document.write( "Question 963649: Please solve this equation exactly. \r
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Algebra.Com's Answer #588776 by Edwin McCravy(20060)\"\" \"About 
You can put this solution on YOUR website!
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document.write( "A little more detail.  It's good to draw triangles\r\n" );
document.write( "on x and y axes for such problems:\r\n" );
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document.write( "sin(arccos(3/5) - arctan(5/12))\r\n" );
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document.write( "Let A = arccos(3/5) and B = arctan(5/12)\r\n" );
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document.write( "The we use the identity:\r\n" );
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document.write( "sin(A-B)= sin(A)cos(B)-cos(A)sin(B)\r\n" );
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document.write( "First we draw A = arccos(3/5).\r\n" );
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document.write( "arccos(3/5) means \r\n" );
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document.write( "\"the angle in the first quadrant whose cosine is 3/5\".\r\n" );
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document.write( "So we draw a triangle in the first quadrant.  Since \r\n" );
document.write( "cosine = adj/hyp = x/r we make the adjacent side, x, the\r\n" );
document.write( "same as the numerator of 3/5, which is x=3 and make \r\n" );
document.write( "the hypotenuse, r, the denominator of 3/5 which is r=5.\r\n" );
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document.write( "Next we draw arctan(5/12).\r\n" );
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document.write( "arctan(5/12) means \r\n" );
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document.write( "\"the angle in the first quadrant whose tangent is 5/12\".\r\n" );
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document.write( "So we draw another right triangle in the first quadrant.  \r\n" );
document.write( "Since tangent = opp/adj = y/x we make the opposite side, y, the\r\n" );
document.write( "same as the numerator of 5/12, which is y=5 and make \r\n" );
document.write( "the adjacent side, x, the denominator of 5/12 which is x=12.\r\n" );
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document.write( "Now we can easily finish:\r\n" );
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document.write( "sin(A-B)= sin(A)cos(B)-cos(A)sin(B)\r\n" );
document.write( "        = (4/5)(12/13)-(3/5)(5/13)\r\n" );
document.write( "        = 48/65 - 15/65\r\n" );
document.write( "        = 33/65\r\n" );
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document.write( "Edwin
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