document.write( "Question 1056788: Find sin(x/2), cos(x/2), and tan(x/2) from the given information
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Algebra.Com's Answer #671854 by ikleyn(52788)\"\" \"About 
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document.write( "The condition \"tan(x)=1; 0° < x < 90° \" implies\r\n" );
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document.write( "x = 45°.\r\n" );
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document.write( "Then x/2 = 22°30'.\r\n" );
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document.write( "To calculate  sin(22°30'), apply the formula of half argument for sines  (see the lesson  Trigonometric functions of half argument in this site):\r\n" );
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document.write( "sin(22°30') = \"sqrt%28%281-cos%28pi%2F4%29%29%2F2%29\" = \"sqrt%28%281-sqrt%282%29%2F2%29%2F2%29\" = \"sqrt%28%282-sqrt%282%29%29%2F4%29\" = \"sqrt%282-sqrt%282%29%29%2F2\".\r\n" );
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document.write( "Similarly, \r\n" );
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document.write( "cos(22°30') = \"sqrt%28%281%2Bcos%28pi%2F4%29%29%2F2%29\" = \"sqrt%28%281%2Bsqrt%282%29%2F2%29%2F2%29\" = \"sqrt%28%282%2Bsqrt%282%29%29%2F4%29\" = \"sqrt%282%2Bsqrt%282%29%29%2F2\".\r\n" );
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document.write( "Hence,\r\n" );
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document.write( "tan(22°30') = sin(22°30')/cos(22°30') = \"sqrt%28%282-sqrt%282%29%29%2F%282%2Bsqrt%282%29%29%29\" =  = \"sqrt%28+%28%282-sqrt%282%29%29%5E2%29%2F%282%5E2-%28sqrt%282%29%29%5E2%29+%29\" = \"%282-sqrt%282%29%29%2Fsqrt%282%29\" = \"sqrt%282%29-1\".\r\n" );
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\n" ); document.write( "\n" ); document.write( "See the lesson  Trigonometric functions of half argument - Examples in this site.\r
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