document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=35486>Question 35486</A>: <I><SPAN STYLE=\"word-break: break-all;\"> find the exact values for the trig functions, with the radian measure: pie/4 </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #21624 by </B><A HREF=/tutors/aboutme.mpl?userid=venugopalramana><B>venugopalramana(3286)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=venugopalramana><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.geocities.com/venugopalramana/FREE_MATHS_ASSISTANCE.html valign=middle><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=21624\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> ANGLE = PI/4..THAT IS IT IS AN ISOCELLES RIGHT ANGLED TRIANGLE.HENCE LET BOTH  SIDES BE 1 EACH. THEN HYPOTENUSE =SQRT(1^2+1^2)=SQRT(2)<BR>\n" );
document.write( "SO <BR>\n" );
document.write( "SIN (PI/4)=1/SQRT(2)=SQRT(2)/2=1.4142/2=0.7071<BR>\n" );
document.write( "COSEC(PI/4)=SQRT(2)=1.4142<BR>\n" );
document.write( "COS(PI/4)= 1/SQRT(2)=SQRT(2)/2=1.4142/2=0.7071<BR>\n" );
document.write( "SEC(PI/4)=SQRT(2)=1.4142<BR>\n" );
document.write( "TAN(PI/4)=1/1=1<BR>\n" );
document.write( "COT(PI/4)=1/1=1 </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );