document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=740819>Question 740819</A>: <I><SPAN STYLE=\"word-break: break-all;\"> For the following, assume that all the given angles are in simplest form, so that if A is in QIV you may assume that 270° < A < 360°.\r<BR>\n" );
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document.write( "If sin B = &#8722; 1/5 with B in QIII, find the following.cot B/2 <BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #451703 by </B><A HREF=/tutors/aboutme.mpl?userid=stanbon><B>stanbon(75887)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=stanbon><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=451703\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> If sin B = &#8722; 1/5 with B in QIII, find the following.<BR>\n" );
document.write( "cot(B/2) = 1/tan(B/2) = (1 + sqrt(1+tan^2(B))/tan(B)<BR>\n" );
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document.write( "Since sin(B) = -1/5, y = -1 and r = 5<BR>\n" );
document.write( "Then x = sqrt(r^2-y^2) = sqrt(25-1) = sqrt(24)<BR>\n" );
document.write( "Note: x is negative in QIII<BR>\n" );
document.write( "So x = -sqrt(24)<BR>\n" );
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document.write( "Therefore tan(B) = y/x = -1/-sqrt(24) = 1/sqrt(24)<BR>\n" );
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document.write( "Finally, cot(B/2) = (1 + sqrt(1+(1/24))/[1/sqrt(24)]<BR>\n" );
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document.write( "= (1+sqrt(25/24))*(sqrt(24))<BR>\n" );
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document.write( "= (1+5/(sqrt(24))(sqrt(24)<BR>\n" );
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document.write( "= sqrt(24)+5<BR>\n" );
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document.write( "Cheers,<BR>\n" );
document.write( "Stan H.\r<BR>\n" );
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