Angles and the Unit Circle

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Angles in Trigonometry

Even though the word trigonometry is derived from the word “triangle”, you’ll see a lot of circles when you work with Trig!

We talked about angle measures in the Introduction to Trigonometry section, and now we’ll see how angles relate to the circumference of a circle.

Again, an angle is made up of two rays. A ray is a line that extends forever starting at a point called a vertex. Think of the initial side ray as the ray where the angle starts, and the terminal side ray as the ray where the angle stops.

An angle is in standard position if its vertex is at the origin \(\left( {0,\,0} \right)\) and its initial side starts at the positive \(x\)-axisQuadrants are the four different areas of the coordinate system, and “Quadrant I” (“Quadrant One” ) starts where both \(x\) and \(y\) are positive. It helps me to remember where the quadrants are if you draw a big “C” across the coordinate system:


So Quadrant I is between and 90°, Quadrant II is between 90° and 180°Quadrant III is between 180° and 270°, and Quadrant IV is between 270° and 360° (back to ). Don’t worry why – someone just decided it would be this way.

Angle measurements start at 0 on the positive \(x\)-axis and go counter-clockwise around the “circle” until they are back at 360, as shown in the graphs below. Remember from Geometry that there are 360 degrees (360°) in a circle. When going clockwise from the positive \(x\)-axis, angles measurements are negative.

Here are some examples. Note that these graphics are extremely important to understand how angles are set up in Trigonometry:

Positive Angles Negative Angles

Degrees, Radians, and Co-terminal Angles


Again, since there are 360° in a circle; do you see why it makes sense that an angle that is formed by rotating the terminal side 360° actually coincides with itself – it’s the same angle? We call an angle like this co-terminal, since they have the same angle measurement, starting from .


One degree happens to be \(\displaystyle \frac{1}{{360}}\) of a complete revolution, and thus, as we saw above, a right angle pointing up is 90°, a straight angle pointing to the left is 180°, and a right angle pointing down is 270°.

Converting Decimal Degrees to/from Degrees, Minutes and Seconds

Sometimes you’ll have to know how to convert degrees in decimals to degrees, minutes, and seconds (a measure dating back to the Babylonians, and is used in latitudes and longitudes).

To do these conversions, we’ll have to multiply or divide by powers of 60, since there are 60 seconds in a minute and 60 minutes in a degree (thus 3600 seconds in 1 degree).

Let’s first see an example where we convert degrees in decimals to degrees, minutes, and seconds:

Convert 48.78° to degrees, minutes, and secondsThe “whole” part of the degrees is 48, so we have 48°We have .78 leftover, which is \(\displaystyle .78\times 60=\,46.{8}’\) (minutes). We have .8 leftover, which is \(.8\times 60=\,48″\) (seconds). So we have \(48.78{}^\circ =48{}^\circ \,46’\,\,48″\).

To do this conversion on the graphing calculator (make sure in mode-DEGREE), type in 48.78, then hit 2nd apps (angle), then hit 4 or scroll to DMS, and hit ENTER:

Degrees In CalculatorConvert \(48{}^\circ \,46’\,\,48″\)  to decimal degrees:  When converting back to degrees in decimal, we need to divide the minutes by 60 and the seconds by 3600 and add to the whole degrees: \(\displaystyle 48{}^\circ \,46’\,48″=48+\frac{{46}}{{60}}+\frac{{48}}{{3600}}=48.78{}^\circ \).

To do this conversion on the graphing calculator (make sure in mode-DEGREE), type in 48 for degrees, then hit 2nd apps (angle) 1. Then type in 46 for minutes and hit 2nd apps (angle) 2. Then type in the 48 for seconds and hit alpha + (seconds), and hit enter: Degrees In Calculator 2


As you get more advanced in trig, you’ll deal with radians instead of degrees, and these can be confusing. Now we’re getting more into the circle part of trig.

First some definitions: if you were to create a circle around the origin, a central angle is the angle formed by having the vertex at the origin, and the arc length of that circle is the measurement around the circumference of that circle (like if you were to take measuring tape around the circle).

The word radian comes from the word radius, and is just another way to measure angles in a “trig circle”. One radian means that the length of the radius of that circle is the same as the arc length; and we like to deal with “trig circles” where the radius is 1.

And remember that the circumference of a circle is \(2\pi r\) so when the radius is 1 in a “trig circle”, all the way around is \(2\pi \left( 1 \right)\) or \(2\pi\) radians. This circle is called a Unit Circle; you will get intimately involved with this circle ;). \(2\pi\) is about 6.28 (2 times 3.14).


Again, the coordinate system consists of four quadrants. Quadrant I is between 0 and \(\displaystyle \frac{\pi }{2}\) radians, Quadrant II is between \(\displaystyle \frac{\pi }{2}\) and \(\pi\) radians, Quadrant III is between \(\pi \) and \(\displaystyle \frac{3\pi }{2}\) radians, and Quadrant IV is between \(\displaystyle \frac{3\pi }{2}\) and \(2\pi\) radians (back to 0 radians).   

The main reason we want to work with radians and not degrees is that degrees are sort of artificial numbers and radians have real meaning since they are tied to the circumference of a circle. There are many other reasons, too, since it’s much easier to do calculations with radians than degrees in Calculus.

But the weird thing about radians is that they really don’t have a unit, like degrees, feet or meters. They are just radians.

Here are some graphics of the concept of radians, which may help:

Concept of Radians

Co-Terminal Angles

We saw earlier that a complete revolution of the “trig circle” is 360° or \(2\pi \) radians.

So if we are given an angle that is greater than either 360° or \(2\pi \) radians (either in positive or negative measurements), we have to keep subtracting (or adding, if we have a negative angle) either 360 or \(2\pi\) until we get an angle between 0 and 360° (or 0 and \(2\pi\) radians). We call the new angle co-terminal with the original angle, since they are the same angle in the “trig circle”.

Let’s do some problems. Notice that when dealing with radians, we have to remember how to get and use Greatest Common Denominators to Add Fractions. Don’t worry; all of this will get easier!

Co-Terminal Problem


Find the angle that is co-terminal (and between 0 and 360°) to the angle 450°. Since 450 > 360, we can just subtract 360 once to get \(450-360=90{}^\circ \).
Find the angle that is co-terminal (and between 0 and \(2\pi \) radians) to the angle \(\displaystyle \frac{{14\pi }}{3}\). Since \(\displaystyle \frac{{14\pi }}{3}\) is positive and in radians (clue: no degree sign, and has a \(\pi \) in it), we can just keep subtracting  \(2\pi \), or \(\displaystyle \frac{{6\pi }}{3}\) (getting common denominator) until we get between 0 and \(2\pi \): \(\displaystyle \frac{{14\pi }}{3}-\frac{{6\pi }}{3}=\frac{{8\pi }}{3}-\frac{{6\pi }}{3}=\frac{{2\pi }}{3}\).
Find the angle that is co-terminal (and between 0 and \(2\pi \) radians) to the angle \(\displaystyle -\frac{{90\pi }}{7}\).


Graphic representation:

Since we’re dealing with a negative angle we can just keep adding \(2\pi \), or \(\displaystyle \frac{{14\pi }}{7}\) (getting common denominator) until we get between 0 and \(2\pi \):


\(\displaystyle -\frac{{90\pi }}{7}+\frac{{14\pi }}{7}=-\frac{{76\pi }}{7}+\frac{{14\pi }}{7}=-\frac{{62\pi }}{7}+\frac{{14\pi }}{7}=-\frac{{48\pi }}{7}+\frac{{14\pi }}{7}\)

\(\displaystyle \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=-\frac{{34\pi }}{7}+\frac{{14\pi }}{7}=-\frac{{20\pi }}{7}+\frac{{14\pi }}{7}=-\frac{{6\pi }}{7}+\frac{{14\pi }}{7}=\frac{{8\pi }}{7}\)



Note that the easier way to do this is to divide 90 by 14 (2 times 7, since we want multiples of \(2\pi \)) to see that the remainder is 6. Since \(\displaystyle -\frac{{90\pi }}{7}\) is negative, we want to go backwards around the circle by \(\displaystyle \frac{{6\pi }}{7}\) (which would be \(\displaystyle -\frac{{6\pi }}{7}\)). If we add \(2\pi \) to \(\displaystyle -\frac{{6\pi }}{7}\), we end up at \(\displaystyle \frac{{8\pi }}{7}\). Tricky!


Here are more types of problems you may see with co-terminal angles:

Co-Terminal Angle Problem Solution
List all co-terminal angles between \(-360\le \theta \le 360\):


(a) –150°      (b) 458°

(a) Since subtracting 360 will make the angle too negative, we need to add multiples of 360 until we get angles in between –360° and 360°: \(-150+360=210\). Since adding 360 again will be too large, the only angle co-terminal to –150° between –360° and 360° is 210°.


(b) Subtract multiples of 360 until we get angles in between –360° and 360°: \(458-360=98-360=-262\). (Subtracting 360 again will be too small). The co-terminal angles of 458° between –360° and 360° are 98° and –262°.

List all co-terminal angles between \(-4\pi \le \theta \le 4\pi \):


(a) \(\displaystyle \frac{{5\pi }}{4}\)        (b) \(\displaystyle -\frac{{2\pi }}{3}\)

(a) Subtract and add multiples of \(2\pi \) until we get angles in between \(-4\pi \) and \(4\pi \), or between \(\displaystyle -\frac{{16\pi }}{4}\) and \(\displaystyle \frac{{16\pi }}{4}\): \(\displaystyle \frac{{5\pi }}{4}-\frac{{8\pi }}{4}=-\frac{{3\pi }}{4}-\frac{{8\pi }}{4}=-\frac{{11\pi }}{4}\);   \(\displaystyle \frac{{5\pi }}{4}+\frac{{8\pi }}{4}=\frac{{13\pi }}{4}\). The co-terminal angles of \(\displaystyle \frac{{5\pi }}{4}\) are \(\displaystyle -\frac{{3\pi }}{4},-\frac{{11\pi }}{4},\) and \(\displaystyle \frac{{13\pi }}{4}\).


(b) Add and subtract multiples of \(2\pi \) until we get angles in between \(-4\pi \) and \(4\pi \), or between \(\displaystyle -\frac{{12\pi }}{3}\) and \(\displaystyle \frac{{12\pi }}{3}\):  \(\displaystyle -\frac{{2\pi }}{3}-\frac{{6\pi }}{3}=-\frac{{8\pi }}{3}\);   \(\displaystyle -\frac{{2\pi }}{3}+\frac{{6\pi }}{3}=\frac{{4\pi }}{3}+\frac{{6\pi }}{3}=\frac{{10\pi }}{3}\). The co-terminal angles of \(\displaystyle -\frac{{2\pi }}{3}\) are \(\displaystyle -\frac{{8\pi }}{3},\frac{{4\pi }}{3},\) and \(\displaystyle \frac{{10\pi }}{3}\).

The following angles terminate in which quadrant?


(a) \(\displaystyle \frac{{17\pi }}{3}\)      (b) –655°

(a) Since \(\displaystyle \frac{{17\pi }}{3}\) is greater than \(2\pi \), we need to subtract multiples of \(2\pi \) until we get an angle between 0 and \(2\pi \), or \(\displaystyle -\frac{{6\pi }}{3}\,\) and \(\displaystyle \frac{{6\pi }}{3}\):  \(\displaystyle \frac{{17\pi }}{3}-\frac{{6\pi }}{3}=\frac{{11\pi }}{3}-\frac{{6\pi }}{3}=\frac{{5\pi }}{3}\). Since \(\displaystyle \frac{{5\pi }}{3}\) is between \(\displaystyle \frac{{3\pi }}{2}\) and \(2\pi \) radians, it’s in Quadrant IV.


(b) Since –655° is less than 0, we have to add multiples of 360 until we get an angle in between 0 and 360°: \(-655+360=-295+360=65{}^\circ \). Since 65°  is between and 90°, it’s in Quadrant I.

Converting Degrees to Radians, and Radians to Degrees

You’ll have to know how to convert back and forth between degrees and radians. I have a somewhat simple trick to remember how to do this. Since we saw above that there are \(2\pi\) radians in 360°, we know that there are \(\pi \) radians in 180°.

So when doing the conversions between degrees and radians, we are always multiplying by either \(\displaystyle \frac{\pi }{180}\,\,\text{or }\frac{180}{\pi }\text{ }\). You can see this by converting 90 degrees to radians by using a Unit Multiplier:    

\(\displaystyle \require{cancel} \frac{{90\text{ }\cancel{{\text{degrees}}}}}{1}\,\times \,\frac{{\text{ }\!\!\pi\!\!\text{ radians}}}{{\text{180 }\cancel{{\text{degrees}}}}}\,=\,\frac{{90\pi \text{ radians}}}{{180}}\,=\,\frac{\pi }{2}\,\,\text{radians}\).

Here is the trick:  If you are converting from degrees to radians, you want a \(\pi\) in the answer, so the \(\pi\) should be on the top of the fraction (so multiply by \(\displaystyle \frac{\pi }{{180}}\)). If you are converting from radians to degrees, you want to get rid of the \(\pi\), so the \(\pi\) should be on the bottom of the fraction (so multiply by \(\displaystyle \frac{{180}}{\pi }\)).

Note that you can convert back and forth from degrees to radians in the graphing calculator. To convert from degrees to radians, put the calculator in RADIAN mode (mode – RADIAN), then type in the angle in degrees. Then hit 2nd apps (angle) 1 (degrees), and hit ENTER:
Degrees In Calculator 3


To convert back to degrees, put the calculator in DEGREE mode (mode – DEGREE), then type in the angle in radians. Then hit 2nd apps (angle) 3 (radians), and hit ENTER:
Degrees In Calculator 4


Let’s do some problems (without using graphing calculator conversions!):

Converting Between Degrees and Radians

The Unit Circle

The Unit Circle is probably the most important tool you’ll use in both Pre-Calculus, and then later (occasionally) in Calculus.

The Unit Circle is basically a visual representation of certain “special angles”, for which the exact values of the trig functions are known. It is called the “unit” circle, since its radius is 1.

The reason you’ll have to “memorize” the Unit Circle is so that you can come up with trig values for these angles quickly and without a calculator. That’s all it is!

So here it is. Embedded Math has good printouts of the Unit Circle here, and also a blank one so you can practice filling it in.

Note:  Sometimes you will see \(\displaystyle \frac{1}{\sqrt{2}}\) instead of \(\displaystyle \frac{\sqrt{2}} {2}\) in the Unit Circle. This is the same number; the former has not been rationalized, and the latter has (see how to rationalize fractions with radicals in the Powers, Exponents, Radicals and Scientific Notation Section here).

So here’s how it works (and we’ll show why below with an example):  For the ordered pairs on the outside, the first number (\(x\) value) is the cosine of the angle, and the second (\(y\) value) is the sine of the angle. If you were to divide the second by the first (second over first), you’ll get the tangent of the angle.

Before I give you hints on how to learn the Unit Circle, let’s see how we arrived at the order pairs by demonstrating Right Triangle Trigonometry.  

We know from either a 30-60-90 triangle (to be discussed later), using our calculator, or using the Pythagorean Theorem, that the sides for the triangle below are 1 for the hypotenuse (since it’s a Unit Circle), \(\displaystyle \frac{1}{2}\) for the shortest side or leg, and \(\displaystyle \frac{{\sqrt{3}}}{2}\) for the longer leg. Thus we can “prove” the sin and cos values for 30°. All the other special angles have similar proofs.

Reciprocal Trig Functions

To get the other three trigonometric values (cosecant or csc, secant or sec, and cotangent or cot), just take the reciprocals (flip) of the sin, cos, and tan values, respectively.

For example, the csc of 30° is \(\displaystyle \frac{1}{{\frac{1}{2}}}=2\), the sec of 30° is \(\displaystyle \frac{1}{\frac{\sqrt{3}}{2}}\,=\,\frac{2}{\sqrt{3}}\,=\,\frac{2\sqrt{3}}{3}\), and the cot of 30° is \(\displaystyle \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}=\sqrt{3}\).

Hints of Learning Unit Circle

I highly recommend that you print out a lot of blank Unit Circle charts (again, here on the Embedded Math website) and fill them out until you can do it without thinking 🙂 .

Also remember to get the other three trigonometric values (csc, sec, and cot values), just take the reciprocals of the sin, cos and tan values, respectively.

Here are more hints to help you learn it:

  • Notice that if you know the (cos, sin) ordered pair values in the first quadrant, you know them in all the quadrants! Look for the mirror images of the ordered pairs, but with the different signs for that quadrant.

I also like to use the mnemonic All Students Take Calculus, as shown above in the Unit Circle chart, to remember the signs. Starting in the first quadrant, All trigonometric values are positive, in the second quadrant, Sin (and thus csc) values are positive, in the third quadrant, Tan (and thus cot) values are positive, and in the fourth quadrant, Cos (and thus sec) values are positive.

  • Notice that the denominators of the three inside radian measurements are “3 4 6”, in every quadrant (with the “6” closest to the horizontal axis, and a \(\pi \) in the numerator). For the first quadrant, there is nothing in the numerator except for the \(\pi \), in the second quadrant the numerator is one less than the denominator, in the third quadrant, the numerator is one more than the denominator, and in the fourth quadrant, the numerator is twice the denominator minus 1.
  • Think of different slices of the circle, like slices of a pie. When dividing up the circle into fourths, just think of going around starting with 0, and adding 90° (or \(\displaystyle \frac{\pi }{2}\)) counter-clockwise. Then remember that if we have \(\left( {0,\,0} \right)\) in the center, we have \(\left( {1,\,0} \right)\), \(\left( {0,\,1} \right)\), \(\left( {-1,\,0} \right)\), and \(\left( {0,\,-1} \right)\) on the axes:

Dividing into Fourths

  • When dividing the circle into sixths, just think about going around the circle counterclockwise, and add 30° (or \(\displaystyle \frac{\pi }{6},\,\,\frac{{2\pi }}{6},\,\frac{{3\pi }}{6},\,\frac{{4\pi }}{6},\,\frac{{5\pi }}{6},\)…?Dividing into Sixths
  • So when dividing the circle into thirds, it will look like this. See, how starting with , we have \(\displaystyle \frac{\pi }{3},\,\,\frac{{2\pi }}{3},\,\frac{{3\pi }}{3},\frac{{4\pi }}{3},\,\frac{{5\pi }}{3},\,\text{and}\,\,\frac{{6\pi }}{3}\)?

Dividing into Thirds

  • Try to visualize the triangles in the Unit Circle, and think where the placement of the (sin, cos) coordinate points are. For example, for 30° in the first quadrant, the \(x\) value is greater than the \(y\) value, so we can remember that the coordinates of this point is \(\displaystyle \left( {\frac{{\sqrt{3}}}{2},\,\,\frac{1}{2}} \right)\) since \(\displaystyle \frac{{\sqrt{3}}}{2}>\frac{1}{2}\). (Of course \(\displaystyle \frac{\sqrt{2}}{2}=\frac{\sqrt{2}}{2}\), so we the coordinates for 45° are the same.) We can use this method to go around the triangle and remember the coordinates of the other special angles by just keeping in mind the signs of the \(x\) values and \(y\) values:

Remembering Unit Circle with Triangles

  • Some choose to remember the Unit Circle coordinates (sin, cos) pairs by remembering 13-22-31. Starting from the top right, if you remember 13-22-31 coming down to the \(x\)-axis, you can remember \(\displaystyle \left( {\frac{1}{2},\,\frac{{\sqrt{3}}}{2}} \right),\,\,\,\left( {\frac{{\sqrt{2}}}{2},\,\frac{{\sqrt{2}}}{2}} \right),\,\,\,\left( {\frac{{\sqrt{3}}}{2},\,\frac{1}{2}} \right)\). See how you have 13-22-31 on the top of the fractions (with square roots), and 2 on the bottom?

Then use the fact that the other coordinates are mirror images, making allowances for the different signs in each quadrant.


  • If you have a negative angle, just go forward to see how far you’ve gone, and then go that much the other way (clockwise) from 0. For example, to get the cos of \(\displaystyle -\frac{{4\pi }}{3}\), you look to see how far forward you get to \(\displaystyle \frac{{4\pi }}{3}\), and then you go backwards (clockwise) that much to end up at \(\displaystyle \frac{2\pi }{3}\). Of course you can always just add \(2\pi \) to \(\displaystyle -\frac{4\pi }{3}\) to get \(\displaystyle \frac{2\pi }{3}\).

Negative Angles on Unit Circle

Finding Exact Values of Angles Using Unit Circle

Here are some problems that you may have to work, using what you know about co-terminal angles and the Unit Circle.

Note that at first, you’ll want to redraw the whole Unit Circle if you have to know these for a test, and later, you can just draw part of the circle, since you’ll know it so well!

Unit Circle Problem Solution
Find the exact values without using a calculator:


\(\displaystyle \text{(a)}\,\,\sin \left( {\frac{{2\pi }}{3}} \right)\)      \(\displaystyle \text{(b)}\,\,\cos \left( {\frac{{-3\pi }}{2}} \right)\)

\(\displaystyle \text{(c)}\,\,\tan \left( {\frac{{9\pi }}{4}} \right)\)      \(\displaystyle \text{(d)}\,\,\cos \left( {\frac{{-8\pi }}{3}} \right)\)

\(\displaystyle \text{(e)}\,\,\sin \left( {\frac{\pi }{3}+2\pi n} \right)\), \(n\) is an integer

Let’s use the Unit Circle and find the \(y\) value for the sin values, the \(x\) value for the cos values, and the \(\displaystyle \frac{y}{x}\) value for the tan. Remember to go counterclockwise from 0 for positive values, and clockwise for negative values.


\(\displaystyle \text{(a)}\,\,\sin \left( {\frac{{2\pi }}{3}} \right)=\frac{{\sqrt{3}}}{2}\)      \(\displaystyle \text{(b)}\,\,\cos \left( {\frac{{-3\pi }}{2}} \right)=\cos \left( {\frac{\pi }{2}} \right)=0\)

\(\displaystyle \text{(c)}\,\,\tan \left( {\frac{{9\pi }}{4}} \right)=\tan \left( {\frac{\pi }{4}} \right)=1\)        \(\displaystyle \text{(d)}\,\,\cos \left( {\frac{{-8\pi }}{3}} \right)=\cos \left( {\frac{{-2\pi }}{3}} \right)=\cos \left( {\frac{{4\pi }}{3}} \right)=-\frac{1}{2}\)

\(\displaystyle \text{(e)}\,\,\sin \left( {\frac{\pi }{3}+2\pi n} \right)=\sin \left( {\frac{\pi }{3}} \right)=\frac{{\sqrt{3}}}{2}\)

For \(-2\pi \le \theta <2\pi \), find 4 angles whose sin is:


\(\displaystyle \text{(a)}-\frac{{\sqrt{3}}}{2}\)        \(\displaystyle \text{(b)}\,\frac{1}{2}\)




For \(-2\pi \le \theta <2\pi \), find all angles whose cos is:


\(\text{(c)}\,-1\)       \(\displaystyle \text{(d)}\,\frac{{\sqrt{2}}}{2}\)

Now we’ll use the Unit Circle to go backwards: we’ll find angles, given a coordinate.


(a)  Since the sin is negative, the angles will be in quadrants III and IV. The angles where sin (\(y\) value of coordinate) is \(\displaystyle -\frac{{\sqrt{3}}}{2}\) between 0 and \(2\pi \) are \(\displaystyle \frac{{4\pi }}{3}\) and \(\displaystyle \frac{{5\pi }}{3}\), and then going backwards from 0 to \(-2\pi \) are \(\displaystyle -\frac{\pi }{3}\) and \(\displaystyle -\frac{{2\pi }}{3}\).


(b)  Since the sin is positive, the angles will be in quadrants I and II. The angles where sin is \(\displaystyle \frac{1}{2}\) between 0 and \(2\pi \) are \(\displaystyle \frac{\pi }{6}\) and \(\displaystyle \frac{{5\pi }}{6}\), and then going backwards from 0 to \(-2\pi \) are \(\displaystyle -\frac{{7\pi }}{6}\) and \(\displaystyle -\frac{{11\pi }}{6}\).


(c)  Since the cos is negative, the angles should be in quadrants II and III. But where cos is  –1  is between these quadrants at \(\pi \). Going backwards from 0 to \(-2\pi \), we get cos is  –1 at \(-\pi \).


(d)  Since the cos is positive, the angles will be in quadrants I and IV. The angles where cos (\(x\) value of coordinate) is \(\displaystyle \frac{{\sqrt{2}}}{2}\) between  0 and \(2\pi \) are \(\displaystyle \frac{\pi }{4}\) and \(\displaystyle \frac{{7\pi }}{4}\), and then going backwards from 0 to \(\displaystyle -2\pi \) are \(\displaystyle -\frac{\pi }{4}\) and \(\displaystyle -\frac{{7\pi }}{4}\).



Now let’s work on getting all six (6) trigonometric values for the following special angles:

Unit Circle Problem Solution
Find the exact values of the 6 trigonometric functions of the following special angles:

(If not defined, say “undefined”).


\(\displaystyle \text{(a)}\,\frac{{5\pi }}{4}\)          \(\displaystyle \text{(b)}\,-\frac{{5\pi }}{2}\)



\(\text{(c)}-405{}^\circ \)         \(\displaystyle \text{(d)}\,\frac{{14\pi }}{3}\)

Note that we may have to add or subtract multiples of 360°  (or \(2\pi \)) to get the co-terminal angle that is between 0 and 360° (0 and \(2\pi \)).


\(\begin{align}\sin \left( {\frac{{5\pi }}{4}} \right)&=-\frac{{\sqrt{2}}}{2};\,\,\,\cos \left( {\frac{{5\pi }}{4}} \right)=-\frac{{\sqrt{2}}}{2};\,\,\,\tan \left( {\frac{{5\pi }}{4}} \right)=1\\\csc \left( {\frac{{5\pi }}{4}} \right)&=-\frac{2}{{\sqrt{2}}}\cdot \frac{{\sqrt{2}}}{{\sqrt{2}}}=-\sqrt{2};\,\,\,\sec \left( {\frac{{5\pi }}{4}} \right)=-\sqrt{2};\,\,\,\cot \left( {\frac{{5\pi }}{4}} \right)=1\end{align}\)



\(\begin{align}\sin \left( {-\frac{{5\pi }}{2}} \right)&=\sin \left( {-\frac{{5\pi }}{2}+\frac{{8\pi }}{2}} \right)=\sin \left( {\frac{{3\pi }}{2}} \right)=-1;\,\,\,\cos \left( {\frac{{3\pi }}{2}} \right)=0;\,\,\tan \left( {\frac{{3\pi }}{2}} \right)=\text{undefined}\\\csc \left( {\frac{{3\pi }}{2}} \right)&=-1;\,\,\,\sec \left( {\frac{{3\pi }}{2}} \right)=\text{undefined};\,\,\,\cot \left( {\frac{{3\pi }}{2}} \right)=0\end{align}\)



\(\begin{align}\sin \left( {-405{}^\circ } \right)&=\sin \left( {315{}^\circ } \right)=-\frac{{\sqrt{2}}}{2};\,\,\,\cos \left( {315{}^\circ } \right)=\frac{{\sqrt{2}}}{2};\,\,\,\tan \left( {315{}^\circ } \right)=-1\\\csc \left( {315{}^\circ } \right)&=-\frac{2}{{\sqrt{2}}}\cdot \frac{{\sqrt{2}}}{{\sqrt{2}}}=-\sqrt{2};\,\,\,\sec \left( {315{}^\circ } \right)=\sqrt{2};\,\,\,\cot \left( {315{}^\circ } \right)=-1\end{align}\)



\(\begin{align}\sin \left( {\frac{{14\pi }}{3}} \right)&=\sin \left( {\frac{{14\pi }}{3}-\frac{{12\pi }}{3}} \right)=\sin \left( {\frac{{2\pi }}{3}} \right)=\frac{{\sqrt{3}}}{2};\,\,\,\cos \left( {\frac{{2\pi }}{3}} \right)=-\frac{1}{2};\,\,\,\tan \left( {\frac{{2\pi }}{3}} \right)=-\sqrt{3}\\\csc \left( {\frac{{2\pi }}{3}} \right)&=\frac{2}{{\sqrt{3}}}\cdot \frac{{\sqrt{3}}}{{\sqrt{3}}}=\frac{{2\sqrt{3}}}{3};\,\,\,\sec \left( {\frac{{3\pi }}{2}} \right)-2;\,\,\,\cot \left( {\frac{{3\pi }}{2}} \right)=-\frac{1}{{\sqrt{3}}}\cdot \frac{{\sqrt{3}}}{{\sqrt{3}}}=-\frac{{\sqrt{3}}}{3}\end{align}\)

Understand these problems, and practice, practice, practice!

Click on Submit (the arrow to the right of the problem) to solve this problem. You can also type in more problems, or click on the 3 dots in the upper right hand corner to drill down for example problems.

If you click on “Tap to view steps”, you will go to the Mathway site, where you can register for the full version (steps included) of the software.  You can even get math worksheets.

You can also go to the Mathway site here, where you can register, or just use the software for free without the detailed solutions.  There is even a Mathway App for your mobile device.  Enjoy!

On to Linear and Angular Speeds, Area of Sectors, and Length of Arcs – you’re ready!