document.write( "Question 852727: The question must be answered in trigonometric form:\r
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Algebra.Com's Answer #513643 by Edwin McCravy(20056)\"\" \"About 
You can put this solution on YOUR website!
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document.write( "\"x%2Biy\"\"%22%22=%22%22\"\"4i+%2F+%28-1%2B+sqrt%283%29i%29\"\r\n" );
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document.write( "First get it in the form x+iy by multiplying numerator\r\n" );
document.write( "and denominator by the conjugate of the denominator: \r\n" );
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document.write( "\"4i%28-1-sqrt%283%29i%29+%2F%28%28-1%2B+sqrt%283%29i%29%28-1-+sqrt%283%29i%29%29\"\r\n" );
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document.write( "\"%28-4i-4sqrt%283%29i%5E2%29%2F%281-3i%5E2%29\"\r\n" );
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document.write( "Change both iČ's to (-1):\r\n" );
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document.write( "\"%28-4i-4sqrt%283%29%28-1%29%29%2F%281-3%28-1%29%29\"\r\n" );
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document.write( "\"%28-4i%2B4sqrt%283%29%29%29%2F%281%2B3%29\"\r\n" );
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document.write( "Factor 4 out of the numerator:\r\n" );
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document.write( "\"%284%28-i%2Bsqrt%283%29%29%29%2F4\"\r\n" );
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document.write( "\"%28cross%284%29%28-i%2Bsqrt%283%29%29%29%2Fcross%284%29\"\r\n" );
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document.write( "\"-i%2Bsqrt%283%29\"\r\n" );
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document.write( "\"sqrt%283%29-i\"\r\n" );
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document.write( "So the rectangular form is: \r\n" );
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document.write( "\"x%2Biy\"\"%22%22=%22%22\"\"sqrt%283%29-1\"\r\n" );
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document.write( "That is represented by the vector (line) connecting the origin to the point \r\n" );
document.write( "(x,y)= (\"sqrt%283%29\",-1), We draw a perpendicular to the x-axis, and indicate\r\n" );
document.write( "the angle \"theta\" by a red arc:\r\n" );
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document.write( "We find the value of r (the hypotenuse) by the Pythagorean theorem:\r\n" );
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document.write( "\"r%5E2\"\"%22%22=%22%22\"\"x%5E2%2By%5E2\"\r\n" );
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document.write( "\"r%5E2\"\"%22%22=%22%22\"\"%28sqrt%283%29%29%5E2%2B%28-1%29%5E2\"\r\n" );
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document.write( "\"r%5E2\"\"%22%22=%22%22\"\"3%2B1\"\r\n" );
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document.write( "\"r%5E2\"\"%22%22=%22%22\"\"4\"\r\n" );
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document.write( "r is always positive so we take the positive square root:\r\n" );
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document.write( "\"r\"\"%22%22=%22%22\"\"2\"\r\n" );
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document.write( "We find \"theta\" by using any trig function, say the cosine;\r\n" );
document.write( "cos(theta)=adjacent/hypotenuse=x/r=sqrt(3)/2}}}\r\n" );
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document.write( "This tells us that the angle has a refence angle of 30°, but since\r\n" );
document.write( "it is in quadrant IV, we subtract from 360° and get \"theta=%22330%B0%22\"\r\n" );
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document.write( "\"r\"\"%22%22=%22%22\"\"2\"\r\n" );
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document.write( "Next we know that\r\n" );
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document.write( "\"adjacent%2Fhypotenuse=x%2Fr=cos%28theta%29\", so \"x=r%2Acos%28theta%29=2%2Acos%28%22330%B0%22%29\"\r\n" );
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document.write( "Also, we know that\r\n" );
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document.write( "\"opposite%2Fhypotenuse=y%2Fr=sin%28theta%29\", so \"y=r%2Asin%28theta%29=+2%2Asin%28%22330%B0%22%29\"\r\n" );
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document.write( "So \r\n" );
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document.write( "x + iy = 2·cos(330°)+i·2·sin(330°) = 2[cos(330°) + i·sin(330°)]\r\n" );
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document.write( "Edwin
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