document.write( "Question 1030548: A is an angle in quadrant 2. TanA = -3/root5. Solve the following and rationalize the denominator to the exact value.
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Algebra.Com's Answer #645398 by ikleyn(52803)\"\" \"About 
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\n" ); document.write( "A is an angle in quadrant 2. TanA = -3/root5. Solve the following and rationalize the denominator to the exact value.
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\n" ); document.write( "c) Cos(A + pi/2)
\n" ); document.write( "d) Sin(A + pi/2)
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document.write( "First of all, we need to find  sin(A)  and  cos(A)  based on the given value of tan(A) and the fact that angle A is in quadrant 2.\r\n" );
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document.write( "For it, we will use well known formulas  of Trigonometry \r\n" );
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document.write( "\"sin%5E2%28A%29\" = \"tan%5E2%28A%29%2F%281+%2B+tan%5E2%28A%29%29\"  and  \"cos%5E2%28A%29\" = \"1%2F%281+%2B+tan%5E2%28A%29%29\".\r\n" );
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document.write( "Since tan(A) = \"-3%2Fsqrt%285%29\", we have \r\n" );
document.write( "sin(A) = \"sqrt%28tan%5E2%28A%29%2F%281+%2B+tan%5E2%28A%29%29%29\" = \"sqrt%28%28-3%2Fsqrt%285%29%29%5E2%2F%281+%2B+%28-3%2Fsqrt%285%29%29%5E2%29%29\" = \"sqrt%28%28%289%2F5%29%29%2F%281+%2B+9%2F5%29%29\" = \"sqrt%28%28%289%2F5%29%29%2F%28%2814%2F5%29%29%29\" = \"sqrt%289%2F14%29\"     and \r\n" );
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document.write( "cos(A) = -\"1%2Fsqrt%281+%2B+tan%5E2%28A%29%29\" = -\"1%2Fsqrt%281+%2B+%28-3%2Fsqrt%285%29%29%5E2%29\" = -\"1%2Fsqrt%281+%2B+9%2F5%29\" = -\"1%2Fsqrt%2814%2F5%29\" = \"-sqrt%285%2F14%29\".\r\n" );
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document.write( "The sign for sin(A) is \"+\" since the angle A is in Q2.  The sign for cos(A) is \"-\" due to the same reason.\r\n" );
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document.write( "Now\r\n" );
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document.write( "a)  sin(2A) = 2*sin(a)*cos(A) = \"2%2Asqrt%289%2F14%29%2A%28-sqrt%285%2F14%29%29\" = \"-2%2Asqrt%28%289%2A5%29%2F14%5E2%29\" = \"-2%2Asqrt%2845%29%2F14\" = \"-sqrt%2845%29%2F7\".\r\n" );
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document.write( "b)  cos(2A) = \"2%2Acos%5E2%28A%29+-+1\" = \"2%2A%285%2F14%29+-+1\" = \"5%2F7+-+1\" = \"-2%2F7\".\r\n" );
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document.write( "    (By the way, the fact that cos(2A) is negative means that the angle 2A is in Q3).\r\n" );
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document.write( "c)  cos(A + pi/2) = -sin(A) = -\"sqrt%289%2F14%29\".\r\n" );
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document.write( "d)  sin(A + pi/2) = cos(A) = \"-sqrt%285%2F14%29\".\r\n" );
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document.write( "The problem is solved. \r\n" );
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