document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1174989>Question 1174989</A>: <I><SPAN STYLE=\"word-break: break-all;\"> 1. Given sinB=5/13 in QII and (6,-8) is on the terminal side of a ,find the exact value of sin (a+B). </SPAN>\n" );
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document.write( "If sin(B) = 5/13, then cos(B) = -12/13. Use the pythagorean theorem to see this. Draw out a right triangle. Note how cosine is negative in quadrant 2.\r<BR>\n" );
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document.write( "The distance from (0,0) to (6,-8) is 10 units. Again you'll need the pythagorean theorem to see why this is the case, or you could use the distance formula (which is a rephrasing of the pythagorean theorem).\r<BR>\n" );
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document.write( "We have a right triangle with hypotenuse 10, and the two legs are 6 and 8 units each. Let's say we had a third point at (6,0). This is where the 90 degree angle is located. The distance from (0,0) to (6,0) is 6 units to form the horizontal leg, while the vertical leg is 8 units because it spans from (6,0) to (6,-8)\r<BR>\n" );
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document.write( "So,<BR>\n" );
document.write( "sin(a) = opposite/hypotenuse<BR>\n" );
document.write( "sin(a) = -8/10<BR>\n" );
document.write( "sin(a) = -4/5<BR>\n" );
document.write( "and<BR>\n" );
document.write( "cos(a) = adjacent/hypotenuse<BR>\n" );
document.write( "cos(a) = 6/10<BR>\n" );
document.write( "cos(a) = 3/5<BR>\n" );
document.write( "In quadrant 4, sine is negative and cosine is positive\r<BR>\n" );
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document.write( "We have these four facts<BR>\n" );
document.write( "sin(a) = -4/5<BR>\n" );
document.write( "cos(a) = 3/5<BR>\n" );
document.write( "sin(B) = 5/13<BR>\n" );
document.write( "cos(B) = -12/13\r<BR>\n" );
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document.write( "Now we use a trig identity to find what we're after<BR>\n" );
document.write( "sin(a+B) = sin(a)*cos(B) + cos(a)*sin(B)<BR>\n" );
document.write( "sin(a+B) = (-4/5)*(-12/13) + (3/5)*(5/13)<BR>\n" );
document.write( "sin(a+B) = 48/65 + 15/65<BR>\n" );
document.write( "sin(a+B) = (48+15)/65<BR>\n" );
document.write( "sin(a+B) = 63/65\r<BR>\n" );
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document.write( "Answer: 63/65<BR>\n" );
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