document.write( "Question 1194315: If tan A = ¾ and cos B = 12/13, find
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document.write( "a. Tan (A - B)
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document.write( "b. c. tan ( B - A )
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document.write( "c. Sin ( A + B)
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document.write( "d. cos 2A \n" );
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Algebra.Com's Answer #826500 by Alan3354(69443)![]() ![]() You can put this solution on YOUR website! If tan A = ¾ and cos B = 12/13, find \n" ); document.write( "a. Tan (A - B) \n" ); document.write( "b. c. tan ( B - A ) \n" ); document.write( "c. Sin ( A + B) \n" ); document.write( "d. cos 2A \n" ); document.write( "---------------- \n" ); document.write( "tan(A) = 3/4 \n" ); document.write( "tan = y/x \n" ); document.write( "r = sqrt(x^2 + y^2) = 5 \n" ); document.write( "----- \n" ); document.write( "sin = y/r \n" ); document.write( "cos = x/r \n" ); document.write( "cot = 1/tan \n" ); document.write( "sec = 1/cos \n" ); document.write( "csc = 1/sin \n" ); document.write( "---- \n" ); document.write( "cos(B) = 12/13 = x/r ---> y = 5 \n" ); document.write( "sin = y/r \n" ); document.write( "cos = x/r \n" ); document.write( "tan = y/x \n" ); document.write( "cot = 1/tan \n" ); document.write( "sec = 1/cos \n" ); document.write( "csc = 1/sin \n" ); document.write( "================= \n" ); document.write( "a. Tan (A - B) \n" ); document.write( "b. c. tan ( B - A ) \n" ); document.write( "c. Sin ( A + B) \n" ); document.write( "d. cos 2A \n" ); document.write( "----- \n" ); document.write( "Look up formulas at Wikipedia \"Half angle formulas\" \n" ); document.write( "---- Email via the TY note for help or to check your work. \n" ); document.write( " |