document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=868352>Question 868352</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Assume that sin u =2/3 and that sin v=1/3 and that u and v are between 0 and pi/2.  Evaluate sin(u+v) </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #523501 by </B><A HREF=/tutors/aboutme.mpl?userid=jim_thompson5910><B>jim_thompson5910(35256)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=jim_thompson5910><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=523501\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Since u and v are between 0 and pi/2, this means that they're both in quadrant I\r<BR>\n" );
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document.write( "If sin(u) = 2/3, then....\r<BR>\n" );
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document.write( "sin^2(u) + cos^2(u) = 1<BR>\n" );
document.write( "(2/3)^2 + cos^2(u) = 1<BR>\n" );
document.write( "4/9 + cos^2(u) = 1<BR>\n" );
document.write( "cos^2(u) = 1 - 4/9<BR>\n" );
document.write( "cos^2(u) = 9/9 - 4/9<BR>\n" );
document.write( "cos^2(u) = 5/9<BR>\n" );
document.write( "cos(u) = sqrt(5/9)  ... ignore the negative square root since cos(u) > 0 in quadrant I<BR>\n" );
document.write( "cos(u) = sqrt(5)/sqrt(9)<BR>\n" );
document.write( "cos(u) = sqrt(5)/3 \r<BR>\n" );
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document.write( "------------------------------<BR>\n" );
document.write( "If sin(v) = 1/3, then....\r<BR>\n" );
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document.write( "sin^2(v) + cos^2(v) = 1<BR>\n" );
document.write( "(1/3)^2 + cos^2(v) = 1<BR>\n" );
document.write( "1/9 + cos^2(v) = 1<BR>\n" );
document.write( "cos^2(v) = 1 - 1/9<BR>\n" );
document.write( "cos^2(v) = 9/9 - 1/9<BR>\n" );
document.write( "cos^2(v) = 8/9<BR>\n" );
document.write( "cos(v) = sqrt(8/9)  ... ignore the negative square root since cos(v) > 0 in quadrant I<BR>\n" );
document.write( "cos(v) = sqrt(8)/sqrt(9)<BR>\n" );
document.write( "cos(v) = sqrt(8)/3 <BR>\n" );
document.write( "cos(v) = sqrt(4*2)/3 <BR>\n" );
document.write( "cos(v) = sqrt(4)*sqrt(2)/3 <BR>\n" );
document.write( "cos(v) = 2*sqrt(2)/3 <BR>\n" );
document.write( "-----------------------------\r<BR>\n" );
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document.write( "From those two parts above, we have found cos(u) = sqrt(5)/3 and cos(v) = 2*sqrt(2)/3 \r<BR>\n" );
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document.write( "So all together, we know this:\r<BR>\n" );
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document.write( "sin(u) = 2/3<BR>\n" );
document.write( "sin(v) = 1/3<BR>\n" );
document.write( "cos(u) = sqrt(5)/3<BR>\n" );
document.write( "cos(v) = 2*sqrt(2)/3 \r<BR>\n" );
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document.write( "We use them to plug them into the formula below (and simplify)\r<BR>\n" );
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document.write( "sin(u+v) = sin(u)*cos(v) + cos(u)*sin(v)<BR>\n" );
document.write( "sin(u+v) = (2/3)*(2*sqrt(2)/3) + (sqrt(5)/3)*(1/3)<BR>\n" );
document.write( "sin(u+v) = 4*sqrt(2)/9 + sqrt(5)/9<BR>\n" );
document.write( "sin(u+v) = (4*sqrt(2) + sqrt(5))/9\r<BR>\n" );
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document.write( "So the final answer is <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=sin%28u%2Bv%29+=+%284%2Asqrt%282%29+%2B+sqrt%285%29%29%2F9 ALIGN=MIDDLE  ALT=\"sin%28u%2Bv%29+=+%284%2Asqrt%282%29+%2B+sqrt%285%29%29%2F9\"   DISABLED_style= \"max-width:90%; height:auto;\" > </SPAN>\n" );
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