document.write( "Question 1056289: if cosx=-12/13 and sin x>0, find tan(2x)
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Algebra.Com's Answer #671461 by ikleyn(52814)\"\" \"About 
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\n" ); document.write( "if cosx=-12/13 and sin x>0, find tan(2x)
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document.write( "0.  The useful notice is: Since cos(x) is negative and sin(x) is positive, the angle \"x\" terminates in QII.\r\n" );
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document.write( "1.  Since cos(x) = \"-12%2F13\", sin(x) = \"sqrt%281-cos%5E2%28x%29%29\" = \"sqrt%281-%28-12%2F13%29%5E2%29\" = \"sqrt%281-144%2F169%29\" = \"sqrt%28%28169-144%29%2F169%29\" = \"sqrt%2825%2F169%29\" = \"5%2F13\".\r\n" );
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document.write( "    The sign is \"+\" at the square root, since sine is positive, according to the condition.\r\n" );
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document.write( "2.  Then \"tan%28x%29\" = \"sin%28x%29%2Fcos%28x%29\" = \"%28%285%2F13%29%29%2F%28%28-12%2F13%29%29\" = \"-5%2F12\".\r\n" );
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document.write( "3.  Finally, \"tan%282x%29\" = \"%282%2Atan%28x%29%29%2F%281-tan%5E2%28x%29%29\" = \"%282%2A%28-5%2F12%29%29%2F%281-%28%28-5%2F12%29%5E2%29%29\" = \"%28%28-10%2F12%29%29%2F%28%281-25%2F144%29%29\" = \"%28%28-10%2F12%29%29%2F%28%28%28144-25%29%2F144%29%29\" = \"%28%28-10%2F12%29%29%2F%28%28119%2F144%29%29\" = \"%28-10%2A144%29%2F%2812%2A119%29\" = \"%28-10%2A12%29%2F119\" = \"-120%2F119\".\r\n" );
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\n" ); document.write( "\n" ); document.write( "For more solved similar problems on calculating trig functions see the lessons \r
\n" ); document.write( "\n" ); document.write( "    - Calculating trigonometric functions of angles\r
\n" ); document.write( "\n" ); document.write( "    - Advanced problems on calculating trigonometric functions of angles\r
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\n" ); document.write( "\n" ); document.write( "Also, you have this free of charge online textbook in ALGEBRA-II in this site\r
\n" ); document.write( "\n" ); document.write( "    - ALGEBRA-II - YOUR ONLINE TEXTBOOK.\r
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\n" ); document.write( "\n" ); document.write( "The referred lessons are the part of this online textbook under the topic \"Trigonometry: Solved problems\". \r
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