document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=855933>Question 855933</A>: <I><SPAN STYLE=\"word-break: break-all;\"> How would I find cos(x-y) if cos x= 3/5 and cos y= 4/5 when the exact value needs to be between 0 and 90 degrees. I have tried to subtract 3/5 and 4/5 to get cos -1/5 but I don't know what to do </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #515637 by </B><A HREF=/tutors/aboutme.mpl?userid=stanbon><B>stanbon(75887)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=stanbon><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=515637\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> How would I find cos(x-y) if cos x= 3/5 and cos y= 4/5 when the exact value needs to be between 0 and 90 degrees.<BR>\n" );
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document.write( "If cos(x) = 3/5, sin(x) = sqrt(5^2-3^2)/5 = 4/5<BR>\n" );
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document.write( "If cos(y) = 4/5, sin(y)= sqrt(5^2-4^2)/5 = 3/5<BR>\n" );
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document.write( "cos(x-y) = cos(x)cos(y)+ sin(x)sin(y) = (3/5)(4/5)+(4/5)(3/5)<BR>\n" );
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document.write( "= (12+12)/25 = 24/25<BR>\n" );
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document.write( "Cheers,<BR>\n" );
document.write( "Stan H. </SPAN>\n" );
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