document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=921332>Question 921332</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Suppose sec t = 2 and csc t > 0. Find each of the following: <BR>\n" );
document.write( "cos t = <BR>\n" );
document.write( "cos (&#8722;t) = <BR>\n" );
document.write( "cos (&#960; + t) = <BR>\n" );
document.write( "sin (&#8722;t) = <BR>\n" );
document.write( "tan (t + &#960;) = <BR>\n" );
document.write( "I think I have this right since sec = 2 it's therefore 2/1 and that's hyp/opp so I use the Pythagorean theorem to find A <BR>\n" );
document.write( "A^2 = 2^2 + 1^2 \r<BR>\n" );
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document.write( "A = &#8730;(5) \r<BR>\n" );
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document.write( "So I have the opposite, adjacent and hypotenuse. I just don't understand what it's asking with -t? &#960;+t? and t+&#960; \r<BR>\n" );
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document.write( "please someone explain this <BR>\n" );
document.write( "Thanks! </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #558953 by </B><A HREF=/tutors/aboutme.mpl?userid=jim_thompson5910><B>jim_thompson5910(35256)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=jim_thompson5910><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=558953\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> cos t = 1/sec(t) \r<BR>\n" );
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document.write( "cos(t) = <font color=\"red\">1/2</font>\r<BR>\n" );
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document.write( "cos (&#8722;t) = cos(t) since cosine is an even function\r<BR>\n" );
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document.write( "cos(-t) = <font color=\"red\">1/2</font>\r<BR>\n" );
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document.write( "cos (&#960; + t) = cos(pi)*cos(t) - sin(pi)*sin(t) ... trig identity\r<BR>\n" );
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document.write( "cos (&#960; + t) = -1*cos(t) - 0*sin(t)\r<BR>\n" );
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document.write( "cos (&#960; + t) = -cos(t)\r<BR>\n" );
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document.write( "cos (&#960; + t) = -(1/2)\r<BR>\n" );
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document.write( "cos (&#960; + t) = <font color=\"red\">-1/2</font>\r<BR>\n" );
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document.write( "sin (&#8722;t) = -sin(t) since sine is an odd function\r<BR>\n" );
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document.write( "sin(-t) = <font color=\"red\">-sqrt(3)/2</font>\r<BR>\n" );
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document.write( "Note: your triangle should have 1/2 as the leg and 1 as the hypotenuse. So a = 1/2, c = 1. The missing side is b for now\r<BR>\n" );
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document.write( "a^2 + b^2 = c^2<BR>\n" );
document.write( "(1/2)^2 + b^2 = 1^2<BR>\n" );
document.write( "1/4 + b^2 = 1<BR>\n" );
document.write( "b^2 = 1 - 1/4<BR>\n" );
document.write( "b^2 = 3/4<BR>\n" );
document.write( "b = sqrt(3/4) ... csc(t) > 0, sin(t) > 0<BR>\n" );
document.write( "b = sqrt(3)/2\r<BR>\n" );
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document.write( "So that's how I got sqrt(3)/2. You can also use the unit circle.\r<BR>\n" );
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document.write( "tan (t + &#960;) = [tan(t) + tan(pi)]/[1 - tan(t)*tan(pi)] ... trig identity\r<BR>\n" );
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document.write( "tan (t + &#960;) = [tan(t) + 0]/[1 - tan(t)*0]\r<BR>\n" );
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document.write( "tan (t + &#960;) = [tan(t) + 0]/[1 - 0]\r<BR>\n" );
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document.write( "tan (t + &#960;) = [tan(t)]/[1]\r<BR>\n" );
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document.write( "tan(t + &#960;) = <font color=\"red\">tan(t)</font> \r<BR>\n" );
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document.write( "This makes sense because tangent has period of pi. It's going to repeat itself every pi units which is exactly what tan(t + &#960;) = <font color=\"red\">tan(t)</font> tells us.\r<BR>\n" );
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document.write( "Let me know if that helps or not. Thanks.\r<BR>\n" );
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document.write( "If you need more help, feel free to email me at <a  rel=nofollow HREF=\"mailto:jim_thompson5910@hotmail.com?Subject=I%20Need%20Algebra%20Help\">jim_thompson5910@hotmail.com</a>\r<BR>\n" );
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