document.write( "Question 551032: Prove that
\n" ); document.write( "sin(n+1)Asin(n+2)A + cos(n+1)Acos(n+2)A = cosA \n" ); document.write( "
Algebra.Com's Answer #359467 by Edwin McCravy(20060)\"\" \"About 
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document.write( "sin[(n+1)A]sin[(n+2)A] + cos[(n+1)A]cos[(n+2)A] = cosA\r\n" );
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document.write( "Use the commutative principles of multiplication and addition\r\n" );
document.write( "to rearrange the left side as\r\n" );
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document.write( "cos[(n+2)A]cos[(n+1)A] + sin[(n+2)A]sin[(n+1)A]\r\n" );
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document.write( "The left side is the right side of the identity \r\n" );
document.write( "                \"cos%28alpha-beta%29=cos%28alpha%29cos%28beta%29%2Bsin%28alpha%29sin%28beta%29\" with \"alpha=%28n%2B2%29A\" and \"beta=%28n%2B1%29A\"\r\n" );
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document.write( "So the left side becomes:\r\n" );
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document.write( "cos[(n+2)A - (n+1)A]\r\n" );
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document.write( "cos[nA + 2A - nA - A]\r\n" );
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document.write( "cos(A)\r\n" );
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document.write( "Edwin
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