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Simplifying Expressions with Trig Functions


Date: 04/04/2001 at 10:56:21
From: Betsey
Subject: Cosine and Cotangent

     4 csc(3pi/4) - cot(-pi/4)

I am home schooling my niece and it has been over 25 years since I 
worked with this kind of stuff. Can you work this out for me so I can 
get a grasp and then be able to explain this to her?


Date: 04/04/2001 at 12:45:15
From: Doctor Peterson
Subject: Re: Cosine and Cotangent

Hi, Betsey.

I find it helpful to draw a rough sketch of the angle and a triangle 
to define the trig functions, and then use my knowledge of the angle 
to determine the values of the relevant sides. Let's do the first part 
of this, the cosecant of 3pi/4. First, what does this angle look like? 
It will be halfway between pi/2 and pi, so placed in standard position 
in a unit circle, it is something like this:

                      ***********
                 *****     |     *****
        (x,y)****          |          ****
           **|             |              **
          *  | \           |                *
        **   |   \   1     |                 **
       *    y|     \       |                   *
      *      |       \     |                    *
      *      |         \   |                    *
     *       |         T \ |                     *
     *-------+-------------+---------------------*
     *             -x      |                     *
      *                    |                    *
      *                    |                    *
       *                   |                   *
        **                 |                 **
          *                |                *
           **              |              **
             ****          |          ****
                 *****     |     *****
                      ***********

The acute angle T is 45 degrees (pi/4 radians); that makes the 
triangle a right isosceles triangle, and I know that the positive 
lengths -x and y are equal. Using the Pythagorean theorem, they're 
equal to sqrt(2)/2.

Now, the cosecant is the reciprocal of the sine; the sine is y/1 
(opposite/hypotenuse), so the cosecant is 1/y. This makes the cosecant 
equal to the reciprocal of sqrt(2)/2, which is sqrt(2). With angles 
like this outside the first quadrant, you have to watch the signs; in 
this case the sine, y, is positive, but the cosine, x, would be 
negative since the point (x,y) is on the left side of the circle.

You can do the same thing to find the cotangent of -pi/4:

                      ***********
                 *****     |     *****
             ****          |          ****
           **              |              **
          *                |                *
        **                 |                 **
       *                   |                   *
      *                    |                    *
      *                    |                    *
     *                     |      x              *
     *---------------------+-------------+-------*
     *                     | \T          |       *
      *                    |   \   1     |      *
      *                    |     \       |      *
       *                   |       \     |-y   *
        **                 |         \   |   **
          *                |           \ |  *
           **              |             |**
             ****          |          ****(x,y)
                 *****     |     *****
                      ***********

Here the cotangent is the reciprocal of the tangent, which is y/x 
(opposite/adjacent). Again, we have a right isoscecles triangle, so 
the absolute values of x and y are equal. But since x is positive and 
y is negative, the tangent is 1-1 = -1, and the cotangent is x/y = -1 
as well.

If you have trouble with any of the steps I took, please write back 
and let me know, so I can give more details where they are needed. 
Also, don't miss our archived answers on high school trigonometry:

   http://mathforum.org/dr.math/tocs/trig.high.html   

Here are a few good introductory pages on finding functions of 
familiar angles:

   Deriving Sines of 30, 45, 60 and 90 Degrees
   http://mathforum.org/dr.math/problems/michael.6.02.99.html   

   Sine, Cosine, and Tangent on a Circle
   http://mathforum.org/dr.math/problems/jongejan.04.14.99.html   

Our FAQ on trig formulas might be overwhelming at first, but there's a 
lot of good stuff there:

   http://mathforum.org/dr.math/faq/formulas/faq.trig.html   

- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Trigonometry

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