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Terminal Side

Date: 04/21/97 at 17:07:21
From: Laura
Subject: sin/cosin

Find cos(A) if the point (2,-3) is on the terminal side of A.
I am totally stuck and I have no idea how to even begin!

Date: 04/27/97 at 01:14:41
From: Doctor Mike
Subject: Re: sin/cosin

Dear Laura,
I'm going to have to try to draw a picture for this. Sorry if the 
picture is not great.   The "terminal" side is the hypotenuse of the 
triangle.  The "initial" side is the triangle leg that goes from the 
origin along the positive x-axis direction. 

          -2      -1        |       +1       +2
   -x <---*--------*--------+--------*--------*------> +x
                            |\  A             |
                            |  \              |
                          -1|    \            |
                            |      \          |
                            |        \        | -3 
                          -2|          \      |
                            |            \    |
                            |              \  |
                          -3|                \|
                            |                  (+2,-3) 
Now, by the Pythagorean theorem, since the lengths of the two legs of 
this right triangle are 2 and 3, you find the length of the hypotenuse 
by taking the square root of 2^2 + 3^2 = 4+9 = 13. So, the hypotenuse 
length is sqrt(13). Note that I have labelled the vertical leg "-3" 
because it is going downwards from the x-axis.     
Now, with all this set-up your question is easy. The cos(A) is the 
"adjacent" over "hypotenuse," or +2/sqrt(13). If you want to get rid 
of the square root in the denominator, multiply both numerator and 
denominator by sqrt(13), like the following: 
            +2        sqrt(13)        2*sqrt(13)
        ---------- * ----------  =  -------------      
         sqrt(13)     sqrt(13)            13     
I hope this helps. Now do the same for sin(A) and tan(A)!   
-Doctor Mike,  The Math Forum
 Check out our web site!   
Associated Topics:
High School Trigonometry

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