document.write( "Question 190622: I need your help on the following problem:
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Algebra.Com's Answer #143093 by jim_thompson5910(35256)\"\" \"About 
You can put this solution on YOUR website!
Note: since sine is an odd function, this means that \"sin%28-x%29=-sin%28x%29\". So \"sin%28-pi%2F3%29=-sin%28pi%2F3%29\"\r
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\n" ); document.write( "\n" ); document.write( "In order to evaluate \"-sin%28pi%2F3%29\", just evaluate \"sin%28pi%2F3%29\" and negate it.\r
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\n" ); document.write( "\n" ); document.write( "Looking at the unit circle:\r
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\n" ); document.write( "\n" ); document.write( "We see that at the theta value \"pi%2F3\", the y-coordinate of the point on the unit circle is \"y=sqrt%283%29%2F2\"\r
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\n" ); document.write( "\n" ); document.write( "So this simply means that \"sin%28pi%2F3%29=sqrt%283%29%2F2\"\r
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\n" ); document.write( "\n" ); document.write( "Now negate to get \"-sin%28pi%2F3%29=-sqrt%283%29%2F2\"\r
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\n" ); document.write( "\n" ); document.write( "Answer:\r
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\n" ); document.write( "\n" ); document.write( "So \"sin%28-pi%2F3%29=-sqrt%283%29%2F2\"
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