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Unit Circle

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The Unit Circle
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The unit circle is a circle of radius one, centered at the origin, that summarizes all the 30-60-90 and 45-45-90 triangle relationships that exist.  When memorized, it is extremely useful for evaluating expressions like \cos (135^\circ)  or \sin \left(-\frac{5 \pi}{3}\right) .  It also helps to produce the parent graphs of sine and cosine

How can you use the unit circle to evaluate \cos (135^\circ)  and \sin \left(-\frac{5 \pi}{3}\right)

Watch This

http://www.youtube.com/watch?v=i56P6xzsB5Y James Sousa: Determine Trigonometric Function Values Using the Unit Circle

Guidance

You already know how to translate between degrees and radians and the triangle ratios for 30-60-90 and 45-45-90 right triangles.  In order to be ready to completely fill in and memorize a unit circle, two triangles need to be worked out.  Start by finding the side lengths of a 30-60-90 triangle and a 45-45-90 triangle each with hypotenuse equal to 1.

30^\circ 60^\circ 90^\circ
x x \sqrt{3} 2x
\frac{1}{2} \frac{\sqrt{3}}{2} 1
45^\circ 45^\circ 90^\circ
x x x \sqrt{2}
\frac{\sqrt{2}}{2} \frac{\sqrt{2}}{2} 1

This is enough information to fill out the important points in the first quadrant of the unit circle.  The values of the x  and y  coordinates for each of the key points are shown below.  Remember that the x  and y  coordinates come from the lengths of the legs of the special right triangles, as shown specifically for the 30^\circ  angle.  Always remember to measure the angle from the positive portion of the x -axis.


Knowing the first quadrant well is the key to knowing the entire unit circle.  Every other point on the unit circle can be found using logic and this quadrant.

Example A

Use your knowledge of the first quadrant of the unit circle to identify the angles and important points of the second quadrant. 

Solution:  The heights are mirrored and equal which correspond to the  y values.  The  x values are all negative.

Example B

Identify a pattern in the heights of the points in the first quadrant to help you remember the points. 

Solution:  The heights of the points in the first quadrant are the y -coordinates which are:  0, \frac{1}{2}, \frac{\sqrt{2}}{2}, \frac{\sqrt{3}}{2}, 1

When rewritten, the pattern becomes clear: \frac{0}{2}, \frac{\sqrt{1}}{2}, \frac{\sqrt{2}}{2}, \frac{\sqrt{3}}{2}, \frac{\sqrt{4}}{2} .

The three points in the middle are the most often mixed up.  This pattern illustrates how they increase in size from small \frac{1}{2} , to medium \frac{\sqrt{2}}{2} , to large \frac{\sqrt{3}}{2} .  When you fill in the unit circle, look for the heights that are small, medium and large and this will tell you were each value should go.  Notice that the heights for these five points in the second quadrant are also 0, \frac{1}{2}, \frac{\sqrt{2}}{2}, \frac{\sqrt{3}}{2}, 1 .

This technique also works for the widths.  This can make memorizing the 16 points of the unit circle a matter of logic and the pattern: \frac{0}{2}, \frac{\sqrt{1}}{2}, \frac{\sqrt{2}}{2}, \frac{\sqrt{3}}{2}, \frac{\sqrt{4}}{2} .

Example C

Evaluate \cos 60^\circ  using the unit circle and right triangle trigonometry. What is the connection between the  x coordinate of the point and the cosine of the angle? 

Solution:  The point on the unit circle for 60^\circ  is \left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)  and the point is one unit from the origin.  This can be represented as a 30-60-90 triangle.


Since cosine is adjacent over hypotenuse, cosine turns out to be exactly the  x coordinate \frac{1}{2} .

Concept Problem Revisited

The  x value of a point along the unit circle corresponds to the cosine of the angle.  The  y value of a point corresponds to the sine of the angle.  When the angles and points are memorized, simply recall the  x or  y coordinate. 

When evaluating \cos (135^\circ)  your thought process should be something like this:

You know 135^\circ  goes with the point \left(-\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right)  and cosine is the  x portion.  So, \cos (135^\circ)=-\frac{\sqrt{2}}{2} .

When evaluating \sin \left(-\frac{5 \pi}{3}\right)  your thought process should be something like this:

You know  -\frac{5 \pi}{3} goes with the point \left(\frac{1}{2},-\frac{\sqrt{3}}{2}\right)  and sine is the  y portion.  So, \sin \left(-\frac{5 \pi}{3}\right)=-\frac{\sqrt{3}}{2} .

Vocabulary

Coterminal Angles are sets of angles such as -10^\circ, 350^\circ, 710^\circ  that start at the positive x -axis and end at the same terminal side.  Since coterminal angles end at identical points along the unit circle, trigonometric expressions involving coterminal angles are equivalent: \sin -10^\circ=\sin 350^\circ = \sin 710^\circ .

Guided Practice

1. Using knowledge of the first quadrant of the unit circle, identify the angles and important points of the third quadrant. 

2. For each of the six trigonometric functions, identify the quadrants where they are positive and the quadrants where they are negative. 

3. Evaluate the following trigonometric expressions using the unit circle. 

a.  \sin \frac{\pi}{2}

b.  \cos 210^\circ

c. \tan 315^\circ

d.  \cot 270^\circ

e.  \sec \frac{11 \pi}{6}

f.  \csc - \frac{5 \pi}{4}

Answers:

1. Both the  x values and  y values are negative and their respective coordinates correspond to those of the other quadrants.

2. In quadrant I, the hypotenuse, adjacent and opposite side are all positive.  Thus all 6 trigonometric functions are positive. 

In quadrant II the hypotenuse and opposite sides are positive and the adjacent side is negative.  This means that every trigonometric expression involving an adjacent side is negative.  Sine and its reciprocal cosecant are the only two trigonometric functions that do not refer to the adjacent side which makes them the only positive ones. 

In quadrant III only the hypotenuse is positive.  Thus the only trigonometric functions that are positive are tangent and its reciprocal cotangent because these functions refer to both adjacent and opposite sides which will both be negative. 

In quadrant IV the hypotenuse and the adjacent sides are positive while the opposite side is negative.  This means that only cosine and its reciprocal secant are positive. 

A mnemonic device to remember which trigonometric functions are positive and which trigonometric functions are negative is “ All S tudents T ake C alculus.”  All refers to all the trigonometric functions are positive in quadrant I.  The letter S refers to sine and its reciprocal cosecant that are positive in quadrant II.  The letter T refers to tangent and its reciprocal cotangent that are positive in quadrant III.  The letter C refers to cosine and its reciprocal secant that are positive in quadrant IV.

3. a. \sin \frac{\pi}{2}=1

b. \cos 210^\circ=-\frac{\sqrt{3}}{2}

c. \tan 315^\circ=-1

d. \cot 270^\circ=0

e. \sec \frac{11 \pi}{6}=\frac{1}{\cos \frac{11 \pi}{6}}=\frac{2}{\sqrt{3}}=\frac{2 \sqrt{3}}{3}

f. \csc-\frac{5 \pi}{4}=\frac{1}{\sin-\frac{5 \pi}{4}}=-\frac{2}{\sqrt{2}}=-\sqrt{2}

Practice

Name the angle between  0^\circ and  360^\circ that is coterminal with…

1. -20^\circ

2. 475^\circ

3. -220^\circ

4. 690^\circ

5. -45^\circ

Use your knowledge of the unit circle to help determine whether each of the following trigonometric expressions is positive or negative.

6.  \tan 143^\circ

7. \cos \frac{\pi}{3}

8. \sin 362^\circ

9. \csc \frac{3 \pi}{4}

Use your knowledge of the unit circle to evaluate each of the following trigonometric expressions.

10.  \cos 120^\circ

11. \sec \frac{\pi}{3}

12. \tan 225^\circ

13. \cot 120^\circ

14. \sin \frac{11 \pi}{6}

15. \csc 240^\circ

16. Find \sin \theta  and \tan \theta  if \cos \theta=\frac{\sqrt{3}}{2}  and \cot \theta > 0 .

17. Find  \cos \theta and  \tan \theta if \sin \theta =-\frac{1}{2}  and \sec \theta < 0 .

18. Draw the complete unit circle (all four quadrants) and label the important points.

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