document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1063203>Question 1063203</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Rewrite with only sin x and cos x.\r<BR>\n" );
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document.write( "cos 2x - sin x </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #678256 by </B><A HREF=/tutors/aboutme.mpl?userid=Boreal><B>Boreal(15235)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Boreal><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=Boreal><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=678256\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> cos 2x - sin x<BR>\n" );
document.write( "cos(2x)=1-2 sin^2 x<BR>\n" );
document.write( "Therefore we have 1-2sin^2x-sinx<BR>\n" );
document.write( "That is -(2sin^2 x+sin x-1)=-(2sin x-1)(sin x+1).<BR>\n" );
document.write( "Not clear if that answers the question or whether both sin x and cos x have to be present.<BR>\n" );
document.write( "cos (2x)=2 cos^2 x -1<BR>\n" );
document.write( "Then it is 2 cos^2x -sin x-1\r<BR>\n" );
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