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Trigonometry: Positive vs. Negative


Date: 5/31/96 at 21:11:4
From: Anonymous
Subject: Trigonometry

This time my question is about positive vs. negative.

The problem is this:

tan x= 5/12 and sec x= -13/12

I found the sine to =   5/13
             cosine =  12/13
              cotan =  12/5
              cosec =  13/5

I was wondering if I did the problem right, and how to determine the 
sign of the answers.

Thanks!


Date: 5/31/96 at 21:56:58
From: Doctor Pete
Subject: Re: Trigonometry

Here's how I like to think about trigonometric functions and their 
signs:

   [A]ll [S]tudents [T]ake [C]alculus

           ^
           |
       S   |   A
           |
   ________|________
           |
           |
       T   |   C
           |

So in the first quadrant, *all* the trigonometric functions 
(sin, cos, tan, etc.) are positive.  In the second quadrant, 
only sin (and its reciprocal, 1/sin = csc), is positive.  
In the third quadrant, only tan (and 1/tan = cot) is positive, 
and in the fourth quadrant, only cos (and 1/cos = sec) is
positive.

So in your example, tan(x) = 5/12.  Think about a right triangle 
with these dimensions:

              /|
             /x|
            /  |
         r /   |
          /    | 12
         /     |
        /      |
       /      _|
      /______|_|
           5

(This is because the tangent is opposite/adjacent.)  So the 
Pythagorean Theorem says r = sqrt(5^2+12^2) = 13.  But sec(x) is 
positive in quadrant 1 and negative in quadrant 3, so x is in 
quadrant 3.  Therefore

        sin(x) = -5/13     csc(x) = -13/5
        cos(x) = -12/13    sec(x) = -13/12
        tan(x) = 5/12      cot(x) = 12/5 .


Now, try this similar problem:

 csc(x) = -25/24
 cos(x) = 7/25

Find sin(x), sec(x), tan(x), cot(x).

Bonus question:  

What is sin(x/2), cos(x/2), tan(x/2), csc(x/2), sec(x/2), cot(x/2)?

-Doctor Pete,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   


Date: 5/31/96 at 22:19:8
From: Anonymous
Subject: Trigonometry

I didn't even think about using the quadrants. 
Okay, the problem you gave me was: csc x = -25/24
                                   cos x =   7/25

I did the formula, (24^2 + 7^2 = 25^2)

I found the triangle to be in the second quadrant, so the sine and 
cosecant are negative.

sin = -24/25
sec = 25/7
tan= 24/7
cot= 7/24

Thanks for the help!  Could you please tell me if I did the problem 
wrong? Thanks!


Date: 6/1/96 at 19:22:27
From: Doctor Pete
Subject: Re: reponse

You're very close; you got the sine and cosines, but the tangent isn't 
quite right; perhaps you're thinking of the quadrants in a different 
way than I am.  

Quadrant 1 is in the upper right corner, as you have it, but I go
counterclockwise, so the upper left corner is quadrant 2, and so on.  
The association tells you which trigonometric functions are *positive, 
and all the others are *negative*.  So the way I see it, angle x is in 
quadrant 4, and so [C], the cosine and its reciprocal, are the *only* 
trig. functions that are positive.  That means the tangent and 
cotangent are negative.

You will find that most people use a "counterclockwise = positive" 
rule in trigonometry, so angles are measured in a counterclockwise 
direction from the positive x-axis.  Thus it makes sense to count the 
quadrants that way, too, and therefore the association follows that 
direction (see my previous message to you), and tells you exactly 
which trig. functions are positive.

Good luck!

-Doctor Pete,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Trigonometry

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