\pi) = \sin \theta = \dfrac{12}{13}|

This gives the same point |P(\theta + 8\pi) = \left(\dfrac{5}{13},\dfrac{12}{13}\right).|

Here are the steps in more detail:

1. Calculate the number of rotations |N| to be subtracted from the given angle |t| by dividing the angle measurement by |2\pi| and rounding the result obtained down to the nearest integer:
   
   |\displaystyle N=\left[\frac{t}{2\pi}\right]|

2. Subtract the angle formed by the number of rotations previously calculated from the given angle (thus, the measurement is |2\pi N|): |t'=t-2\pi N|

3. Locate the result of the previous subtraction, which is a number between |0| and |2\pi,| on the unit circle and indicate its coordinates.

BE CAREFUL: When the angle has a negative value, the value of |N| will also be negative. In this case, subtract a negative multiple of |2\pi,| which amounts to adding rotations to find the result in the interval |[0,2\pi].|

Determine the position of the point on the unit circle |P(\frac{31\pi}{6})| and indicate its coordinates.

1. Calculate the number of rotations
   
   |\displaystyle N= \left[t\div2\pi\right]|
   |\displaystyle N=\left[\frac{31\pi}{6}\div2\pi \right]|
   |N=[2{.}583...]|
   |N=2|

2. Subtract the number of rotations
   
   |t'=t-N\times2\pi|
   |t'=\frac{31\pi}{6}-2\times2\pi|
   |t'=\frac{31\pi}{6} -\frac{24\pi}{6}|
   |t'=\frac{7\pi}{6}|

3. Locate the point
   
   The point coincides with the point |P(\frac{7\pi}{6}) = (-\frac{\sqrt{3}}{2},-\frac{1}{2})| so its coordinates are: ||P\left(\frac{31\pi}{6}\right)=\left(-\frac{\sqrt{3}}{2},-\frac{1}{2}\right)||

Determine the position of the point on the unit circle |P(\frac{-26\pi}{3})| and indicate its coordinates.

1. Calculate the number of rotations
   
   |N=[t\div2\pi]|
   |\displaystyle N= \left[\frac{-26\pi}{3}\div2\pi \right]|
   |N=[-4{,}33...]|
   |N=-5|

2. Subtract the number of rotations
   
   |t'=t-N\times2\pi|
   |t'=\dfrac{-26\pi}{3}-(-5)\times2\pi|
   |t'=\dfrac{-26\pi}{3} -(-10\pi)|
   |t'=\dfrac{4\pi}{3}|

3. Locate the point
   
   The point coincides with the point |P(\frac{4\pi}{3})=(-\frac{1}{2},-\frac{\sqrt{3}}{2})| so its coordinates are: ||P\left(\frac{-26\pi}{3}\right)=\left(-\frac{1}{2},-\frac{\sqrt{3}}{2}\right)||

Here are the steps in more detail:

1. On the unit circle, locate the point whose coordinates are known and find the value of |t’| associated with the point in the interval |[0,2\pi].|

2. Calculate the number of rotations to add to |t’| using the formula |\displaystyle N=\left[\frac{b-t'}{2\pi}\right]|, where the square brackets denote rounding down to the nearest integer and where |b| represents the upper boundary of the requested interval.

3. Add to |t’| the number of rotations calculated previously using the formula: |t=t' + 2\pi N|

Determine the value of the angle |t| in radians if |P(t) = \left(\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2}\right)| and if |t \in [9\pi,11\pi].|

1. Location of the point on the unit circle
   
   Between |0| and |2\pi|, the angle at the centre corresponding to the point with coordinates |\left(\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2}\right)| is |\frac{7\pi}{4}.|
   Note the point |t’.|

2. Calculation of the number of rotations
   
   |\displaystyle N=\left[\frac{b-t'}{2\pi}\right]|
   |\displaystyle N=\left[\frac{11\pi-\frac{7\pi}{4}}{2\pi}\right]|
   |\displaystyle N=\left[\frac{\frac{37\pi}{4}}{2\pi}\right]|
   |\displaystyle N=\left[\frac{37}{8}\right]|
   |N=[4{.}625]|
   |N=4|

3. Add the number of rotations
   
   |t=t'+N\times2\pi|
   |t=\dfrac{7\pi}{4}+4\times2\pi|
   |t=\dfrac{7\pi}{4}+8\pi|
   |t=\dfrac{39\pi}{4}|

In the interval |[9\pi,11\pi]| , the point |P\left(\dfrac{39\pi}{4}\right)| has the coordinates |\left(\dfrac{\sqrt{2}}{2},-\dfrac{\sqrt{2}}{2}\right).|

Determine the value of the angle |t| in radians if |P(t)=(\frac{\sqrt{3}}{2},\frac{1}{2})| and if |t \in [-6\pi,-4\pi].|

1. Location of the point on the unit circle
   
   Between |0| and |2\pi|, the central angle corresponding to the point with coordinates |(\frac{\sqrt{3}}{2},\frac{1}{2})| is |\dfrac{\pi}{6}.| Note the point |t'.|

2. Calculation of the number of rotations
   
   |\displaystyle N=\left[\frac{b-t'}{2\pi}\right]|
   |\displaystyle N=\left[\frac{-4\pi-\frac{\pi}{6}}{2\pi} \right]|
   |\displaystyle N=\left[\frac{\frac{-25\pi}{6}}{2\pi}\right]|
   |\displaystyle N=\left[\frac{-25}{12} \right]|
   |N=[-2{.}083...]|
   |N=-3|

3. Add the number of rotations
   
   |t=t'+N\times2\pi|
   |t=\dfrac{\pi}{6}+-3\times2\pi|
   |t=\dfrac{\pi}{6}-6\pi|
   |t=\dfrac{-35\pi}{6}|

In the interval |[-6\pi,-4\pi]|, the point |P\left(\dfrac{-35\pi}{6}\right)| has the coordinates |\left(\dfrac{\sqrt{3}}{2},\dfrac{1}{2}\right).|