Terminal SideDate: 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. +y ^ +1| | -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! http://mathforum.org/dr.math/ |
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