document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=443330>Question 443330</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Prove that each is an identity\r<BR>\n" );
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document.write( "(sin^3 x + cos^3 x)/ (sin x + cos x) = 1 - (sin 2 x/ 2) </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #306253 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=306253\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Prove that each is an identity <BR>\n" );
document.write( "(sin^3 x + cos^3 x)/ (sin x + cos x) = 1 - (sin 2 x/ 2)<BR>\n" );
document.write( "..<BR>\n" );
document.write( "Start with left side<BR>\n" );
document.write( "Use sum of cubes<BR>\n" );
document.write( "[(sinx+cosx)(sin^2x-sinxcosx+cos^2x)]/(sin x + cos x)<BR>\n" );
document.write( "cancel (sin x + cos x) and use identity sin^2+cos^2=1<BR>\n" );
document.write( "=1-sinxcosx*(2/2)<BR>\n" );
document.write( "=1-(2sinxcosx)/2=1-(sin 2x/2)<BR>\n" );
document.write( "left side=right side </SPAN>\n" );
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