document.write( "Question 1126750: What is the value of cos(2α) , if sin(3α) =2 sin(α)? \n" ); document.write( "
Algebra.Com's Answer #743080 by ikleyn(52788)\"\" \"About 
You can put this solution on YOUR website!
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\n" ); document.write( "\n" ); document.write( "                The solution by the tutor @MathLover1 is not exactly correct,\r
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document.write( "Start from this basic identity\r\n" );
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document.write( "    sin(3a) = 3*sin(a) - 4*sin^3(a).\r\n" );
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document.write( "We are given that\r\n" );
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document.write( "    3*sin(a) - 4*sin^3(a) = 2*sin(a).\r\n" );
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document.write( "It implies\r\n" );
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document.write( "    \"%284%2Asin%5E2%28a%29+-+1%29%2Asin%28a%29\" = 0.\r\n" );
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document.write( "So, either  sin(a) = 0   or   \"4%2Asin%5E2%28a%29+-+1\" = 0;  the last is equivalent to  \"sin%5E2%28a%29\" = \"1%2F4\".\r\n" );
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document.write( "Case 1.  If  sin(a) = 0,  then  cos(2a) = \"1+-+2%2Asin%5E2%28a%29\" = 1.\r\n" );
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document.write( "Case 2.  If  \"sin%5E2%28a%29\" = \"1%2F4\",  then  cos(2a) = \"1+-+2%2Asin%5E2%28a%29%29\" = \"1+-+2%2A%281%2F4%29\" = \"1%2F2\".\r\n" );
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document.write( "Answer.  If  sin(3a) =2*sin(a),  then EITHER  cos(2a) = 1  OR  cos(2a) = \"1%2F2\".\r\n" );
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