Question

Which of the following is true in triangle ABC.
A. \[\left( {b + c} \right)\sin \dfrac{{B - C}}{2} = 2a\cos \dfrac{A}{2}\]
B. \[\left( {b + c} \right)\cos \dfrac{A}{2} = 2a\sin \dfrac{{B - C}}{2}\]
C. \[\left( {b - c} \right)\cos \dfrac{A}{2} = a\sin \dfrac{{B - C}}{2}\]
D. \[\left( {b - c} \right)\sin \dfrac{{B - C}}{2} = 2a\cos \dfrac{A}{2}\]

Answer 1

Hint: We will use sine law to find the value of \[\dfrac{{b - c}}{a}\]. Then we will apply the formula of the difference between two sin angles, the half angle formula of sine, and the sum of angles of a triangle to get the desired result.

Formula used:
Double angle formula
\[\sin 2\theta = 2\sin \theta \cos \theta \]
Difference in sin angle
\[\sin C - \sin D = 2\cos \dfrac{{C - D}}{2}\sin \dfrac{{C + D}}{2}\]
Sine Law
\[\dfrac{{\sin A}}{a} = \dfrac{{\sin B}}{b} = \dfrac{{\sin C}}{c}\]

Complete step by step solution:
We will find the value of \[\dfrac{{b - c}}{a}\].
We know sine law \[\dfrac{{\sin A}}{a} = \dfrac{{\sin B}}{b} = \dfrac{{\sin C}}{c} = \dfrac{1}{k}\]
\[a = k\sin A\], \[b = k\sin B\], and \[c = k\sin C\]
Now substitute \[a = k\sin A\], \[b = k\sin B\], and \[c = k\sin C\] in \[\dfrac{{b - c}}{a}\]
\[\dfrac{{k\sin B - k\sin C}}{{k\sin A}}\]
\[ = \dfrac{{\sin B - \sin C}}{{\sin A}}\]
Now applying the formula \[\sin C - \sin D = 2\cos \dfrac{{C + D}}{2}\sin \dfrac{{C - D}}{2}\] in the numerator
\[ = \dfrac{{2\cos \dfrac{{B + C}}{2}\sin \dfrac{{B - C}}{2}}}{{\sin A}}\]
Now applying double angle formula in denominator
\[ = \dfrac{{2\cos \dfrac{{B + C}}{2}\sin \dfrac{{B - C}}{2}}}{{2\sin \dfrac{A}{2}\cos \dfrac{A}{2}}}\]
We know that the sum of all angles of a triangle is \[\pi \]. Thus \[A + B + C = \pi \]. This implies \[\dfrac{{B + C}}{2} = \dfrac{\pi }{2} - \dfrac{A}{2}\].
Putting \[\dfrac{{B + C}}{2} = \dfrac{\pi }{2} - \dfrac{A}{2}\] in the above expression
\[ = \dfrac{{2\cos \left( {\dfrac{\pi }{2} - \dfrac{A}{2}} \right)\sin \dfrac{{B - C}}{2}}}{{2\sin \dfrac{A}{2}\cos \dfrac{A}{2}}}\]
\[ = \dfrac{{2\sin \dfrac{A}{2}\sin \dfrac{{B - C}}{2}}}{{2\sin \dfrac{A}{2}\cos \dfrac{A}{2}}}\]
Cancel out \[2\sin \dfrac{A}{2}\] from the denominator and numerator
\[ = \dfrac{{\sin \dfrac{{B - C}}{2}}}{{\cos \dfrac{A}{2}}}\]
Therefore, \[\dfrac{{b - c}}{a} = \dfrac{{\sin \dfrac{{B - C}}{2}}}{{\cos \dfrac{A}{2}}}\]
\[ \Rightarrow \left( {b - c} \right)\cos \dfrac{A}{2} = a\sin \dfrac{{B - C}}{2}\]
Hence option C is the correct option.

Note: Some students often make mistakes in the formula of difference of two sin angles. They used \[\sin C - \sin D = 2\sin \dfrac{{C + D}}{2}\cos \dfrac{{C - D}}{2}\] which is incorrect. The correct formula is \[\sin C - \sin D = 2\cos \dfrac{{C + D}}{2}\sin \dfrac{{C - D}}{2}\].