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Trig Functions, Reference Angles, and the Unit Circle

Date: 07/02/2004 at 11:22:09
From: Karen
Subject: Trig Functions & Angles greater than 90 Degrees

Hello Dr. Math,

I'm confused with the trigonometric functions.  Why is a reference 
angle used to calculate an angle greater than 90 degrees?  From the
definition of the trigonometric functions, they are used for acute
angles.  If an angle is greater than 90 degrees, how is a reference
angle used? 

Thanks in advance.


Date: 07/02/2004 at 13:00:51
From: Doctor Peterson
Subject: Re: Trig Functions & Angles greater than 90 Degrees

Hi, Karen.

Trigonometry can be taught in several different ways, depending on 
the order in which the teacher wants to introduce concepts.  It sounds 
like your class has started with the right triangle version, and is 
now moving into working with angles greater than 90 degrees, which 
requires extending those ideas.  This can be a little misleading, 
because it suggests that trig functions are defined only for acute 
angles, and then you have to use some tricks to find functions of 
larger angles.

What I prefer is to motivate the idea of the trig functions using 
right triangles, but quickly extend that idea so that our actual 
definition of the trig functions is based on the "unit circle" 
concept.  Then there is no trouble defining the functions for any 
angle; but you find that there are rules that allow you to find, say, 
the sine of a large angle in terms of the sine of a specific acute 
angle, called its reference angle.

Let's take the sine.  In a right triangle, this is the "opposite over 
the hypotenuse":

        c  /  |
        /     |a     sin(A) = a/c
     /A       |

If we take this same angle and put it at the origin of a coordinate 
plane, and then change c to 1 (making a similar triangle), we have

           ooo     |     ooo
         oo        |        oo
        o          |          o
       o           |           + (x,y)
      o            |        /  |o
     o             |     /     |yo
     o             |  / A      | o

Now the opposite side is just y, the y-coordinate of the point on the 
unit circle at the given angle A.  The hypotenuse is 1, so the sine of 
angle A is y/1 or y.

Now we extend the idea: for ANY angle A, define the sine of that angle 
to be the y-coordinate of the corresponding point on the unit circle.

How, then, are sines of non-acute angles related to those of acute 
angles?  For angles in the second quadrant, we find that the sine is 
the same as that of the supplementary angle, which is acute:

           ooo     |     ooo
         oo        |        oo
        o          |          o
 (x,y) +-----------+-----------+
      o|  \        |           |o
     o |     \     |           |yo
     o |        \  B           | o

Here B is the angle from the positive x-axis to our point (x,y), 
which is about 150 degrees.  But the sine is the same as that of our 
original angle (30 degrees), because both points are the same height 
above the x-axis.  The rule is

  sin(180 - A) = sin(A)

You'll also notice that the cosine of our angle (x) is negative here, 
but has the same absolute value as that of the original angle; so

  cos(180 - A) = -cos(A)

Other rules are:

  sin(-A) = -sin(A)

  cos(-A) = cos(A)

  sin(A + 360) = sin(A)

  cos(A + 360) = cos(A)

These rules allow you to express the sine or cosine of any angle as 
equal to, or the negative of, some acute angle.

Here are some pages that may help:

  Signs of Sines (and Other Trigonometric Ratios) 

  The Unit Circle 

If you have any further questions, feel free to write back.

- Doctor Peterson, The Math Forum 

Date: 07/02/2004 at 15:34:00
From: Karen
Subject: Thank you (Trig Functions & Angles greater than 90 Degrees)

That demystifies a lot.  The fact that the sine of 30 and 150 degrees
are the same, because those angles have the same y-coordinate on the
unit circle, explains it all.

Thank you very much, Dr. Math!
Associated Topics:
High School Trigonometry

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