document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=889804>Question 889804</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Use the power-reducing formulas as many times as possible to rewrite the expression in terms of the first power of the cosine.<BR>\n" );
document.write( "sin^2(4x)cos^2(4x)\r<BR>\n" );
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document.write( "I am confused about how to reduce this problem.  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #538446 by </B><A HREF=/tutors/aboutme.mpl?userid=charu91><B>charu91(7)</B></A><img src=\"/images/stars/stars-05.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=charu91><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=538446\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> cos2A=2cos^2(A)-1<BR>\n" );
document.write( "cos^(A)=(cos2A+1)/2<BR>\n" );
document.write( "so cos^2(4x) can be written as cos^2(4x)=(1+cos8x)/2<BR>\n" );
document.write( "cos2A=1-2sin^2(A)<BR>\n" );
document.write( "sin^2(A)=(1-cos2A)/2<BR>\n" );
document.write( "sin^2(4x)=(1-cos8x)/2<BR>\n" );
document.write( "substituting these in eqn sin^2(4x)cos^2(4x) we get ((1-cos8x)/2)((1+cos8x)/2)=(1-cos^2(8x))/4<BR>\n" );
document.write( "                                               =(1-(cos16x+1)/2)/4<BR>\n" );
document.write( "                                               =1/4-(1/8*cos16x)-1/8<BR>\n" );
document.write( "                                               =1/8-(1/8*cos16x)<BR>\n" );
document.write( "                                               =1/8*(1-cos16x)<BR>\n" );
document.write( "answer to the question<BR>\n" );
document.write( " sin^2(4x)cos^2(4x) = 1/8*(1-cos16x)<BR>\n" );
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