document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=17612>Question 17612</A>: <I><SPAN STYLE=\"word-break: break-all;\"> cosx+cosy-cos(x+y)=3/2<BR>\n" );
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document.write( "this is only possible if x=y=60<BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #8505 by </B><A HREF=/tutors/aboutme.mpl?userid=venugopalramana><B>venugopalramana(3286)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=venugopalramana><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.geocities.com/venugopalramana/FREE_MATHS_ASSISTANCE.html valign=middle><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=8505\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> cosx+cosy-cos(x+y)=3/2..<BR>\n" );
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document.write( "[actually from here you can straight away put the step<BR>\n" );
document.write( " 2cos((x+y)/2)cos((x-y)/2)-2[cos((x+y)/2))]^2-0.5=0...using cos 2a=2(cos a)^2-1,<BR>\n" );
document.write( "and proceed further as shown below.but i am showing another elaboration ,in case ,this problem appears as one to prove in any triangle xyz,where sum of angles x+y+z=180...]<BR>\n" );
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document.write( "cosx+cosy+cos(180-x-y)=3/2...<BR>\n" );
document.write( "2cos((x+y)/2)cos((x-y)/2)+1-2[sin((180-x-y)/2)]^2-3/2=0<BR>\n" );
document.write( "2cos((x+y)/2)cos((x-y)/2)-2[sin(90-((x+y)/2))]^2-0.5=0...<BR>\n" );
document.write( "2cos((x+y)/2)cos((x-y)/2)-2[cos((x+y)/2))]^2-0.5=0...<BR>\n" );
document.write( " for ease of simplification put cos((x+y)/2))=p and cos((x-y)/2))=b..we get<BR>\n" );
document.write( "-2p^2+2pb-0.5=0..or dividing with -2 throughput p^2-pb+0.25=0<BR>\n" );
document.write( "..taking this as a quadratic in p we find that the discriminant is b^2-1 which is to be positive to get real solution for p.but the highest value of........... b= cos((x-y)/2)) is 1 so the only possibility for getting a real solution is for b=1 or b^2-1=0..in such a case we get p=b/2.hence we have<BR>\n" );
document.write( "b= cos((x-y)/2)=1 or (x-y)/2=0..or..x=y and<BR>\n" );
document.write( "p=b/2=1/2=cos((x+y)/2)..putting x=y as obtained above,we get<BR>\n" );
document.write( "coa((x+x)/2)=cos(x)=1/2...or x=60..so..x=y=60<BR>\n" );
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