document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=236887>Question 236887</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Find (2x) given that sin(x)=2/5 and x is in quadrant II. Give exact value for sin(2x) </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #174244 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=174244\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Sine is positive in quadrants 1 and 2.\r<BR>\n" );
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document.write( "sin2x is one of our double trig identities.\r<BR>\n" );
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document.write( "sin2x = 2sinxcosx\r<BR>\n" );
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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.\r<BR>\n" );
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document.write( "Let A = adjacent side\r<BR>\n" );
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document.write( "A^2 + 2^2 = 5^2\r<BR>\n" );
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document.write( "A^2 + 4 = 25\r<BR>\n" );
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document.write( "A^2 = 25 - 4\r<BR>\n" );
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document.write( "A^2 = 21\r<BR>\n" );
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document.write( "A = sqrt{21}.\r<BR>\n" );
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document.write( "Now that we know the value of the adjacent side, we can find cosx.\r<BR>\n" );
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document.write( "cosx = adjacent side over hypotenuse.\r<BR>\n" );
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document.write( "cosx = sqrt{21}/5\r<BR>\n" );
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document.write( "Of course, cosine is negative in quadrant 2.\r<BR>\n" );
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document.write( "So, the correct answer for cosx = -sqrt{21}/5\r<BR>\n" );
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document.write( "We can now plug the values for sinx and cosx into the double trig identity for sin2x and simplify.\r<BR>\n" );
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document.write( "sin2x = 2((2/5)(-sqrt{21}/5)\r<BR>\n" );
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document.write( "sin2x = -4(sqrt{21})/25 </SPAN>\n" );
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