document.write( "Question 1182618: Evaluate the six trigonometric function of: s=2π/3. Show your solution.
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Algebra.Com's Answer #812708 by Edwin McCravy(20060)\"\" \"About 
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Evaluate the six trigonometric function of: s=2π/3. Show your solution.
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document.write( "\"2pi%2F3\" is 120o, has a referent angle of 60o and is in the second quadrant, \r\n" );
document.write( "so we draw a 30-60-90 triangle with hypotenuse=r=2, adjacent=x=1, and\r\n" );
document.write( "opposite=y=√3\r\n" );
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document.write( "However, x goes left in QII, so we must make x negative, so we have x=-1\r\n" );
document.write( "instead of 1 in the graph below.  y is positive because it goes upward.\r\n" );
document.write( "r, the hypotenuse or radius vector is always taken positive in all quadrants.\r\n" );
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document.write( "adjacent = x = -1,  opposite = y = √3, hypotenuse = r = 2\r\n" );
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document.write( "\"sin%28s%29+=+opposite%2Fhypotenuse+=+y%2Fr+=+sqrt%283%29%2F2\"\r\n" );
document.write( "\"cos%28s%29+=+adjacent%2Fhypotenuse+=+x%2Fr+=+%28-1%29%2F2+=-1%2F2\"\r\n" );
document.write( "\"tan%28s%29+=+opposite%2Fadjacent+=+y%2Fx+=+sqrt%283%29%2F%28-1%29+=+-sqrt%283%29\"\r\n" );
document.write( "\"sec%28s%29+=+hypotenuse%2Fadjacent+=+r%2Fx+=+2%2F%28-1%29=-2\"\r\n" );
document.write( "\"csc%28s%29+=+hypotenuse%2Fopposite+=+r%2Fy+=+2%2F%28sqrt%283%29%29+=+2sqrt%283%29%2F3\" \r\n" );
document.write( "\"cot%28s%29+=+adjacent%2Fopposite+=+x%2Fy+=+%28-1%29%2Fsqrt%283%29+=+-sqrt%283%29%2F3\" \r\n" );
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document.write( "Edwin
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