Question

Simplify \[\dfrac{{\cos \dfrac{1}{2}\left( {B - C} \right)}}{{\sin \dfrac{1}{2}A}}\] for \[\Delta ABC\].
A. \[\dfrac{{b - c}}{a}\]
B. \[\dfrac{{b + c}}{a}\]
C. \[\dfrac{a}{{b - c}}\]
d. \[\dfrac{a}{{b + c}}\]

Answer 1

Hint: We will multiply \[\sin \dfrac{{B + C}}{2}\]with the denominator and numerator of the given expression and apply trigonometry identity and the relation of the sum of all angles of the triangle to simplify the given expression. Then we will apply the sine law to find the value of \[\sin A\], \[\sin B\], \[\sin C\] and substitute it in the given expression to find the value of the given expression.

Formula used:
Trigonometry identity:
\[2\sin A\cos B = \sin \left( {A + B} \right) + \sin \left( {A - B} \right)\]
\[\sin 2\theta = 2\sin \theta \cos \theta \]
Sine Law:
\[\dfrac{{\sin A}}{a} = \dfrac{{\sin B}}{b} = \dfrac{{\sin C}}{c}\]

Complete step by step solution:
Given expression is \[\dfrac{{\cos \dfrac{1}{2}\left( {B - C} \right)}}{{\sin \dfrac{1}{2}A}}\]
Now multiply \[\sin \dfrac{{B + C}}{2}\] with denominator and numerator
\[ = \dfrac{{\sin \dfrac{{B + C}}{2}\cos \dfrac{1}{2}\left( {B - C} \right)}}{{\sin \dfrac{{B + C}}{2}\sin \dfrac{1}{2}A}}\]
Now multiply 2 with denominator and numerator
\[ = \dfrac{{2\sin \dfrac{{B + C}}{2}\cos \dfrac{1}{2}\left( {B - C} \right)}}{{2\sin \dfrac{{B + C}}{2}\sin \dfrac{1}{2}A}}\]
Apply \[2\sin A\cos B = \sin \left( {A + B} \right) + \sin \left( {A - B} \right)\] in the numerator
\[ = \dfrac{{\sin \left( {\dfrac{{B + C}}{2} + \dfrac{{B - C}}{2}} \right) + \sin \left( {\dfrac{{B + C}}{2} - \dfrac{{B - C}}{2}} \right)}}{{2\sin \dfrac{{B + C}}{2}\sin \dfrac{1}{2}A}}\]
\[ = \dfrac{{\sin B + \sin C}}{{2\sin \dfrac{{B + C}}{2}\sin \dfrac{1}{2}A}}\]
We know the sum angles of triangle is \[{180^ \circ }\] or \[\pi \].
\[A + B + C = \pi \]
\[ \Rightarrow B + C = \pi - A\]
Putting \[B + C = \pi - A\] in the denominator
\[ = \dfrac{{\sin B + \sin C}}{{2\sin \dfrac{{\pi - A}}{2}\sin \dfrac{1}{2}A}}\]
\[ = \dfrac{{\sin B + \sin C}}{{2\sin \left( {\dfrac{\pi }{2} - \dfrac{A}{2}} \right)\sin \dfrac{1}{2}A}}\]
Apply the formula \[\sin \left( {\dfrac{\pi }{2} - \theta } \right) = \cos \theta \].
\[ = \dfrac{{\sin B + \sin C}}{{2\cos \dfrac{A}{2}\sin \dfrac{1}{2}A}}\]
Now applying the formula \[\sin 2\theta = 2\sin \theta \cos \theta \] in the denominator:
\[ = \dfrac{{\sin B + \sin C}}{{\sin A}}\] ….(i)
We know the sine law:
\[\dfrac{{\sin A}}{a} = \dfrac{{\sin B}}{b} = \dfrac{{\sin C}}{c} = k\left( {say} \right)\]
\[\sin A = ak\], \[\sin B = bk\], and \[\sin C = ck\]
Substitute the value of \[\sin A\], \[\sin B\], \[\sin C\] in (i)
\[ = \dfrac{{bk + ck}}{{ak}}\]
\[ = \dfrac{{k\left( {b + c} \right)}}{{ak}}\]
Cancel k from the denominator and numerator
\[ = \dfrac{{b + c}}{a}\]
Hence option B is the correct option.

Note: Students often confused with formula \[2\sin A\cos B = \sin \left( {A + B} \right) + \sin \left( {A - B} \right)\] and \[2\sin A\cos B = \sin \left( {A + B} \right) - \sin \left( {A - B} \right)\]. If A > B , then \[2\sin A\cos B = \sin \left( {A + B} \right) + \sin \left( {A - B} \right)\]. If B > A, then \[2\sin A\cos B = \sin \left( {A + B} \right) - \sin \left( {A - B} \right)\].