document.write( "Question 898845: Find the remaining trigonometric ratios.
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Algebra.Com's Answer #545021 by Edwin McCravy(20059)\"\" \"About 
You can put this solution on YOUR website!
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document.write( "\"sin%5E2%28x%29%2Bcos%5E2%28x%29=1\"\r\n" );
document.write( "\"sin%5E2%28x%29%2B%28-1%2F3%29%5E2=1\"\r\n" );
document.write( "\"sin%5E2%28x%29%2B1%2F9=1\"\r\n" );
document.write( "\"9sin%5E2%28x%29%2B1=9\"\r\n" );
document.write( "\"9sin%5E2%28x%29=8\"\r\n" );
document.write( "\"sin%5E2%28x%29=8%2F9\"\r\n" );
document.write( "\"sin%28x%29=%22%22+%2B-+sqrt%288%2F9%29\"\r\n" );
document.write( "\"sin%28x%29=%22%22+%2B-+sqrt%288%29%2F3\"\r\n" );
document.write( "\"sin%28x%29=%22%22+%2B-+sqrt%284%2A2%29%2F3\"\r\n" );
document.write( "\"sin%28x%29=%22%22+%2B-+2sqrt%282%29%2F3\"\r\n" );
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document.write( "Since π < x < 3π/2 is quadrant III, \r\n" );
document.write( "we take the negative answer\r\n" );
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document.write( "\"sin%28x%29=-2sqrt%282%29%2F3\"\r\n" );
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document.write( "Edwin
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