document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=686909>Question 686909</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Dear Sir,\r<BR>\n" );
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document.write( "Q: If sin A=x^2-2x+2, Find Cos A\r<BR>\n" );
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document.write( "I have tried this way\r<BR>\n" );
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document.write( "sin^2 A = (x^2-2x+2)^2<BR>\n" );
document.write( "1-cos^2 A= (x^2-2x+2)^2<BR>\n" );
document.write( "Cos^2 A= 1-(x^2-2x+2)^2<BR>\n" );
document.write( "Cos^2 A= 1^2-(x^2-2x+2)^2 = [1+(x^2-2x+2)][1-(x^2-2x+2)]\r<BR>\n" );
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document.write( "is it correct or am going somewhere wrong?\r<BR>\n" );
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document.write( "Kindly help me to solve this.\r<BR>\n" );
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document.write( "Regards,<BR>\n" );
document.write( "AN </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #424928 by </B><A HREF=/tutors/aboutme.mpl?userid=vleith><B>vleith(2983)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=vleith><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=424928\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Just take the sqrt of both side to get CosA. You could expand more, and possibly simplify, but I wouldn't. In fact, I would have stopped one step earlier and just taken then sqrt of 1-(x^2-2x+2)^2 </SPAN>\n" );
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