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Question

In $triangle A B C comma left parenthesis left parenthesis b minus c right parenthesis divided by a right parenthesis sin invisible function application left parenthesis left parenthesis B plus C right parenthesis divided by 2 right parenthesis equals$

$\cos ((B - C) / 2)$
$\sin ((B - C) / 2)$
$\sin ((A - B) / 2)$
$\cos ((A - B) / 2)$

hint Hint:

In this question first we have to use Sine rule $fraction numerator a over denominator sin open parentheses A close parentheses end fraction equals fraction numerator b over denominator sin open parentheses B close parentheses end fraction equals fraction numerator c over denominator sin open parentheses C close parentheses end fraction equals k$ . Then we will use some trigonometric formula $sin open parentheses C close parentheses minus sin open parentheses D close parentheses space equals space 2 cos open parentheses fraction numerator C plus D over denominator 2 end fraction close parentheses sin open parentheses fraction numerator C minus D over denominator 2 end fraction close parentheses$ and $sin open parentheses 2 x close parentheses equals 2 sin open parentheses x close parentheses cos open parentheses x close parentheses$ and then we will do some rearranging of the expression.

The correct answer is: $sin invisible function application left parenthesis left parenthesis B minus C right parenthesis divided by 2 right parenthesis$

In this question we have to find the value of $open parentheses fraction numerator b minus c over denominator a end fraction close parentheses sin open parentheses fraction numerator B plus C over denominator 2 end fraction close parentheses$
By using Sine Rule
$fraction numerator a over denominator sin open parentheses A close parentheses end fraction equals fraction numerator b over denominator sin open parentheses B close parentheses end fraction equals fraction numerator c over denominator sin open parentheses C close parentheses end fraction equals k$
=> $open parentheses fraction numerator k sin open parentheses B close parentheses minus k sin open parentheses C close parentheses over denominator k sin open parentheses A close parentheses end fraction close parentheses sin open parentheses fraction numerator B plus C over denominator 2 end fraction close parentheses$
=> $open parentheses fraction numerator k sin open parentheses B close parentheses minus k sin open parentheses C close parentheses over denominator k sin open parentheses A close parentheses end fraction close parentheses sin open parentheses fraction numerator B plus C over denominator 2 end fraction close parentheses$
By using trigonometric formula
$sin open parentheses C close parentheses minus sin open parentheses D close parentheses space equals space 2 cos open parentheses fraction numerator C plus D over denominator 2 end fraction close parentheses sin open parentheses fraction numerator C minus D over denominator 2 end fraction close parentheses$ and $sin open parentheses 2 x close parentheses equals 2 sin open parentheses x close parentheses cos open parentheses x close parentheses$
=> $open parentheses fraction numerator 2 cos open parentheses begin display style fraction numerator B plus C over denominator 2 end fraction end style close parentheses sin open parentheses begin display style fraction numerator B minus C over denominator 2 end fraction end style close parentheses over denominator sin open parentheses A close parentheses end fraction close parentheses sin open parentheses fraction numerator B plus C over denominator 2 end fraction close parentheses$
We know that $A plus B plus C space equals space pi$
$fraction numerator B plus C over denominator 2 end fraction space equals space pi over 2 minus A over 2$
$sin open parentheses fraction numerator B plus C over denominator 2 end fraction close parentheses equals sin open parentheses straight pi over 2 minus A over 2 close parentheses$
=> $cos open parentheses A over 2 close parentheses$
Similarly $cos open parentheses fraction numerator B plus C over denominator 2 end fraction close parentheses equals sin open parentheses A over 2 close parentheses$
=> $open parentheses fraction numerator 2 sin open parentheses begin display style A over 2 end style close parentheses sin open parentheses begin display style fraction numerator B minus C over denominator 2 end fraction end style close parentheses over denominator 2 sin open parentheses begin display style A over 2 end style close parentheses cos open parentheses begin display style A over 2 end style close parentheses end fraction close parentheses cos open parentheses A over 2 close parentheses$
=> $sin open parentheses fraction numerator B minus C over denominator 2 end fraction close parentheses$

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