document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=387970>Question 387970</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Given sin(theta) = -3/7 and tan(theta) > 0, what are all 6 exact trig values and what is the corresponding quadrant and point on the circle?   </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #274287 by </B><A HREF=/tutors/aboutme.mpl?userid=jsmallt9><B>jsmallt9(3758)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=jsmallt9><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=274287\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> sin is negative in the 3rd and 4th quadrants. tan is positive in the 1st and 3rd quadrants. So for theta to have a negative sin and a positive tan, it must terminate in the 3rd quadrant. And in the third quadrant sin, cos, sec and csc are all negative while tan and cot are positive.<br><BR>\n" );
document.write( "We can use the Pythagorean Theorem or cos^2(theta) = 1 - sin^2(theta) to find cos(theta):<BR>\n" );
document.write( "cos^2(theta) = 1 - (-3/7)^2<BR>\n" );
document.write( "cos^2(theta) = 1 - 9/49<BR>\n" );
document.write( "cos^2(theta) = 40/49<BR>\n" );
document.write( "Since cos is negative in the 3rd quadrant we will use only the negative square root:<BR>\n" );
document.write( "cos(theta) = <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=-sqrt%2840%2F49%29+=+-sqrt%2840%29%2Fsqrt%2849%29+=+-%28sqrt%284%29%2Asqrt%2810%29%29%2Fsqrt%2849%29+=+-2sqrt%2810%29%2F7 ALIGN=MIDDLE   DISABLED_style= \"max-width:90%; height:auto;\" ><br><BR>\n" );
document.write( "csc(theta) = 1/sin(theta) = <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=1%2F%28-3%2F7%29+=+-7%2F3 ALIGN=MIDDLE  ALT=\"1%2F%28-3%2F7%29+=+-7%2F3\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "sec(theta) = 1/cos(theta) = <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=1%2F%28-2sqrt%2810%29%2F7%29+=+-7%2F2sqrt%2810%29+=+%28-7%2F2sqrt%2810%29%29%28sqrt%2810%29%2Fsqrt%2810%29%29+=+-7sqrt%2810%29%2F20 ALIGN=MIDDLE   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "tan(theta) - sin(theta)/cos(theta) = <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28-3%2F7%29%2F%28-2sqrt%2810%29%2F7%29+=+3%2F2sqrt%2810%29+=+%283%2F2sqrt%2810%29%29%28sqrt%2810%29%2Fsqrt%2810%29%29+=+3sqrt%2810%29%2F20 ALIGN=MIDDLE   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "cot(theta) = cos(theta)/sin(theta) = <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28-2sqrt%2810%29%2F7%29%2F%28-3%2F7%29+=+2sqrt%2810%29%2F3 ALIGN=MIDDLE  ALT=\"%28-2sqrt%2810%29%2F7%29%2F%28-3%2F7%29+=+2sqrt%2810%29%2F3\"   DISABLED_style= \"max-width:90%; height:auto;\" ><br><BR>\n" );
document.write( "The \"point on the circle\" is (cos(theta), sin(theta) = (<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=-2sqrt%2810%29%2F7 ALIGN=MIDDLE  ALT=\"-2sqrt%2810%29%2F7\"   DISABLED_style= \"max-width:90%; height:auto;\" >, <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=-3%2F7 ALIGN=MIDDLE  ALT=\"-3%2F7\"   DISABLED_style= \"max-width:90%; height:auto;\" >) </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );