document.write( "Question 1197232: If sec x= 3/2 and cosec y= 3; determine cos (x+y)\r
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Algebra.Com's Answer #830418 by Alan3354(69443)\"\" \"About 
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If sec x= 3/2 and cosec y= 3; determine cos (x+y)
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\n" ); document.write( "sec = 3/2 ---> cos(x) = 2/3
\n" ); document.write( "\"sin%28x%29+=+sqrt%281+-cos%5E2%28x%29%29+=+sqrt%285%29%2F3\"
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\n" ); document.write( "csc = 3 ----> sin(y) = 1/3
\n" ); document.write( "\"cos%28y%29+=+sqrt%281+-+sin%5E2%28y%29%29+=+sqrt%288%2F9%29+=+2sqrt%282%29%2F3\"
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\n" ); document.write( "cos(x+y) = cos(x)*cos(y) - sin(x)*sin(y) ------ I'm pretty sure.
\n" ); document.write( "Look it up at \"half-angle formulas\" on Wikipedia.
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