A Different View

<BODY LANG="EN" > <HR><center> <table border=1> <td><A HREF="node34.html"><img height=36 width=36 alt="PREVIOUS" src="/UCESIcons/tools/previous.gif"></A></td> <td><A HREF="node31.html"><img height=36 width=36 alt="UP" src="/UCESIcons/tools/up.gif"></A></td> <td><A HREF="node36.html"><img height=36 width=36 alt="NEXT" src="/UCESIcons/tools/next.gif"></A></td> <td><a href="contents-node35.html"><img height=36 width=36 alt="CONTENTS" src="/UCESIcons/tools/toc.gif"></a></td> </table> </center> <HR> <H3>A Different View</H3> <P> Consider the following problem. Let <IMG WIDTH=6 HEIGHT=12 ALIGN=BOTTOM ALT="theta" SRC="img1.gif"> be an angle between 90° and 180°. Assume <IMG WIDTH=61 HEIGHT=29 ALIGN=MIDDLE ALT="sin(theta) = 1/5" SRC="img83.gif">. Find the values of <IMG WIDTH=30 HEIGHT=12 ALIGN=BOTTOM ALT="cos(theta)" SRC="img84.gif"> and <IMG WIDTH=33 HEIGHT=12 ALIGN=BOTTOM ALT="tan(theta)" SRC="img85.gif">. <P> At first glance, you might say, ``But I cannot find the values of the other trig functions because I do not know what the angle <IMG WIDTH=6 HEIGHT=12 ALIGN=BOTTOM ALT="theta" SRC="img1.gif"> equals.'' You do not need to know the actual value of <IMG WIDTH=6 HEIGHT=12 ALIGN=BOTTOM ALT="theta" SRC="img1.gif"> in order to find <IMG WIDTH=30 HEIGHT=12 ALIGN=BOTTOM ALT="cos(theta)" SRC="img84.gif"> and <IMG WIDTH=33 HEIGHT=12 ALIGN=BOTTOM ALT="tan(theta)" SRC="img85.gif">. All you need to do is to draw the diagram corresponding to the information given in the statement of the problem. <P> First, because you know that <IMG WIDTH=6 HEIGHT=12 ALIGN=BOTTOM ALT="theta" SRC="img1.gif"> is an angle between 90° and 180°, you know that the terminal side of <IMG WIDTH=6 HEIGHT=12 ALIGN=BOTTOM ALT="theta" SRC="img1.gif"> lies in Quadrant II. Second, since <IMG WIDTH=61 HEIGHT=29 ALIGN=MIDDLE ALT="sin(theta) = 1/5" SRC="img83.gif">, you see that you can write in the values of both the hypotenuse of the right triangle in question as well as the opposite leg since <IMG WIDTH=29 HEIGHT=12 ALIGN=BOTTOM ALT="sin(theta)" SRC="img86.gif"> is defined as <P> <center> <IMG WIDTH=92 HEIGHT=33 ALIGN=BOTTOM ALT="sin(theta)=opp/hyp" SRC="img87.gif"> </CENTER> <P> Thus, your drawing should look something like this: <P> <CENTER> <IMG SRC="p8-8.gif" ALT="[diagram]" WIDTH=142 HEIGHT=237> </CENTER> <P> Now, all you need is the length of the adjacent leg. Remember that, since your point has a negative value for <I>x</I>, we will label this with a negative sign. How do you find the length of this third leg? Answer: The Pythagorean Theorem! If we call the length of the adjacent leg <I>x</I>, then we have <nobr><I>x</I><sup>2</SUP> + 1<SUP>2</SUP> = 5<SUP>2</SUP></NOBR> from the Pythagorean Theorem. Hence, we have <nobr><I>x</I><sup>2</SUP> + 1 = 25</NOBR> or <nobr><I>x</I><sup>2</SUP> = 24</NOBR>. Therefore, we know <IMG WIDTH=75 HEIGHT=30 ALIGN=MIDDLE ALT="x = +/-sqrt(24)" SRC="img91.gif">. In this case, we choose <IMG WIDTH=75 HEIGHT=30 ALIGN=MIDDLE ALT="x = -sqrt(24)" SRC="img92.gif"> or <IMG WIDTH=75 HEIGHT=30 ALIGN=MIDDLE ALT="x = -2 sqrt(6)" SRC="img93.gif"> since, as we noted above, we need to label this leg with a negative value. Therefore, our diagram now can be completed as follows:<P> <CENTER> <IMG SRC="p8-9.gif" ALT="[diagram]" WIDTH=142 HEIGHT=237> </CENTER><P> From this we see that <P> <CENTER> <IMG WIDTH=169 HEIGHT=81 ALIGN=BOTTOM ALT="[trig values]" SRC="img94.gif"> </CENTER> <P> <HR><center> <table border=1> <td><A HREF="node34.html"><img height=36 width=36 alt="PREVIOUS" src="/UCESIcons/tools/previous.gif"></A></td> <td><A HREF="node31.html"><img height=36 width=36 alt="UP" src="/UCESIcons/tools/up.gif"></A></td> <td><A HREF="node36.html"><img height=36 width=36 alt="NEXT" src="/UCESIcons/tools/next.gif"></A></td> <td><a href="contents-node35.html"><img height=36 width=36 alt="CONTENTS" src="/UCESIcons/tools/toc.gif"></a></td> </table> </center> <HR><P><ADDRESS> <I><A HREF="http://www.math.psu.edu/sellersj/">James A. Sellers</A> </I> </ADDRESS> </BODY>