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1.20: Trigonometric Functions of Angles Greater than 360 Degrees

Difficulty Level: At Grade Created by: CK-12
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Practice Trigonometric Functions of Angles Greater than 360 Degrees
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While out at the local amusement park with friends, you take a ride on the Go Karts. You ride around a circular track in the carts three and a half times, and then stop at a "pit stop" to rest. While waiting for your Go Kart to get more fuel, you are talking with your friends about the ride. You know that one way of measuring how far something has gone around a circle (or the trig values associated with it) is to use angles. However, you've gone more than one complete circle around the track.

Is it still possible to find out what the values of sine and cosine are for the change in angle you've made?

When you complete this Concept, you'll be able to answer this question by computing the trig values for angles greater than \begin{align*}360^\circ\end{align*}360

Watch This

The Unit Circle


Consider the angle \begin{align*}390^\circ\end{align*}390. As you learned previously, you can think of this angle as a full 360 degree rotation, plus an additional 30 degrees. Therefore \begin{align*}390^\circ\end{align*}390 is coterminal with \begin{align*}30^\circ\end{align*}30. As you saw above with negative angles, this means that \begin{align*}390^\circ\end{align*}390 has the same ordered pair as \begin{align*}30^\circ\end{align*}30, and so it has the same trig values. For example,

\begin{align*}\cos 390^\circ = \cos 30^\circ = \frac{\sqrt{3}}{2}\end{align*}


In general, if an angle whose measure is greater than \begin{align*}360^\circ\end{align*}360 has a reference angle of \begin{align*}30^\circ\end{align*}30, \begin{align*}45^\circ\end{align*}45, or \begin{align*}60^\circ\end{align*}60, or if it is a quadrantal angle, we can find its ordered pair, and so we can find the values of any of the trig functions of the angle. Again, determine the reference angle first.

Example A

Find the value of the expression: \begin{align*}\sin 420^\circ\end{align*}sin420


\begin{align*}\sin 420^\circ = \frac{\sqrt{3}}{2}\end{align*}sin420=32

\begin{align*}420^\circ\end{align*}420 is a full rotation of 360 degrees, plus an additional 60 degrees. Therefore the angle is coterminal with \begin{align*}60^\circ\end{align*}60, and so it shares the same ordered pair, \begin{align*}\left ( \frac{1}{2}, \frac{\sqrt{3}}{2} \right )\end{align*}(12,32). The sine value is the \begin{align*}y-\end{align*}ycoordinate.

Example B

Find the value of the expression: \begin{align*}\tan 840^\circ\end{align*}tan840


\begin{align*}\tan 840^\circ = -\sqrt{3}\end{align*}tan840=3

\begin{align*}840^\circ\end{align*}840 is two full rotations, or 720 degrees, plus an additional 120 degrees:

\begin{align*}840 = 360 + 360 + 120\end{align*}


Therefore \begin{align*}840^\circ\end{align*}840 is coterminal with \begin{align*}120^\circ\end{align*}120, so the ordered pair is \begin{align*}\left ( -\frac{1}{2}, \frac{\sqrt{3}}{2} \right )\end{align*}(12,32). The tangent value can be found by the following:

\begin{align*}\tan 840^\circ = \tan 120^\circ = \frac{y}{x} = \frac{\frac{\sqrt{3}}{2}}{-\frac{1}{2}} = \frac{\sqrt{3}}{2} \times -\frac{2}{1} = -\sqrt{3}\end{align*}


Example C

Find the value of the expression: \begin{align*}\cos 540^\circ\end{align*}cos540


\begin{align*}\cos 540^\circ = -1\end{align*}cos540=1

\begin{align*}540^\circ\end{align*}540 is a full rotation of 360 degrees, plus an additional 180 degrees. Therefore the angle is coterminal with \begin{align*}180^\circ\end{align*}, and the ordered pair is (-1, 0). So the cosine value is -1.

Guided Practice

1. Find the value of the expression: \begin{align*}\sin 570^\circ\end{align*}

2. Find the value of the expression: \begin{align*}\cos 675^\circ\end{align*}

3. Find the value of the expression: \begin{align*}\sin 480^\circ\end{align*}


1. Since \begin{align*}570^\circ\end{align*} has the same terminal side as \begin{align*}210^\circ\end{align*}, \begin{align*}\sin 570^\circ = \sin 210^\circ = \frac{\frac{-1}{2}}{1} = \frac{-1}{2}\end{align*}

2. Since \begin{align*}675^\circ\end{align*} has the same terminal side as \begin{align*}315^\circ\end{align*}, \begin{align*}\cos 675^\circ = \cos 315^\circ = \frac{\frac{\sqrt{2}}{2}}{1} = \frac{\sqrt{2}}{2}\end{align*}

3. Since \begin{align*}480^\circ\end{align*} has the same terminal side as \begin{align*}120^\circ\end{align*}, \begin{align*}\sin 480^\circ = \sin 120^\circ = \frac{\frac{\sqrt{3}}{2}}{1} = \frac{\sqrt{3}}{2}\end{align*}

Concept Problem Solution

Since you've gone around the track 3.5 times, the total angle you've traveled is \begin{align*}360^\circ \times 3.5 = 1260^\circ\end{align*}. However, as you learned in this unit, this is equivalent to \begin{align*}180^\circ\end{align*}. So you can use that value in your computations:

\begin{align*} \sin 1260^\circ = \sin 180^\circ = 0\\ \cos 1260^\circ = \cos 180^\circ = -1\\ \end{align*}

Explore More

Find the value of each expression.

  1. \begin{align*}\sin 405^\circ\end{align*}
  2. \begin{align*}\cos 810^\circ\end{align*}
  3. \begin{align*}\tan 630^\circ\end{align*}
  4. \begin{align*}\cot 900^\circ\end{align*}
  5. \begin{align*}\csc 495^\circ\end{align*}
  6. \begin{align*}\sec 510^\circ\end{align*}
  7. \begin{align*}\cos 585^\circ\end{align*}
  8. \begin{align*}\sin 600^\circ\end{align*}
  9. \begin{align*}\cot 495^\circ\end{align*}
  10. \begin{align*}\tan 405^\circ\end{align*}
  11. \begin{align*}\cos 630^\circ\end{align*}
  12. \begin{align*}\sec 810^\circ\end{align*}
  13. \begin{align*}\csc 900^\circ\end{align*}
  14. \begin{align*}\tan 600^\circ\end{align*}
  15. \begin{align*}\sin 585^\circ\end{align*}
  16. \begin{align*}\tan 510^\circ\end{align*}
  17. Explain how to evaluate a trigonometric function for an angle greater than \begin{align*}360^\circ\end{align*}.

Answers for Explore More Problems

To view the Explore More answers, open this PDF file and look for section 1.20. 




Two angles are coterminal if they are drawn in the standard position and both have terminal sides that are at the same location.

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Difficulty Level:

At Grade


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Date Created:

Sep 26, 2012

Last Modified:

Feb 26, 2015
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