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    1. 8 sin invisible function application fraction numerator A over denominator 2 end fraction sin invisible function application fraction numerator B over denominator 2 end fraction sin invisible function application fraction numerator C over denominator 2 end fraction  
    2. 8 cos invisible function application fraction numerator A over denominator 2 end fraction cos invisible function application fraction numerator B over denominator 2 end fraction cos invisible function application fraction numerator C over denominator 2 end fraction  
    3. 8 tan invisible function application fraction numerator A over denominator 2 end fraction tan invisible function application fraction numerator B over denominator 2 end fraction tan invisible function application fraction numerator C over denominator 2 end fraction  
    4. 8 cot invisible function application fraction numerator A over denominator 2 end fraction cot invisible function application fraction numerator B over denominator 2 end fraction cot invisible function application fraction numerator C over denominator 2 end fraction  

    The correct answer is: 8 sin invisible function application fraction numerator A over denominator 2 end fraction sin invisible function application fraction numerator B over denominator 2 end fraction sin invisible function application fraction numerator C over denominator 2 end fraction

      fraction numerator sin invisible function application 2 A plus sin invisible function application 2 B plus sin invisible function application 2 C over denominator sin invisible function application A plus sin invisible function application B plus sin invisible function application C end fraction equals fraction numerator 4 sin invisible function application A sin invisible function application B sin invisible function application C over denominator 4 cos invisible function application fraction numerator A over denominator 2 end fraction cos invisible function application fraction numerator B over denominator 2 end fraction cos invisible function application fraction numerator C over denominator 2 end fraction end fraction equals 8 sin invisible function application fraction numerator A over denominator 2 end fraction sin invisible function application fraction numerator B over denominator 2 end fraction sin invisible function application fraction numerator C over denominator 2 end fraction open square brackets because sin invisible function application A equals 2 sin invisible function application fraction numerator A over denominator 2 end fraction cos invisible function application fraction numerator A over denominator 2 end fraction close square brackets

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