document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=876835>Question 876835</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Sin+cos=1,find sin-cos ? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #529031 by </B><A HREF=/tutors/aboutme.mpl?userid=richard1234><B>richard1234(7193)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=richard1234><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/cgi-bin/show-face.mpl?user=richard1234><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=529031\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Sine and cosine need an argument; the expressions sin = 1/2 or sin+cos = 1 make no sense.\r<BR>\n" );
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document.write( "Assuming you mean sin x + cos x = 1, sin x - cos x = ?, we have (sin x + cos x)^2 = 1 + 2 sin x cos x = 1 (using Pythagorean identity) so sin x cos x = 0.\r<BR>\n" );
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document.write( "Also (sin x - cos x)^2 = 1 - 2 sin x cos x = 1 - 2*0 = 1. So sin x - cos x could equal 1 or -1. </SPAN>\n" );
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