document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=416683>Question 416683</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Use a sum or difference formula to find the exact value of cot 11pie/12 </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #292613 by </B><A HREF=/tutors/aboutme.mpl?userid=lwsshak3><B>lwsshak3(11628)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=lwsshak3><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=292613\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Use a sum or difference formula to find the exact value of cot 11pie/12\r<BR>\n" );
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document.write( "I don't see the cot formula in my book, so I will use the addition formula for tan then take its reciprocal for the cot.<BR>\n" );
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document.write( "tan (s+t)=tan(s)+tan(t)/1-tan(s)tan(t)<BR>\n" );
document.write( "given 11pi/12 can be divided into 2pi/12 and 9pi/12 or pi/6 and 3pi/4<BR>\n" );
document.write( "pi/6 is a 30 degree reference angle(s) in the first quadrant.<BR>\n" );
document.write( "3pi/4 is a 45 degree reference angle(t) in the second quadrant.<BR>\n" );
document.write( "In the tan addition formula, let s=pi/6 and t=3pi/4<BR>\n" );
document.write( "tan(11pi/12)=tan(s+t)=tan(pi/6+3pi/4)=sqrt((3)/3)-1/1-sqrt((3)/3)*(-1)<BR>\n" );
document.write( "tan(11pi/12)=sqrt((3)/3)-1/sqrt((3)/3)+1)=.577-1/.577+1=-.423/1.577=-.268<BR>\n" );
document.write( "ans:Therefore, cot(11pi/12) is 1/-.268=-3.731 </SPAN>\n" );
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