document.write( "Question 1011732: cot(x)*cot(2 x) = 1 \n" ); document.write( "
Algebra.Com's Answer #627474 by Edwin McCravy(20056)\"\" \"About 
You can put this solution on YOUR website!
\"cot%28x%29cot%282x%29\"\"%22%22=%22%22\"\"1\"
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document.write( "We normally aren't given double angle formulas for cotangents.\r\n" );
document.write( "We usually only are given double angle formulas for tangents.\r\n" );
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document.write( "Since we know that \"cot%28theta%29=1%2Ftan%28theta%29\", let's\r\n" );
document.write( "convert to tangents:\r\n" );
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document.write( "\"cot%28x%29cot%282x%29\"\"%22%22=%22%22\"\"1\"\r\n" );
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document.write( "\"%281%2Ftan%28x%29%29%281%2Ftan%282x%29%29\"\"%22%22=%22%22\"\"1\"\r\n" );
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document.write( "\"1%2F%28tan%28x%29tan%282x%29%29\"\"%22%22=%22%22\"\"1\"\r\n" );
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document.write( "We take reciprocals of both sides, and since the reciprocal\r\n" );
document.write( "of 1 is 1, we have\r\n" );
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document.write( "\"tan%28x%29tan%282x%29\"\"%22%22=%22%22\"\"1\"\r\n" );
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document.write( "We use the double-angle formula that we are always given:\r\n" );
document.write( "which is \"tan%282theta%29=%282tan%28theta%29%29%2F%281-tan%5E2%28theta%29%29\"\r\n" );
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document.write( "\"tan%28x%29%28%282tan%28x%29%29%2F%281-tan%5E2%28x%29%29%29\"\"%22%22=%22%22\"\"1\"\r\n" );
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document.write( "\"%28tan%28x%29%2F1%5E%22%22%29%28%282tan%28x%29%29%2F%281-tan%5E2%28x%29%29%29\"\"%22%22=%22%22\"\"1\"\r\n" );
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document.write( "\"%282tan%5E2%28x%29%29%2F%281-tan%5E2%28x%29%29\"\"%22%22=%22%22\"\"1\"\r\n" );
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document.write( "Multiply both sides by the denominator on the left:\r\n" );
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document.write( "\"2tan%5E2%28x%29\"\"%22%22=%22%22\"\"1-tan%5E2%28x%29\"\r\n" );
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document.write( "Solve for tan2(x)\r\n" );
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document.write( "\"3tan%5E2%28x%29\"\"%22%22=%22%22\"\"1\"\r\n" );
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document.write( "\"3tan%5E2%28x%29\"\"%22%22=%22%22\"\"1\"\r\n" );
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document.write( "Take square roots \r\n" );
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document.write( "\"tan%28x%29\"\"%22%22=%22%22\"\"%22%22+%2B-+sqrt%281%2F3%29\"\r\n" );
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document.write( "\"tan%28x%29\"\"%22%22=%22%22\"\"%22%22+%2B-+1%2Fsqrt%283%29\"\r\n" );
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document.write( "Remembering the special 30°-60°-90° right triangle:\r\n" );
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document.write( "we know by the ± that the angle can be in any quadrant\r\n" );
document.write( "with a 30° reference angle, so the solutions are\r\n" );
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document.write( "30°, 150°, 210°, 330°  plus any integer n times 360°.\r\n" );
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document.write( "or in radians\r\n" );
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document.write( "\"pi%2F6\",\"5pi%2F6\",\"7pi%2F6\",\"11pi%2F6\" plus any integer n times \"2pi\".\r\n" );
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document.write( "Edwin
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