document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1184287>Question 1184287</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The line joining A(bCosx, bSinx) and B(aCosy, aSiny), where a and b have different values is produced to the point M(x, y) so that AM:MB = b:a then find the value of uCos((x+y/2)) + vSin((x+y)/2)) and uCos(x+y) + vSin(x+y)? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #849860 by </B><A HREF=/tutors/aboutme.mpl?userid=CPhill><B>CPhill(1959)</B></A><img src=\"/images/stars/stars-12.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=CPhill><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=CPhill><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=849860\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let A = (b cos x, b sin x) and B = (a cos y, a sin y).<BR>\n" );
document.write( "Let M = (u, v) be the point such that AM:MB = b:a.<BR>\n" );
document.write( "Using the section formula for the coordinates of M, we have:<BR>\n" );
document.write( "$$u = \frac{b(a \cos y) + a(b \cos x)}{b+a} = \frac{ab(\cos y + \cos x)}{a+b}$$<BR>\n" );
document.write( "$$v = \frac{b(a \sin y) + a(b \sin x)}{b+a} = \frac{ab(\sin y + \sin x)}{a+b}$$<BR>\n" );
document.write( "We are asked to find the values of:<BR>\n" );
document.write( "1. $u \cos\left(\frac{x+y}{2}\right) + v \sin\left(\frac{x+y}{2}\right)$<BR>\n" );
document.write( "2. $u \cos(x+y) + v \sin(x+y)$\r<BR>\n" );
document.write( "\n" );
document.write( "1. $u \cos\left(\frac{x+y}{2}\right) + v \sin\left(\frac{x+y}{2}\right) = \frac{ab}{a+b} \left[ (\cos y + \cos x) \cos\left(\frac{x+y}{2}\right) + (\sin y + \sin x) \sin\left(\frac{x+y}{2}\right) \right]$<BR>\n" );
document.write( "Using the sum-to-product formulas:<BR>\n" );
document.write( "$\cos x + \cos y = 2 \cos\left(\frac{x+y}{2}\right) \cos\left(\frac{x-y}{2}\right)$<BR>\n" );
document.write( "$\sin x + \sin y = 2 \sin\left(\frac{x+y}{2}\right) \cos\left(\frac{x-y}{2}\right)$<BR>\n" );
document.write( "Substituting these into the expression:<BR>\n" );
document.write( "$= \frac{ab}{a+b} \left[ 2 \cos\left(\frac{x+y}{2}\right) \cos\left(\frac{x-y}{2}\right) \cos\left(\frac{x+y}{2}\right) + 2 \sin\left(\frac{x+y}{2}\right) \cos\left(\frac{x-y}{2}\right) \sin\left(\frac{x+y}{2}\right) \right]$<BR>\n" );
document.write( "$= \frac{2ab}{a+b} \cos\left(\frac{x-y}{2}\right) \left[ \cos^2\left(\frac{x+y}{2}\right) + \sin^2\left(\frac{x+y}{2}\right) \right]$<BR>\n" );
document.write( "$= \frac{2ab}{a+b} \cos\left(\frac{x-y}{2}\right)$\r<BR>\n" );
document.write( "\n" );
document.write( "2. $u \cos(x+y) + v \sin(x+y) = \frac{ab}{a+b} \left[ (\cos y + \cos x) \cos(x+y) + (\sin y + \sin x) \sin(x+y) \right]$<BR>\n" );
document.write( "$= \frac{ab}{a+b} \left[ 2 \cos\left(\frac{x+y}{2}\right) \cos\left(\frac{x-y}{2}\right) \cos(x+y) + 2 \sin\left(\frac{x+y}{2}\right) \cos\left(\frac{x-y}{2}\right) \sin(x+y) \right]$<BR>\n" );
document.write( "$= \frac{2ab}{a+b} \cos\left(\frac{x-y}{2}\right) \left[ \cos\left(\frac{x+y}{2}\right) \cos(x+y) + \sin\left(\frac{x+y}{2}\right) \sin(x+y) \right]$<BR>\n" );
document.write( "$= \frac{2ab}{a+b} \cos\left(\frac{x-y}{2}\right) \cos\left(x+y - \frac{x+y}{2}\right) = \frac{2ab}{a+b} \cos\left(\frac{x-y}{2}\right) \cos\left(\frac{x+y}{2}\right)$\r<BR>\n" );
document.write( "\n" );
document.write( "Final Answer: The final answer is $\boxed{2ab/(a+b)cos((x-y)/2)}$<BR>\n" );
document.write( " </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );