1.17: Trigonometric Functions and Angles of Rotation
You've been working hard in your math class, and are getting to be quite the expert on trig functions. Then one day your friend, who is a year ahead of you in school, approaches you.
"So, you're doing pretty well in math? And you're good with trig functions?" he asks with a smile.
"Yes," you reply confidently. "I am."
"Alright, then what's the sine of
"What? That doesn't make sense. No right triangle has an angle like that, so there's no way to define that function!" you say.
Your friend laughs. "As it turns out, it is quite possible to have trig functions of angles greater than
Is your friend just playing a joke on you, or does he mean it? Can you actually calculate
At the conclusion of this Concept, you'll be able to answer this question.
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James Sousa: Determine Trigonometric Function Values Using the Unit Circle
Guidance
Just as it is possible to define the six trigonometric functions for angles in right triangles, we can also define the same functions in terms of angles of rotation. Consider an angle in standard position, whose terminal side intersects a circle of radius
The point
And, we can extend these functions to include nonacute angles.
Consider an angle in standard position, such that the point
This circle is called the unit circle. With
Notice that in the unit circle, the sine and cosine of an angle are the
We can use the figure above to determine values of the trig functions for the quadrantal angles. For example,
Example A
The point (3, 4) is a point on the terminal side of an angle in standard position. Determine the values of the six trigonometric functions of the angle.
Solution:
Notice that the angle is more than 90 degrees, and that the terminal side of the angle lies in the second quadrant. This will influence the signs of the trigonometric functions.
Notice that the value of
Example B
Use the unit circle above to find the value of
Solution:
The ordered pair for this angle is (0, 1). The cosine value is the
Example C
Use the unit circle above to find the value of
Solution:
The ordered pair for this angle is (1, 0). The ratio
Guided Practice
Use this figure:
to answer the following questions.
1. Find
2. Find
3. Find
Solutions:
1. We can see from the "x" and "y" axes that the "x" coordinate is
2. We know that
3. We know that
Concept Problem Solution
Since you now know that it is possible to apply trigonometric functions to angles greater than
Therefore,
Explore More
Find the values of the six trigonometric functions for each angle below.

0∘ 
90∘ 
180∘ 
270∘  Find the sine of an angle that goes through the point \begin{align*}(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2})\end{align*}
(2√2,2√2) .  Find the cosine of an angle that goes through the point \begin{align*}(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2})\end{align*}.
 Find the tangent of an angle that goes through the point \begin{align*}(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2})\end{align*}.
 Find the secant of an angle that goes through the point \begin{align*}(\frac{\sqrt{3}}{2},\frac{1}{2})\end{align*}.
 Find the cotangent of an angle that goes through the point \begin{align*}(\frac{\sqrt{3}}{2},\frac{1}{2})\end{align*}.
 Find the cosecant of an angle that goes through the point \begin{align*}(\frac{\sqrt{3}}{2},\frac{1}{2})\end{align*}.
 Find the sine of an angle that goes through the point \begin{align*}(\frac{1}{2},\frac{\sqrt{3}}{2})\end{align*}.
 Find the cosine of an angle that goes through the point \begin{align*}(\frac{\sqrt{3}}{2},\frac{1}{2})\end{align*}.
 The sine of an angle in the first quadrant is \begin{align*}0.25\end{align*}. What is the cosine of this angle?
 The cosine of an angle in the first quadrant is \begin{align*}0.8\end{align*}. What is the sine of this angle?
 The sine of an angle in the first quadrant is \begin{align*}0.15\end{align*}. What is the cosine of this angle?
Image Attributions
Description
Learning Objectives
Here you'll learn how to use trig functions on angles of rotation about a circle.
Difficulty Level:
At GradeAuthors:
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Concept Nodes:
Date Created:
Sep 26, 2012Last Modified:
Feb 26, 2015Vocabulary
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