document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=236884>Question 236884</A>: <I><SPAN STYLE=\"word-break: break-all;\"> sin (2x)=2/5 </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #174249 by </B><A HREF=/tutors/aboutme.mpl?userid=nyc_function><B>nyc_function(2741)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=nyc_function><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=nyc_function><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=174249\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> This question is incomplete as written.\r<BR>\n" );
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document.write( "However, are you the same student who asked this same question in more detail regarding finding the value of sin2x?\r<BR>\n" );
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document.write( "If you are the same student, then see reply below.\r<BR>\n" );
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document.write( "Sine is positive in quadrants 1 and 2. <BR>\n" );
document.write( "sin2x is one of our double trig identities. <BR>\n" );
document.write( "sin2x = 2sinxcosx <BR>\n" );
document.write( "Given sinx = 2/5, we know that opposite side of right triangle in quadrant 2 is 2 and the hypotenuse is 5. We also know that sine = opposite/hypotenuse. We need to find the adjacent side. <BR>\n" );
document.write( "Let A = adjacent side <BR>\n" );
document.write( "A^2 + 2^2 = 5^2 <BR>\n" );
document.write( "A^2 + 4 = 25 <BR>\n" );
document.write( "A^2 = 25 - 4 <BR>\n" );
document.write( "A^2 = 21 <BR>\n" );
document.write( "A = sqrt{21}. <BR>\n" );
document.write( "Now that we know the value of the adjacent side, we can find cosx. <BR>\n" );
document.write( "cosx = adjacent side over hypotenuse. <BR>\n" );
document.write( "cosx = sqrt{21}/5 <BR>\n" );
document.write( "Of course, cosine is negative in quadrant 2. <BR>\n" );
document.write( "So, the correct answer for cosx = -sqrt{21}/5 <BR>\n" );
document.write( "We can now plug the values for sinx and cosx into the double trig identity for sin2x and simplify. <BR>\n" );
document.write( "sin2x = 2((2/5)(-sqrt{21}/5) <BR>\n" );
document.write( "sin2x = -4(sqrt{21})/25 <BR>\n" );
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