document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=811471>Question 811471</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Ive been trying to work this out for a while now and can't get it, please help!<BR>\n" );
document.write( "Prove:<BR>\n" );
document.write( "cos2x+2sin^2x=1\r<BR>\n" );
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document.write( "Thanks!\r<BR>\n" );
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document.write( " </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #488752 by </B><A HREF=/tutors/aboutme.mpl?userid=htmentor><B>htmentor(1343)</B></A><img src=\"/images/stars/stars-12.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=htmentor><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=htmentor><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=488752\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Using the trig identity cos(2x) = 2cos^2(x) - 1 we can write the LHS of the equation as<BR>\n" );
document.write( "2cos^2(x) - 1 + 2sin^2(x) -> 2(cos^2(x) + sin^2(x)) - 1<BR>\n" );
document.write( "Since sin^2(x) + cos^2(x) = 1, the LHS becomes 2 - 1 = 1 </SPAN>\n" );
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