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fraction numerator s i n invisible function application 3 theta over denominator 1 plus 2 c o s invisible function application 2 theta end fraction equals

  1. c o s invisible function application theta    
  2. negative s i n invisible function application theta    
  3. negative c o s invisible function application theta    
  4. s i n invisible function application theta    

The correct answer is: c o s invisible function application theta


    We should find the value of fraction numerator s i n invisible function application 3 theta over denominator 1 plus 2 c o s invisible function application 2 theta end fraction equals
    L H S space equals space fraction numerator sin left parenthesis 3 theta right parenthesis over denominator 1 plus 2 cos left parenthesis 2 theta right parenthesis end fraction
equals fraction numerator 3 sin left parenthesis theta right parenthesis minus 4 sin cubed left parenthesis theta right parenthesis over denominator 1 plus 2 left square bracket 1 minus 2 sin squared left parenthesis theta right parenthesis right square bracket end fraction
equals fraction numerator sin left parenthesis theta right parenthesis left square bracket 3 minus 4 sin squared left parenthesis theta right parenthesis right square bracket over denominator 3 minus 4 sin squared left parenthesis theta right parenthesis end fraction space equals sin left parenthesis theta right parenthesis

    Hence Choice 4 is correct

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