document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=759256>Question 759256</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Write cos^4(3x) in terms of cosines all to the 1st power. FIND LCD & SIMPLIFY!\r<BR>\n" );
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document.write( "I GOT cos^4(3x) as (cos^2(3x))^2. By solving cos(2x) = 2cos^2(x) - 1 for cos^2(x), we get cos^2(x) = 1/2+1/2*cos(2x). Thus, cos^2(3x) = 1/2+1/2*cos(6x), and (cos^2(3x))^2 = 1/4 + 1/2*cos(6x) + 1/4*cos^2(6x). Next, cos^2(6x) = 1/2+1/2*cos(12x), so 1/4 + 1/2*cos(6x) + 1/4*cos^2(6x) = 1/4 + 1/2*cos(6x) + 1/8 + 1/8*cos(12x) = 3/8 + 1/2*cos(6x) + 1/8*cos(12x).\r<BR>\n" );
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document.write( "BUT IM NOT SURE PLEASE HELP!!! </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #461953 by </B><A HREF=/tutors/aboutme.mpl?userid=Alan3354><B>Alan3354(69443)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Alan3354><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=Alan3354><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=461953\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Write cos^4(3x) in terms of cosines all to the 1st power.<BR>\n" );
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document.write( "= <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%283+%2B+4cos%286x%29+%2B+cos%2812x%29%29%2F8 ALIGN=MIDDLE  ALT=\"%283+%2B+4cos%286x%29+%2B+cos%2812x%29%29%2F8\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "from Wikipedia half-angle formula<BR>\n" );
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