document.write( "Question 1202005: if \"cos%28x%29+=+-5%2F13\" and \"pi+%3C+x+%3C+3pi%2F2\", find \"cos%28x%2F2%29\" using half angle identities \n" ); document.write( "
Algebra.Com's Answer #836619 by Edwin McCravy(20065)\"\" \"About 
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document.write( "The identity to use is \"cos%28x%2F2%29\"\"%22%22=%22%22\"\"%22%22+%2B-+sqrt%28%281%2Bcos%28x%29%29%2F2%29\"\r\n" );
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document.write( "Since we are finding \"cos%28x%2F2%29\" we divide all three sides of the domain for\r\n" );
document.write( "x by 2, so we'll have the domain for \"x%2F2\".\r\n" );
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document.write( "\"pi+%3C+x+%3C+3pi%2F2\", divide through by 2: \r\n" );
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document.write( "\"pi%2F2+%3C+x%2F2+%3C+3pi%2F4\", which in degrees is between 90o and 135o.\r\n" );
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document.write( "which means x/2 is in QII.  The cosine is negative in QII, so we use the\r\n" );
document.write( "negative sign\r\n" );
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document.write( "\"cos%28x%2F2%29\"\"%22%22=%22%22\"\"-sqrt%28%281%2Bcos%28x%29%29%2F2%29%29\"\r\n" );
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document.write( "\"cos%28x%2F2%29\"\"%22%22=%22%22\"\"-sqrt%28%28%281%2B%28%22-5%2F13%22%29%29%2F2%29%29\"\r\n" );
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document.write( "\"cos%28x%2F2%29\"\"%22%22=%22%22\"\"-sqrt%28%281-%225%2F13%22%29%2F2%29\"\r\n" );
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document.write( "Multiply inside the square root radical by 13/13\r\n" );
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document.write( "\"cos%28x%2F2%29\"\"%22%22=%22%22\"\"-sqrt%28expr%28%28%281-%225%2F13%22%29%2F2%29%29%2813%2F13%29%29\"\r\n" );
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document.write( "\"cos%28x%2F2%29\"\"%22%22=%22%22\"\"-sqrt%28%2813-5%29%2F26%29\"\r\n" );
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document.write( "\"cos%28x%2F2%29\"\"%22%22=%22%22\"\"-sqrt%288%2F26%29\"\r\n" );
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document.write( "\"cos%28x%2F2%29\"\"%22%22=%22%22\"\"-sqrt%284%2F13%29\"\r\n" );
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document.write( "\"cos%28x%2F2%29\"\"%22%22=%22%22\"\"-2%2Fsqrt%2813%29\"\r\n" );
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document.write( "You can rationalize the denominator of that if you like.\r\n" );
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document.write( "Edwin
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