document.write( "Question 1039196: use the cosine of a sum and cosine of a difference identities to find cos(s+t) and cos(s-t).\r
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Algebra.Com's Answer #653949 by ikleyn(52832)\"\" \"About 
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\n" ); document.write( "use the cosine of a sum and cosine of a difference identities to find cos(s+t) and cos(s-t).\r
\n" ); document.write( "\n" ); document.write( "sin s= -4/5 in Q IV
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\n" ); document.write( "\n" ); document.write( "what is cos(s+t)?
\n" ); document.write( "what is cos(s-t)?\r
\n" ); document.write( "\n" ); document.write( "thank you in advance!!!
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document.write( "Use the formulas \r\n" );
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document.write( "cos(s+t) = cos(s)*cos(t) - sin(s)*sin(t)    (1)\r\n" );
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document.write( "   and\r\n" );
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document.write( "cos(s-t) = cos(s)*cos(t) + sin(s)*sin(t)    (2)\r\n" );
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document.write( "Regarding these formulas, see the lesson Addition and subtraction formulas in this site.\r\n" );
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document.write( "In addition to the given  sin(s) = \"-4%2F5\"  and  sin(t) = \"12%2F13\", you need to know  cos(s) and cos(t).\r\n" );
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document.write( "1.  cos(s) = \"sqrt%281-sin%5E2%28s%29%29\" = \"sqrt%281+-+%28-4%2F5%29%5E2%29\" = \"sqrt%281+-+16%2F25%29\" = \"sqrt%2825-16%29%2F25%29\" = \"sqrt%289%2F25%29\" = \"3%2F5\".\r\n" );
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document.write( "   The sign \"+\" was chosen for the square root because cos(s) is positive in Q4.\r\n" );
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document.write( "2.  cos(t) = = -\"sqrt%281-sin%5E2%28t%29%29\" = -\"sqrt%281+-+%2812%2F13%29%5E2%29\" = -\"sqrt%281+-+144%2F169%29\" = -\"sqrt%28169-144%29%2F169%29\" = -\"sqrt%2825%2F169%29\" =\"-5%2F13\".\r\n" );
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document.write( "   The sign \"-\" was chosen for the square root because cos(t) is negative in Q2.\r\n" );
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document.write( "Now all you need to do is to substitute everything into the formulas (1) and (2) and make the calculations.\r\n" );
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document.write( "cos(s-t) = \"%283%2F5%29%2A%28-5%2F13%29+%2B+%28-4%2F5%29%2A%2812%2F13%29\" = \"-15%2F65+-+48%2F65\" = \"%28-15-48%29%2F65\" = \"-63%2F65\",   and\r\n" );
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document.write( "cos(s+t) = \"%283%2F5%29%2A%28-5%2F13%29+-+%28-4%2F5%29%2A%2812%2F13%29\" = \"-15%2F65+%2B+48%2F65\" = \"%28-15%2B48%29%2F65\" = \"33%2F65\".\r\n" );
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